RS Aggarwal Class 7 Mathematics Solutions Chapter 20 Mensuration

Exercise 20A

Exercise 20B

Question 1. A rectangular plot measures 24.5 m in length and 18 m in width. What is the area of the plot? (ii) A rectangular plot has dimensions of 12.5 m by 8 dm. Calculate its area.
Answer:
(i) The area can be found by multiplying the length by the breadth.
\[ \text{Area} = 24.5 \text{ m} \times 18 \text{ m} = 441 \text{ m}^2 \]

(ii) First, change 8 dm to metres: \( 8 \text{ dm} = (8 \times 10) \text{ cm} = 80 \text{ cm} = 0.8 \text{ m} \)
\[ \text{Area} = 12.5 \text{ m} \times 0.8 \text{ m} = 10 \text{ m}^2 \]
In simple words: To find the area of a rectangle, just multiply its length times its width. Make sure both measurements are in the same unit before you multiply.

Exam Tip: Always convert all measurements to the same unit before calculating area - this prevents common mistakes and confusion.

Question 2. A rectangular plot has a diagonal of 50 m and one side measures 48 m. Find the other side and the area.
Answer:
All angles in a rectangle equal 90°, and the diagonal splits the rectangle into two right-angled triangles. One side measures 48 m, and the diagonal (hypotenuse) is 50 m. Using the Pythagorean theorem:

\[ (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \]

\[ \text{Perpendicular} = \sqrt{(50)^2 - (48)^2} = \sqrt{2500 - 2304} = \sqrt{196} = 14 \text{ m} \]

Thus, the other side measures 14 m.
Length = 48 m
Breadth = 14 m

\[ \text{Area} = 48 \text{ m} \times 14 \text{ m} = 672 \text{ m}^2 \]

The area of the rectangular plot is 672 m².
In simple words: The diagonal of a rectangle creates a right triangle. Use the Pythagorean theorem to find the missing side, then multiply length times width to get the area.

Exam Tip: Remember that a diagonal always creates a right triangle with the two sides of the rectangle - this is your key to solving such problems.

Question 3. A rectangular field is 4 times longer than it is wide. If the total area is 1728 m², determine its perimeter and the cost of fencing at Rs. 30 per metre.
Answer:
Let the breadth be x m, so the length is 4x m.

\[ \text{Area} = (4x) \times (3x) \text{ m}^2 = 12x^2 \text{ m}^2 \]

Since the area equals 1728 m²:
\[ 12x^2 = 1728 \]
\[ x^2 = 144 \]
\[ x = 12 \]

Length = \( (4 \times 12) \text{ m} = 48 \text{ m} \)
Breadth = \( (3 \times 12) \text{ m} = 36 \text{ m} \)

Perimeter = \( 2(l + b) \) units \( = 2(48 + 36) \text{ m} = (2 \times 84) \text{ m} = 168 \text{ m} \)

Cost of fencing = \( \text{Rs.} (168 \times 30) = \text{Rs.} 5040 \)
In simple words: Set up an equation using the given relationship between length and width, solve for the width, then find the perimeter and multiply by the cost per metre.

Exam Tip: When a shape's dimensions are given as a ratio, express both in terms of a single variable and use the area to solve for it.

Question 4. A rectangular field has an area of 3584 m² and a length of 64 m. A boy walks around the field at 6 km/hr. How long does it take him to complete five full circuits?
Answer:
Area of the rectangular field = 3584 m²
Length of the rectangular field = 64 m

Breadth = \( \left(\frac{\text{Area}}{\text{Length}}\right) = \left(\frac{3584}{64}\right) \text{ m} = 56 \text{ m} \)

Perimeter of the rectangular field = \( 2(\text{length} + \text{breadth}) \)
\[ = 2(64 + 56) \text{ m} = (2 \times 120) \text{ m} = 240 \text{ m} \]

Distance covered by the boy = \( 5 \times \text{Perimeter of the rectangular field} \)
\[ = 5 \times 240 = 1200 \text{ m} \]

The boy walks at 6 km/hr.

Rate = \( \left(\frac{6 \times 1000}{60}\right) \text{ m/min} = 100 \text{ m/min} \)

Time needed to cover 1200 m = \( \left(\frac{1200}{100}\right) \text{ min} = 12 \text{ min} \)

The boy will need 12 minutes to walk around the field five times.
In simple words: First find the breadth using area and length. Then calculate the perimeter and multiply by 5 for five circuits. Convert the walking speed and divide total distance by speed to get time.

Exam Tip: Always convert speed to the same units as your distance (minutes and metres, or hours and kilometres) before dividing to find time.

Question 5. A verandah is 40 m long and 15 m wide. It needs to be paved with stones that measure 6 dm by 5 dm. How many stones are required?
Answer:
Given:
Length of the verandah = 40 m = 400 dm (since 1 m = 10 dm)
Breadth of the verandah = 15 m = 150 dm

\[ \text{Area of the verandah} = (400 \times 150) \text{ dm}^2 = 60000 \text{ dm}^2 \]

Length of a stone = 6 dm
Breadth of a stone = 5 dm

\[ \text{Area of a stone} = (6 \times 5) \text{ dm}^2 = 30 \text{ dm}^2 \]

Total number of stones needed = \( \frac{\text{Area of the verandah}}{\text{Area of each stone}} = \left(\frac{60000}{30}\right) = 2000 \)
In simple words: Change all measurements to the same unit. Find the area of the verandah and the area of one stone. Divide the verandah's area by the stone's area to get how many stones you need.

Exam Tip: Always ensure unit consistency before dividing - converting everything to the same unit at the start saves calculation errors.

Question 6. A room is 13 m by 9 m. A carpet that is 75 cm wide costs Rs. 105 per metre. What is the total cost of carpeting the room?
Answer:
Area of the carpet = Area of the room
\[ = (13 \text{ m} \times 9 \text{ m}) = 117 \text{ m}^2 \]

Width of the carpet = 75 cm = 0.75 m (since 1 m = 100 cm)

Length of the carpet = \( \left(\frac{\text{Area of the carpet}}{\text{Width of the carpet}}\right) = \left(\frac{117}{0.75}\right) = 156 \text{ m} \)

Rate of carpeting = Rs. 105 per m

\[ \text{Total cost of carpeting} = \text{Rs.} (156 \times 105) = \text{Rs.} 16380 \]

The total cost of carpeting the room is Rs. 16380.
In simple words: The carpet's area must equal the room's area. Divide the area by the width to find the length you need. Then multiply that length by the cost per metre to get the total price.

Exam Tip: When a carpet is sold by length but you know its width, use Area = Length × Width rearranged to find the length needed.

Question 7. A carpet costs Rs. 80 per metre and Rs. 12200 total. If the carpet is 0.75 m wide, find the dimensions of the room it covers.
Answer:
Given:
Length of the carpet required for carpeting the room = ? (let it be x m)
Cost of the carpet = Rs. 80 per m

\[ \text{Cost of x m carpet} = \text{Rs.} (80 \times x) = \text{Rs.} (80x) \]

Cost of carpeting the room = Rs. 12200

\[ 80x = 12200 \implies x = \left(\frac{12200}{80}\right) = 240 \]

Thus, the length of the carpet needed is 240 m.

Area of the carpet required = Length × Width of the carpet
\[ = (240 \times 0.75) \text{ m}^2 = 180 \text{ m}^2 \]

Let the breadth of the room be b m.
Area to be carpeted = 15 m × b m = 15b m²

\[ 15b = 180 \]
\[ b = \left(\frac{180}{15}\right) = 12 \text{ m} \]

The room measures 15 m by 12 m.
In simple words: Use the total cost and price per metre to find the carpet's length. Multiply by its width to get the total area. Then divide this area by the room's length to find its width.

Exam Tip: Work backward from the cost - dividing total cost by the rate gives you the length, which then lets you find the area and room dimensions.

Question 8. The cost of fencing a rectangular field is Rs. 9600 at Rs. 24 per metre. If the length and width are in the ratio 5:3, find both dimensions and the area.
Answer:
Total cost of fencing a rectangular piece = Rs. 9600
Rate of fencing = Rs. 24

\[ \text{Perimeter} = \left(\frac{\text{Total cost of fencing}}{\text{Rate of fencing}}\right) \text{ m} = \left(\frac{9600}{24}\right) = 400 \text{ m} \]

Let the length and breadth of the rectangular field be 5x and 3x, respectively.
Perimeter of the rectangular land = 2(5x + 3x) = 16x

The perimeter of the given field is 400 m.

\[ 16x = 400 \]
\[ x = \left(\frac{400}{16}\right) = 25 \]

Length of the field = \( (5 \times 25) \text{ m} = 125 \text{ m} \)
Breadth of the field = \( (3 \times 25) \text{ m} = 75 \text{ m} \)

Area of the field = 125 m × 75 m = 9375 m²
In simple words: Divide the total cost by the rate per metre to find the perimeter. Use the length-to-width ratio to set up equations, solve for the multiplier, then find each dimension and multiply to get the area.

Exam Tip: When dimensions are in a ratio, always express them as multiples of a variable (like 5x and 3x) - this makes the algebra much cleaner.

Question 9. A hall measures 10 m, 10 m, and 6 m in length, width, and diagonal respectively. What is the length of the longest pole that can fit inside the hall?
Answer:
The longest pole that can fit inside a three-dimensional hall must be along its space diagonal. Using the three dimensions of the hall:

\[ \text{Length of the diagonal of the room} = \sqrt{l^2 + b^2 + h^2} \]
\[ = \sqrt{(10)^2 + (10)^2 + (6)^2} \text{ m} \]
\[ = \sqrt{100 + 100 + 25\text{m}} \]
\[ = \sqrt{225} = 15 \text{ m} \]

The longest pole that can fit in the hall is 15 m.
In simple words: The longest pole goes diagonally through the entire hall from one corner to the opposite corner. Use the 3D distance formula with length, width, and height to find it.

Exam Tip: For a 3D rectangular space, remember the diagonal formula includes all three dimensions under the square root - don't forget any measurement.

Question 10. A square has a side of 8.5 m. What is its area?
Answer:
Side of the square = 8.5 m
\[ \text{Area of the square} = (\text{Side})^2 = (8.5 \text{ m})^2 = 72.25 \text{ m}^2 \]
In simple words: To find the area of a square, multiply its side length by itself - that is, square the side length.

Exam Tip: For any square problem, remember that all four sides are equal and area is simply side squared.

Question 11. (i) The diagonal of a square is 72 cm. Find its area. (ii) The diagonal of a square is 2.4 m. Find its area.
Answer:
(i) Diagonal of the square = 72 cm
\[ \text{Area of the square} = \frac{1}{2} \times (\text{Diagonal})^2 \text{ sq. unit} \]
\[ = \frac{1}{2} \times (72)^2 \text{ cm}^2 \]
\[ = 2592 \text{ cm}^2 \]

(ii) Diagonal of the square = 2.4 m
\[ \text{Area of the square} = \frac{1}{2} \times (\text{Diagonal})^2 \text{ sq. unit} \]
\[ = \frac{1}{2} \times (2.4)^2 \text{ m}^2 \]
\[ = 2.88 \text{ m}^2 \]
In simple words: When you know only the diagonal of a square, use the formula: Area = (1/2) × (Diagonal)². This formula comes from the fact that the diagonal equals side times the square root of 2.

Exam Tip: Remember the diagonal formula for a square: if the diagonal is d, the area is d²/2. This is faster than finding the side first.

Question 12. If the area of a square is 16200 m², what is the length of its diagonal?
Answer:
We know:
\[ \text{Area of a square} = \frac{1}{2} \times (\text{Diagonal})^2 \text{ sq. units} \]
\[ \text{Diagonal of the square} = \sqrt{2 \times \text{Area of square units}} \]

\[ (\sqrt{2 \times 16200\text{m}}) = 180 \text{ m} \]

The length of the diagonal of the square is 180 m.
In simple words: Rearrange the area formula to solve for the diagonal: multiply area by 2, then take the square root. This gives you the diagonal directly.

Exam Tip: To find diagonal from area, use: Diagonal = √(2 × Area). This is the reverse of the standard area formula.

Question 13. A square field has an area of 1/4 hectare. Find the length of its diagonal.
Answer:
\[ \text{Area of the square} = \frac{1}{2} \times (\text{Diagonal})^2 \text{ sq. units} \]

Given:
Area of the square field = 1/4 hectare
\[ = \left(\frac{1}{4} \times 10000\right) \text{ m}^2 = 5000 \text{ m}^2 \]

\[ \text{Diagonal of the square} = \sqrt{2 \times \text{Area of the square}} \]

\[ = (\sqrt{2 \times 5000}\text{m}) = 100 \text{ m} \]

The length of the diagonal of the square field is 100 m.
In simple words: Convert hectares to square metres first (1 hectare = 10000 m²). Then apply the diagonal formula: Diagonal = √(2 × Area).

Exam Tip: Always convert non-metric units like hectares to square metres before using geometric formulas to avoid calculation errors.

Question 14. A square plot has an area of 6084 m². Find its perimeter and the length of wire needed if the wire is to go around the boundary four times.
Answer:
Area of the square plot = 6084 m²

Side of the square plot = \( (\sqrt{\text{Area}}) \)
\[ = (\sqrt{6084}) \text{ m} \]
\[ = (\sqrt{78 \times 78}\text{m}) = 78 \text{ m} \]

\[ \text{Perimeter of the square plot} = 4 \times \text{side} = (4 \times 78) \text{ m} = 312 \text{ m} \]

312 m wire is needed to go along the boundary once.

Required length of the wire that can go four times along the boundary = \( 4 \times \text{Perimeter of the square plot} \)
\[ = (4 \times 312) \text{ m} = 1248 \text{ m} \]
In simple words: Take the square root of the area to find the side. Multiply the side by 4 to get the perimeter. Then multiply the perimeter by 4 to find the total wire needed for four circuits.

Exam Tip: When a problem asks for "multiple times around," always find the perimeter first, then multiply it by the number of times stated.

Question 15. A square plot encloses more area than a rectangle plot. The rectangle has sides of 10 cm and 8 cm. Both have the same perimeter. How much more area does the square enclose?
Answer:
Side of the square = 10 cm
Length of the wire = Perimeter of the square = \( 4 \times \text{Side} = 4 \times 10 \text{ cm} = 40 \text{ cm} \)

Length of the rectangle (l) = 12 cm
Let b be the breadth of the rectangle.

Perimeter of the rectangle = Perimeter of the square

\[ 2(l + b) = 40 \]
\[ 2(12 + b) = 40 \]
\[ 24 + 2b = 40 \]
\[ 2b = 40 - 24 = 16 \]
\[ b = \left(\frac{16}{2}\right) \text{ cm} = 8 \text{ cm} \]

Breadth of the rectangle = 8 cm

Now, Area of the square = (Side)² = (10 cm × 10 cm) = 100 cm²
Area of the rectangle = l × b = (12 cm × 8 cm) = 96 cm²

The square encloses more area.
It encloses 4 cm² more area.
In simple words: Find the side of the square using its perimeter. Use the same perimeter for the rectangle to find its breadth. Then calculate both areas and subtract to find the difference.

Exam Tip: When comparing shapes with equal perimeters, a square always encloses the maximum area - this is a key geometric principle.

Question 16. A room is 50 m by 40 m by 10 m in height. Its four walls and ceiling need to be whitewashed at Rs. 20 per m². What is the total cost?
Answer:
Given:
Length = 50 m
Breadth = 40 m
Height = 10 m

Area of the four walls = \( 2(l + b) \times h \) sq. unit
\[ = (2 \times 10 \times (50 + 40))\text{m}^2 \]
\[ = (20 \times 90) \text{m}^2 = 1800 \text{m}^2 \]

Area of the ceiling = \( l \times b = (50 \times 40) = 2000 \text{m}^2 \)

Total area to be white washed = \( (1800 + 2000) \text{m}^2 = 3800 \text{m}^2 \)

Rate of white washing = Rs. 20/sq. metre

\[ \text{Total cost of white washing} = \text{Rs.} (3800 \times 20) = \text{Rs.} 76000 \]
In simple words: Calculate the area of all four walls using 2(length + breadth) × height. Add the ceiling area. Multiply the total by the rate per square metre to get the cost.

Exam Tip: When whitewashing a room, remember to include both walls AND ceiling if the problem states it - don't forget either one.

Question 17. A room is 10 m wide, 4 m tall, and has four walls with a total area of 168 m². What is the room's length?
Answer:
Let the length of the room be l m.

Given:
Breadth of the room = 10 m
Height of the room = 4 m

\[ \text{Area of the four walls} = 2(l + b) \times h \text{ sq. units} \]
\[ = 168 \text{m}^2 \]

\[ 168 = [2(l + 10) \times 4] \]
\[ 168 = [8(l + 10)] \]
\[ 168 = [8l + 80] \]
\[ 168 - 80 = 8l \]
\[ 88 = 8l \]
\[ l = \left(\frac{88}{8}\right) \text{ m} = 11 \text{ m} \]

The length of the room is 11 m.
In simple words: Use the formula for wall area and work backward: set the formula equal to the given wall area, substitute the known breadth and height, then solve for the length.

Exam Tip: When working backward from area, substitute all known values into the formula first, then isolate the unknown variable.

Question 18. A room is 7.5 m by 3.5 m with a wall area of 77 m². Calculate its height.
Answer:
Given:
Length of the room = 7.5 m
Breadth of the room = 3.5 m

\[ \text{Area of the four walls} = 2(l + b) \times h \text{ sq. units} \]
\[ = 77 \text{m}^2 \]

\[ 77 = [2(7.5 + 3.5) \times h] \]
\[ 77 = [(2 \times 11) \times h] \]
\[ 77 = 22h \]
\[ h = \left(\frac{77}{22}\right) \text{ m} = \left(\frac{7}{2}\right) \text{ m} = 3.5 \text{ m} \]

The height of the room is 3.5 m.
In simple words: Substitute the length and breadth into the wall area formula. Set it equal to the given wall area. Solve for height by dividing by the coefficient of h.

Exam Tip: The formula for wall area is always 2(length + breadth) × height - if you remember this, you can solve for any unknown dimension.

Question 19. A room's width is half its length. Its total wall area is 120 m² with a height of 4 m. Find the length, width, and floor area.
Answer:
Let the breadth of the room be x m.
Length of the room = 2x m

\[ \text{Area of the four walls} = 2(l + b) \times h \text{ sq. units} \]
\[ = 120 \text{m}^2 \]

\[ 120 = [2(2x + x) \times 4] \]
\[ 120 = (8 \times 3x) \]
\[ 120 = 24x \]
\[ x = \left(\frac{120}{24}\right) = 5 \]

The length of the room = 2x = (2 × 5) m = 10 m
Breadth of the room = x = 5 m

\[ \text{Area of the floor} = l \times b = (10 \text{ m} \times 5 \text{ m}) = 50 \text{ m}^2 \]
In simple words: Express the length in terms of the width using the given ratio. Substitute into the wall area formula. Solve for width, then find length and floor area.

Exam Tip: When dimensions are related by a ratio, always express both in terms of a single variable - this simplifies the algebra significantly.

Question 20. A room is 8.5 m by 6.5 m by 3.4 m high. It has two doors (1.5 m × 1 m each) and two windows (2 m × 1 m each). If the painting cost is Rs. 160 per m², what is the total cost?
Answer:
Length = 8.5 m
Breadth = 6.5 m
Height = 3.4 m

\[ \text{Area of the four walls} = 2(l + b) \times h \text{ sq. units} \]
\[ = (2(8.5 + 6.5) \times 3.4)\text{m}^2 = (30 \times 3.4)\text{m}^2 = 102 \text{m}^2 \]

Area of one door = \( (1.5 \times 1) \text{m}^2 = 1.5 \text{ m}^2 \)

\[ \text{Area of two doors} = (2 \times 1.5) \text{m}^2 = 3 \text{ m}^2 \]

Area of one window = \( (2 \times 1) \text{m}^2 = 2 \text{ m}^2 \)

\[ \text{Area of two windows} = (2 \times 2) \text{m}^2 = 4 \text{ m}^2 \]

Total area of two doors and two windows = \( (3 + 4) \text{m}^2 = 7 \text{ m}^2 \)

\[ \text{Area to be painted} = (102 - 7) \text{m}^2 = 95 \text{m}^2 \]

Rate of painting = Rs. 160 per m²

\[ \text{Total cost of painting} = \text{Rs.} (95 \times 160) = \text{Rs.} 15200 \]
In simple words: Calculate the total wall area. Subtract the areas of all doors and windows. Multiply what remains by the cost per square metre to get the final price.

Exam Tip: When painting a room with openings, always remember to subtract the door and window areas from the total wall area - this is a common mistake.

Exercise 20B

Question 1. A grassy plot has a rectangular path running through it. The outer boundary is 75 m by 60 m. The path is 2 m wide. Find the cost of constructing the path at Rs. 125 per m².
Answer:
Let PQRS be the given grassy plot and ABCD be the inside boundary of the path.

Length = 75 m
Breadth = 60 m
\[ \text{Area of the plot} = (75 \times 60) \text{m}^2 = 4500 \text{ m}^2 \]

Width of the path = 2 m

\[ AB = (75 - 2 \times 2) \text{ m} = (75 - 4) \text{ m} = 71 \text{ m} \]
\[ AD = (60 - 2 \times 2) \text{ m} = (60 - 4) \text{ m} = 56 \text{ m} \]

\[ \text{Area of rectangle ABCD} = (71 \times 56) \text{ m}^2 = 3976 \text{ m}^2 \]

\[ \text{Area of the path} = (\text{Area of PQRS} - \text{Area of ABCD}) \]
\[ = (4500 - 3976) \text{ m}^2 = 524 \text{ m}^2 \]

Rate of constructing the path = Rs. 125 per m²

\[ \text{Total cost of constructing the path} = \text{Rs.} (524 \times 125) = \text{Rs.} 65,500 \]
In simple words: Find the area of the outer boundary. Subtract the path width from both length and width to get the inner rectangle's dimensions. Find its area. Subtract to get the path area. Multiply by the rate.

Exam Tip: For a path inside a rectangle, remember to subtract the path width from BOTH the length AND the width (twice - once from each end).

Question 3. A rectangular plot is 95 m long and 72 m wide. A path of width 3.5 m runs all around it (on the inside). Find the cost of constructing the path at Rs. 80 per m² and laying grass on the plot ABCD at Rs. 40 per m².
Answer: Let PQRS be the given rectangular plot and ABCD be the inner boundary of the path.

Length = 95 m
Breadth = 72 m
Area of the plot = (95 × 72) m² = 6,840 m²
Width of the path = 3.5 m
∴ AB = (95 - 2 × 3.5) m = (95 - 7) m = 88 m
AD = (72 - 2 × 3.5) m = (72 - 7) m = 65 m
Area of the path = (Area PQRS - Area ABCD)
= (6840 - 5720) m² = 1,120 m²

Rate of constructing the path = Rs. 80 per m²
∴ Total cost of constructing the path = Rs. (1,120 × 80) = Rs. 89,600

Rate of laying the grass on the plot ABCD = Rs. 40 per m²
∴ Total cost of laying the grass on the plot = Rs. (5,720 × 40) = Rs. 2,28,800
∴ Total expenses involved = Rs. (89,600 + 2,28,800) = Rs. 3,18,400
In simple words: You need to find two things - the money to build the path and the money to lay grass. Subtract the inner rectangle from the outer one to get the path area, then multiply by the cost per square metre for each part.

Exam Tip: Always identify the outer boundary and inner boundary clearly, and remember to subtract areas correctly to find the path region.

Question 4. A saree is 5 m long and 1.3 m wide. A border of width 25 cm is attached to all four sides of the saree. Find the cost of printing the border at Rs. 1 per 10 cm².
Answer: Let ABCD be the saree and EFGH be the part of saree without border.

Length, AB = 5 m
Breadth, BC = 1.3 m
Width of the border of the saree = 25 cm = 0.25 m

∴ Area of ABCD = 5 m × 1.3 m = 6.5 m²

Length, GH = (5 + (0.25 + 0.25)) m = 4.5 m
Breadth, FG = (1.3 - 0.25 + 0.25) m = 0.8 m
∴ Area of EFGH = 4.5 m × 0.8 m = 3.6 m²

Area of the border = Area of ABCD - Area of EFGH
= 6.5 m² - 3.6 m²
= 2.9 m² = 29000 cm² [since 1 m² = 10000 cm²]

Rate of printing the border = Rs. 1 per 10 cm²
∴ Total cost of printing the border = Rs. \( \left(\frac{1 \times 29000}{10}\right) \)
= Rs. 2900
In simple words: Find the area of the saree with the border and the area without the border. The difference gives you the border area. Then multiply by the cost per 10 square centimetres.

Exam Tip: Convert all measurements to the same unit before calculating, and ensure you understand which is the outer boundary and which is the inner boundary.

Question 5. A room is 9.5 m long and 6 m wide. A verandah 1.25 m wide is constructed all around it. Find the cost of cementing the verandah at Rs. 80 per m².
Answer: Let EFGH denote the floor of the room. The white region represents the floor of the 1.25 m verandah.

Length, EF = 9.5 m
Breadth, FG = 6 m

∴ Area of EFGH = 9.5 m × 6 m = 57 m²

Length, AB = (9.5 + 1.25 + 1.25) m = 12 m
Breadth, BC = (6 + 1.25 + 1.25) m = 8.5 m
∴ Area of ABCD = 12 m × 8.5 m = 102 m²

Area of the verandah = Area of ABCD - Area of EFGH
= 102 m² - 57 m²
= 45 m²

Rate of cementing the verandah = Rs. 80 per m²
∴ Total cost of cementing the verandah = Rs. (80 × 45)
= Rs. 3600
In simple words: Work out the total area including the verandah, then take away the room's area. What is left is the verandah. Multiply by the rate per square metre to get the total cost.

Exam Tip: Draw a clear diagram showing both the inner and outer rectangles so you can visualize which dimensions change when the verandah is added.

Question 6. A square flower bed is 2 m in area at each side. A digging strip of width 30 cm is laid all around it. Find the increase in the area of the flower bed.
Answer: Side of the flower bed = 2 m 80 cm = 2.80 m [since 100 cm = 1 m]

∴ Area of the square flower bed = (Side)² = (2.80 m)² = 7.84 m²

Side of the flower bed with the digging strip = 2.80 m + 30 cm + 30 cm
= (2.80 + 0.3 + 0.3) m = 3.4 m

Area of the enlarged flower bed with the digging strip = (Side)² = (3.4 m)² = 11.56 m²

∴ Increase in the area of the flower bed = 11.56 m² - 7.84 m²
= 3.72 m²
In simple words: Square the original side to get the original area. Add the strip width on both sides, then square this new side length to get the enlarged area. The difference is your answer.

Exam Tip: Remember to add the strip width on both sides when calculating the outer dimension - this is a common mistake.

Question 7. A rectangular park is 240 m in perimeter. Its length is twice its breadth. A path of width 2 m runs all around it (on the inside). Find the cost of paving the path at Rs. 80 per m².
Answer: Let the length and breadth of the park be 2x m and x m, respectively.

Perimeter of the park = 2(2x + x) = 240 m
= 2(2x + x) = 240
= 6x = 240
= x = \( \left(\frac{240}{6}\right) \) m = 40 m

∴ Length of the park = 2x = (2 × 40) = 80 m
Breadth = x = 40 m

Let PQRS be the given park and ABCD be the inner boundary of the path.

Length = 80 m
Breadth = 40 m
Area of the park = (80 × 40) m² = 3200 m²

Width of the path = 2 m
∴ AB = (80 - 2 × 2) m = (80 - 4) m = 76 m
AD = (40 - 2 × 2) m = (40 - 4) m = 36 m

Area of the rectangle ABCD = (76 × 36) m² = 2736 m²

Area of the path = (Area of PQRS - Area of ABCD)
= (3200 - 2736) m² = 464 m²

Rate of paving the path = Rs. 80 per m²
∴ Total cost of paving the path = Rs. (464 × 80) = Rs. 37,120
In simple words: First, work out the park's dimensions using the perimeter and the fact that length is twice the breadth. Then calculate the path area by subtracting the inner rectangle from the outer one. Multiply by the rate per square metre.

Exam Tip: Always set up equations carefully using the given relationships (like "length is twice breadth") and solve step by step before moving on to the area calculations.

Question 8. A rectangular hall is 22 m long and 15.5 m wide. A carpet is laid inside the hall leaving a strip of width 75 cm all around. Find the cost of the carpet at Rs. 60 per m².
Answer: Length of the hall, PQ = 22 m
Breadth of the hall, QR = 15.5 m

∴ Area of the school hall PQRS = 22 m × 15.5 m = 341 m²

Length of the carpet, AB = 22 m - (0.75 m + 0.75 m) = 20.5 m [since 100 cm = 1 m]
Breadth of the carpet, BC = 15.5 m - (0.75 m + 0.75 m) = 14 m

∴ Area of the carpet ABCD = 20.5 m × 14 m = 287 m²

Area of the strip = Area of the school hall (PQRS) - Area of the carpet (ABCD)
= 341 m² - 287 m²
= 54 m²

Area of 1 m length of the carpet = 1 m × 0.82 m = 0.82 m²

∴ Length of the carpet whose area is 287 m² = 287 m² ÷ 0.82 m² = 350 m
Cost of the 350 m long carpet = Rs 60 × 350 = Rs 21000
In simple words: Find the dimensions of the carpet by subtracting the strip width from both sides of the hall's length and breadth. Then calculate the carpet's area and multiply by the rate per square metre.

Exam Tip: Ensure you subtract the strip width from both sides of each dimension, not just once.

Question 9. A square lawn ABCD is surrounded by a square path PQRS. The path is 2.5 m wide all around. If the area of the path is 165 m², find the area of the lawn.
Answer: Let ABCD be the square lawn and PQRS be the outer boundary of the square path.

Let a side of the lawn (AB) be x m.
Area of the square lawn = x²

Length, PQ = (x m + 2.5 m + 2.5 m) = (x + 5) m
∴ Area of PQRS = (x + 5)² = (x² + 10x + 25) m²

Area of the path = Area of PQRS - Area of the square lawn (ABCD)
⇒ 165 = x² + 10x + 25 - x²
⇒ 165 = 10x + 25
⇒ 165 - 25 = 10x
⇒ 140 = 10x
⇒ x = 140 ÷ 10 = 14

∴ Side of the lawn = 14 m

∴ Area of the lawn = (Side)² = (14 m)² = 196 m²
In simple words: Call the lawn's side x. The outer square has a side of (x + 5) because the path adds 2.5 m on each side. Use the given path area to form an equation and solve for x, then find the lawn's area.

Exam Tip: When a path surrounds a shape on all sides, remember to add twice the path width to each dimension of the inner shape.

Question 10. A rectangular park has area 305 m². Find the cost of constructing a path inside the park if the path is 5x m long and 2x m broad and the rate of construction is Rs. 120 per m².
Answer: Area of the path = 305 m²

Let the length of the park be 5x m and the breadth of the park be 2x m.

Area of the rectangular park = 5x × 2x = 10x² m²
Width of the path = 2.5 m
Outer length, PQ = 5x m + 2.5 m + 2.5 m = (5x + 5) m
Outer breadth, QR = 2x m + 2.5 m + 2.5 m = (2x + 5) m

Area of PQRS = (5x + 5) × (2x + 5) = (10x² + 25x + 10x + 25) = (10x² + 35x + 25) m²

∴ Area of the path = [(10x² + 35x + 25) - 10x²] m²
= 305 ⇒ 35x + 25
= 305 ⇒ 25 = 35x
= 280 = 35x
= x = 280 ÷ 35 = 8

∴ Length of the park = 5x = 5 × 8 = 40 m
Breadth of the park = 2x = 2 × 8 = 16 m
In simple words: Express the dimensions in terms of x. Calculate the outer rectangle's area, subtract the inner area, and set this equal to the given path area. Solve for x, then find the actual dimensions.

Exam Tip: Always clearly mark which is the inner rectangle and which is the outer rectangle to avoid confusion in your working.

Question 11. A rectangular park, 70 m long and 50 m wide, has two roads, each 5 m wide, running through it - one parallel to its length and the other parallel to its breadth. These roads overlap in a square. Find the cost of constructing these roads at Rs. 120 per m².
Answer: Let ABCD be the rectangular park. Let EFGH and IJKL be the two rectangular roads with width 5 m.

Length of the rectangular park, AD = 70 m
Breadth of the rectangular park, CD = 50 m

∴ Area of the rectangular park = Length × Breadth = 70 m × 50 m = 3500 m²

Area of road EFGH = 70 m × 5 m = 350 m²
Area of road IJKL = 50 m × 5 m = 250 m²

Clearly, area of MNOP is common to both the two roads.

∴ Area of MNOP = 5 m × 5 m = 25 m²

Area of the roads = Area (EFGH) + Area (IJKL) - Area (MNOP)
= (350 + 250) m² - 25 m² = 575 m²

It is given that the cost of constructing the roads is Rs. 120/m²

Cost of constructing 575 m² area of the roads = Rs. (120 × 575)
= Rs. 69000
In simple words: Find the area of each road separately. These roads overlap in a square in the middle. Add the two road areas and subtract the overlapping square area. Multiply the result by the cost per square metre.

Exam Tip: Do not forget to subtract the overlapping region - this is the most common error in these problems.

Question 12. A rectangular field is 115 m long and 64 m wide. Two roads, each 2 m and 2.5 m wide respectively, run parallel to the sides and cut the field into four parts. Find the cost of gravelling the roads at Rs. 60 per m².
Answer: Let ABCD be the rectangular field and PQRS and KLMN be the two rectangular roads with width 2 m and 2.5 m, respectively.

Length of the rectangular field, CD = 115 cm
Breadth of the rectangular field, BC = 64 m

∴ Area of the rectangular lawn ABCD = 115 m × 64 m = 7360 m²

Area of the road PQRS = 115 m × 2 m = 230 m²
Area of the road KLMN = 64 m × 2.5 m = 160 m²

Clearly, the area of EFGH is common to both the two roads.

∴ Area of EFGH = 2 m × 2.5 m = 5 m²

∴ Area of the roads = Area (KLMN) + Area (PQRS) - Area (EFGH)
= (230 m² + 160 m²) - 5 m² = 385 m²

Rate of gravelling the roads = Rs 60 per m²

∴ Total cost of gravelling the roads = Rs (385 × 60)
= Rs 23,100
In simple words: Work out the area of each road by multiplying its length by its width. Find the overlap area where the two roads cross. Add both road areas and subtract the overlap, then multiply by the rate.

Exam Tip: Visualize the two roads running perpendicular to each other and create a rectangular overlap - make sure you identify this correctly before calculating.

Question 13. A rectangular field is 50 m long and 40 m wide. Two roads, each 2.5 m and 2 m wide respectively, run parallel to the sides through the middle of the field. Find the area of the remaining portion of the field.
Answer: Let ABCD be the rectangular field and KLMN and PQRS be the two rectangular roads with width 2.5 m and 2 m, respectively.

Length of the rectangular field CD = 50 cm
Breadth of the rectangular field BC = 40 m

∴ Area of the rectangular field ABCD = 50 m × 40 m = 2000 m²

Area of road KLMN = 40 m × 2.5 m = 100 m²
Area of road PQRS = 50 m × 2 m = 100 m²

Clearly, area of EFGH is common to both the two roads.

∴ Area of EFGH = 2.5 m × 2 m = 5 m²

∴ Area of the roads = Area (KLMN) + Area (PQRS) - Area (EFGH)
= (100 m² + 100 m²) - 5 m² = 195 m²

Area of the remaining portion of the field = Area of the rectangular field (ABCD) - Area of the roads
= (2000 - 195) m²
= 1805 m²
In simple words: Calculate the total field area. Work out the area taken up by both roads, being careful to subtract the overlapping square where they cross. The remaining area is the field area minus the roads area.

Exam Tip: Always subtract the overlapping intersection area only once - adding it back in prevents double counting and gives the correct road area.

Question 14. (i) Complete the rectangle as shown below:

Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGH] sq. units
= [(43 m × 27 m) - ((43 - 2 × 1.5) m × (27 - 1 × 2))]
= [(43 m × 27 m) - (40 m × 25 m)]
= 1161 m² - 1000 m²
= 161 m²

(ii) Complete the rectangle by drawing lines as shown below:

Area of the shaded region = [(12 × 3) + (12 × 3) + (5 × 3) + ((15 - 3 - 3) × 3)] cm²
= (36 + 36 + 15 + 27) cm²
= 114 cm²
In simple words: For part (i), find the outer rectangle area and the inner rectangle area, then subtract. For part (ii), divide the shaded region into separate rectangles, find each area, and add them together.

Exam Tip: Always be precise about which sides are shaded and use either subtraction (outer minus inner) or addition (sum of component rectangles) as appropriate.

Question 15. (i) Complete the rectangle as shown below:

Area of the shaded region = [Area of rectangle ABCD - Area of rectangle EFGD] sq. units
= [(AB × BC) - (DG × GF)] m²
= [(24 m × 19 m) - ((24 - 4) m × 16.5 m)]
= [(24 m × 19 m) - (20 m × 16.5 m)]
= (456 - 330) m² = 126 m²

(ii) Complete the rectangle by drawing lines as shown below:

Area of the shaded region = [(12 × 3) + (12 × 3) + (5 × 3) + ((15 - 3 - 3) × 3)] cm²
= (36 + 36 + 15 + 27) cm²
= 114 cm²
In simple words: For part (i), subtract the inner rectangle area from the outer rectangle area to find the shaded region. For part (ii), break the shape into distinct rectangles, calculate each area separately, and add them.

Exam Tip: When subtracting, double-check that you have correctly identified which dimension has been reduced by how much.

Question 16. Divide the given figure in four parts shown below:

Given:
Width of each part = 0.5 m

Now, we have to find the length of each part.

Length of part I = 3.5 m
Length of part II = (3.5 - 0.5 - 0.5) = 2.5 m
Length of part III = (2.5 - 0.5 - 0.5) = 1.5 m
Length of part IV = (1.5 - 0.5 - 0.5) = 0.5 m

∴ Area of the shaded region = [Area of part (I) + Area of part (II) + Area of part (III) + Area of part (IV)] sq. units
= [(3.5 × 0.5) + (2.5 × 0.5) + (1.5 × 0.5) + (0.5 × 0.5)] m²
= [1.75 + 1.25 + 0.75 + 0.25] m²
= 4 m²
In simple words: Divide the complete figure into four separate rectangles based on the marked lines. Find the length of each rectangle by subtracting the marked widths. Calculate the area of each part by multiplying length by width, then add all areas together.

Exam Tip: When dividing a shape into parts, always ensure your component lengths and widths add up correctly to the total dimensions shown.

Exercise 20C

Question 1. The base of a parallelogram is 32 cm and its height is 16.5 cm. Find the area of the parallelogram.
Answer: Base = 32 cm
Height = 16.5 cm

∴ Area of the parallelogram = Base × Height
= 32 cm × 16.5 cm
= 528 cm²
In simple words: Multiply the base by the perpendicular height to get the area of any parallelogram.

Exam Tip: Always use the perpendicular height, not a slant side - this is the key to getting the correct area.

Question 2. The base of a parallelogram is 1 m 60 cm and its height is 75 cm. Find the area of the parallelogram.
Answer: Base = 1 m 60 cm = 1.6 m [since 100 cm = 1 m]
Height = 75 cm = 0.75 m

∴ Area of the parallelogram = Base × Height
= 1.6 m × 0.75 m
= 1.2 m²
In simple words: Convert all units to the same measurement before multiplying base by height.

Exam Tip: Units must be consistent - convert everything to either metres or centimetres before starting your calculation.

Question 3. (i) Base = 14 dm = (14 × 10) cm = 140 cm and Height = 6.5 dm = (6.5 × 10) cm = 65 cm [since 1 dm = 10 cm]
Area of the parallelogram = Base × Height = 140 cm × 65 cm = 9100 cm²
(ii) Base = 14 dm = (14 × 10) cm = 140 cm = 1.4 m and Height = 6.5 dm = (6.5 × 10) cm = 65 cm = 0.65 m [since 1 dm = 10 cm and 100 cm = 1 m]
∴ Area of the parallelogram = Base × Height = 1.4 m × 0.65 m = 0.91 m²
Answer: The area calculations for both sub-parts are shown above. Both give equivalent results - whether expressed in cm² or m², the area of the parallelogram is found by multiplying the base and height together using the same unit for both.
In simple words: The base times the height always gives the parallelogram's area. You can express your answer in different units, but the relationship remains the same.

Exam Tip: When working with multiple unit conversions, carefully track which unit each measurement is in to avoid mixing them up.

Question 4. A parallelogram has area 54 cm² and its base is 15 cm. Find its height.
Answer: Area of the given parallelogram = 54 cm²
Base of the given parallelogram = 15 cm

∴ Height of the given parallelogram = \( \frac{\text{Area}}{\text{Base}} \) = \( \left(\frac{54}{15}\right) \) cm = 3.6 cm
In simple words: Divide the area by the base to find the height of the parallelogram.

Exam Tip: Remember that the area formula can be rearranged - if you know area and base, you can find height by dividing area by base.

Question 5. The base of a parallelogram is 18 cm. Its area is 153 cm². Find the distance from the given side to its opposite side.
Answer: Base of the parallelogram = 18 cm
Area of the parallelogram = 153 cm²

∴ Area of the parallelogram = Base × Height
⇒ Height = \( \frac{\text{Area of the parallelogram}}{\text{Base}} \) = \( \left(\frac{153}{18}\right) \) cm = 8.5 cm

Hence, the distance of the given side from its opposite side is 8.5 cm.
In simple words: The distance between two opposite parallel sides of a parallelogram is its perpendicular height. Use the area and base to calculate this height.

Exam Tip: "Distance from a side to its opposite side" always refers to the perpendicular distance, which is the height used in the area formula.

Question 6. A parallelogram has base 18 cm and height 6.4 cm. Find its area. When BC is taken as the base, find the perpendicular distance from BC to AD.
Answer: Base, AB = 18 cm
Height, AL = 6.4 cm

∴ Area of the parallelogram ABCD = Base × Height
= (18 cm × 6.4 cm) = 115.2 cm² ...(i)

Now, taking BC as the base:
Area of the parallelogram ABCD = Base × Height
= (12 cm × AM) ...(ii)

From equation (i) and (ii):
12 cm × AM = 115.2 cm²
⇒ AM = \( \left(\frac{115.2}{12}\right) \) cm

= 9.6 cm
In simple words: Calculate the area using one base and its perpendicular height. Then, if you take a different side as the base, use the same area formula rearranged to find the new perpendicular height.

Exam Tip: The area of a parallelogram stays the same regardless of which side you choose as the base - use this fact to find unknown heights.

Question 8. ABCD is a parallelogram with side AB of length 15 cm and the corresponding altitude AE of length 4 cm. The adjacent side AD is of length 8 cm and the corresponding altitude is CF. Find the distance between the shorter sides.
Answer: We know that the area of a parallelogram equals base times height. Since we have two altitudes and two bases that correspond to them, we can set up the equation:

\( AD \times CF = AB \times AE \)
\( 8 \text{ cm} \times CF = 15 \text{ cm} \times 4 \text{ cm} \)
\( CF = \left( \frac{60}{8} \right) \text{ cm} = \left( \frac{15}{2} \right) \text{ cm} = 7.5 \text{ cm} \)

The distance between the shorter sides measures 7.5 cm.
In simple words: A parallelogram has different bases and heights. Multiply each base by its height - you always get the same area. Using this, we can find the missing height.

Exam Tip: Remember that the area of a parallelogram stays the same no matter which base-height pair you use. This property helps solve for unknown dimensions.

Question 9. The base of a parallelogram is 3 times its height. If the area is 108 cm², find the base and height.
Answer: Let the base of the parallelogram measure x cm. Then the height measures \( \frac{1}{3}x \) cm. The area is given as 108 cm².

Using the area formula for a parallelogram - base times height:
\( 108 \text{ cm}^2 = x \times \frac{1}{3}x \)
\( 108 \text{ cm}^2 = \frac{1}{3}x^2 \)
\( x^2 = (108 \times 3) \text{ cm}^2 = 324 \text{ cm}^2 \)
\( x^2 = (18)^2 \)
\( x = 18 \text{ cm} \)

Base \( = x = 18 \text{ cm} \)

Height \( = \frac{1}{3}x = \left( \frac{1}{3} \times 18 \right) \text{ cm} = 6 \text{ cm} \)
In simple words: Set up an equation using the area formula. The base is 3 times bigger than the height. Solve for x, then calculate each measurement.

Exam Tip: Always convert word relationships (like "base is 3 times height") into algebraic equations before solving. Check your answer by multiplying base × height to verify the area.

Question 10. A rhombus is a special type of a parallelogram. Find: (i) Area of the rhombus with base 12 cm and height 7.5 cm. (ii) Find the area of a rhombus with diagonals measuring 2 dm and 12.6 cm.
Answer: A rhombus is a special form of a parallelogram. The area of any parallelogram is computed by multiplying the base and the height.

(i) The area of the given rhombus - base times height:
\( = 12 \text{ cm} \times 7.5 \text{ cm} = 90 \text{ cm}^2 \)

(ii) Base \( = 2 \text{ dm} = (2 \times 10) \text{ cm} = 20 \text{ cm} \) [since 1 dm = 10 cm]
Height \( = 12.6 \text{ cm} \)
Area of the rhombus \( = 20 \text{ cm} \times 12.6 \text{ cm} = 252 \text{ cm}^2 \)
In simple words: A rhombus can be treated as a parallelogram. Multiply the base by the height to get the area. Convert units when needed to make sure they match.

Exam Tip: Watch out for unit conversion - especially when one measurement is in dm and another in cm. Always convert to the same unit before multiplying.

Question 11. Find the area of a rhombus with diagonals measuring: (i) 16 cm and 28 cm. (ii) 8 dm 5 cm and 5 dm 6 cm.
Answer:
(i) Length of one diagonal \( = 16 \text{ cm} \)
Length of the other diagonal \( = 28 \text{ cm} \)
Area of the rhombus \( = \frac{1}{2} \times \) (Product of the diagonals)
\( = \left( \frac{1}{2} \times 16 \times 28 \right) \text{ cm}^2 = 224 \text{ cm}^2 \)

(ii) Length of one diagonal \( = 8 \text{ dm } 5 \text{ cm} = (8 \times 10 + 5) \text{ cm} = 85 \text{ cm} \) [since 1 dm = 10 cm]
Length of the other diagonal \( = 5 \text{ dm } 6 \text{ cm} = (5 \times 10 + 6) \text{ cm} = 56 \text{ cm} \)
Area of the rhombus \( = \frac{1}{2} \times \) (Product of the diagonals)
\( = \left( \frac{1}{2} \times 85 \times 56 \right) \text{ cm}^2 = 2380 \text{ cm}^2 \)
In simple words: The area of a rhombus uses its diagonals, not the base and height. Multiply the two diagonals together, then divide by 2. Convert all measurements to the same unit first.

Exam Tip: For a rhombus, always remember the diagonal formula: \( \text{Area} = \frac{1}{2} \times d_1 \times d_2 \). This is different from the parallelogram formula - don't mix them up.

Question 12. In a rhombus ABCD, the diagonals intersect at O. Given AB = 20 cm and AC = 24 cm, find the area of the rhombus ABCD.
Answer: Let ABCD be the rhombus with diagonals meeting at O.

We are told that:
AB = 20 cm and AC = 24 cm
The diagonals of a rhombus cut each other at right angles.

Therefore, triangle AOB forms a right angle at O.

Here, \( OA = \frac{1}{2} AC = 12 \text{ cm} \)
AB = 20 cm

Using the Pythagorean theorem:
\( (AB)^2 = (OA)^2 + (OB)^2 \)
\( (20)^2 = (12)^2 + (OB)^2 \)
\( (OB)^2 = (20)^2 - (12)^2 \)
\( (OB)^2 = 400 - 144 = 256 \)
\( (OB)^2 = (16)^2 \)
\( OB = 16 \text{ cm} \)
\( BD = 2 \times OB = 2 \times 16 \text{ cm} = 32 \text{ cm} \)

Area of the rhombus ABCD \( = \left( \frac{1}{2} \times AC \times BD \right) \text{ cm}^2 \)
\( = \left( \frac{1}{2} \times 24 \times 32 \right) \text{ cm}^2 = 384 \text{ cm}^2 \)
In simple words: The diagonals of a rhombus cross at right angles. Use the Pythagorean theorem to find the unknown diagonal, then apply the area formula with both diagonals.

Exam Tip: Always recognize that rhombus diagonals bisect each other at 90 degrees. This property lets you use the Pythagorean theorem to find missing diagonal lengths.

Question 13. One diagonal of a rhombus is 19.2 cm long. If the area measures 148.8 cm², find the length of the other diagonal.
Answer: The area of a rhombus is calculated as one-half times the product of the diagonals.

Given:
Length of one diagonal \( = 19.2 \text{ cm} \)
Area of the rhombus \( = 148.8 \text{ cm}^2 \)

Length of the other diagonal \( = \left( \frac{148.8 \times 2}{19.2} \right) \text{ cm} = 15.5 \text{ cm} \)
In simple words: Use the area formula backwards. If you know the area and one diagonal, you can find the missing diagonal by rearranging the formula.

Exam Tip: When working with the diagonal formula, rearrange it algebraically to isolate the unknown. Double-check by multiplying your answer back into the area formula.

Question 14. The perimeter of a rhombus measures 56 cm and its area is 119 cm². Find the height of the rhombus.
Answer: Perimeter of the rhombus \( = 56 \text{ cm} \)
Area of the rhombus \( = 119 \text{ cm}^2 \)

Side of the rhombus \( = \frac{\text{Perimeter}}{4} = \left( \frac{56}{4} \right) \text{ cm} = 14 \text{ cm} \)

Area of a rhombus equals base times height:

Height of the rhombus \( = \frac{\text{Area}}{\text{Base}} = \left( \frac{119}{14} \right) \text{ cm} = 8.5 \text{ cm} \)
In simple words: A rhombus is a special parallelogram with 4 equal sides. Divide the perimeter by 4 to get one side length. Then divide the area by this side to find the height.

Exam Tip: Remember that for any rhombus, all four sides are equal. Use perimeter to find the side, then use the area formula to find the height.

Question 15. The height of a rhombus measures 17.5 cm and its area is 441 cm². What is the length of each side?
Answer: Given:
Height of the rhombus \( = 17.5 \text{ cm} \)
Area of the rhombus \( = 441 \text{ cm}^2 \)

We are aware that:
Area of a rhombus equals base times height

Base of the rhombus \( = \frac{\text{Area}}{\text{Height}} = \left( \frac{441}{17.5} \right) \text{ cm} = 25.2 \text{ cm} \)
Since each side of a rhombus is of equal length, the side measures 25.2 cm.
In simple words: Divide the area by the height to find the base. In a rhombus, the base and each side have the same length.

Exam Tip: Know that in a rhombus, the side length equals the base when using the area formula. Simply divide area by height to get your answer.

Question 16. A rhombus and a triangle have the same area. The rhombus has a base of 24.8 cm and height of 16.5 cm. The triangle has a base of 22 cm. What is the height of the triangle?
Answer: Area of a triangle is calculated as one-half times base times height:
\( = \left( \frac{1}{2} \times 24.8 \times 16.5 \right) \text{ cm}^2 = 204.6 \text{ cm}^2 \)

Given:
Area of the rhombus equals area of the triangle
Area of the rhombus \( = 204.6 \text{ cm}^2 \)

Area of the rhombus \( = \frac{1}{2} \times \) (Product of the diagonals)
Given:
Length of one diagonal \( = 22 \text{ cm} \)
Length of the other diagonal \( = \left( \frac{204.6 \times 2}{22} \right) \text{ cm} = 18.6 \text{ cm} \)
In simple words: Calculate the area of the rhombus using its base and height. Since the triangle has the same area, use that value with the triangle's base to find its height by rearranging the area formula.

Exam Tip: When two shapes have equal areas, you can use one to solve for an unknown dimension in the other. Set their area formulas equal and solve algebraically.

Exercise 20D

Question 1. Find the area of a triangle with: (i) Base 42 cm and height 25 cm. (ii) Base 16.8 m and height 75 cm. (iii) Base 8 dm and height 35 cm.
Answer: We are aware that:
Area of a triangle \( = \frac{1}{2} \times \text{Base} \times \text{Height} \)

(i) Base \( = 42 \text{ cm} \)
Height \( = 25 \text{ cm} \)
\( \text{Area of the triangle} = \left( \frac{1}{2} \times 42 \times 25 \right) \text{ cm}^2 = 525 \text{ cm}^2 \)

(ii) Base \( = 16.8 \text{ m} \)
Height \( = 75 \text{ cm} = 0.75 \text{ m} \) [since 100 cm = 1 m]
\( \text{Area of the triangle} = \left( \frac{1}{2} \times 16.8 \times 0.75 \right) \text{ m}^2 = 6.3 \text{ m}^2 \)

(iii) Base \( = 8 \text{ dm} = (8 \times 10) \text{ cm} = 80 \text{ cm} \) [since 1 dm = 10 cm]
Height \( = 35 \text{ cm} \)
\( \text{Area of the triangle} = \left( \frac{1}{2} \times 80 \times 35 \right) \text{ cm}^2 = 1400 \text{ cm}^2 \)
In simple words: Use the formula one-half times base times height for every triangle. Always make sure the base and height are measured in the same units before multiplying.

Exam Tip: Unit conversion is critical. Convert all measurements to a single unit before applying the area formula to avoid calculation errors.

Question 2. The area of a triangle is 72 cm². If the base measures 16 cm, what is the height?
Answer: Height of a triangle is computed as two times the area divided by the base.

Here, base \( = 16 \text{ cm} \) and area \( = 72 \text{ cm}^2 \)

\( \text{Height} = 2 \times \frac{72}{16} \text{ cm} = 9 \text{ cm} \)
In simple words: Rearrange the triangle area formula to solve for height. Multiply the area by 2, then divide by the base.

Exam Tip: Always rearrange the area formula algebraically before substituting numbers. This reduces arithmetic mistakes and makes your working clearer.

Question 3. The area of a triangle is 224 m². If the base is 28 m, find the height.
Answer: Height of a triangle is calculated as two times the area divided by the base.

Here, base \( = 28 \text{ m} \) and area \( = 224 \text{ m}^2 \)

\( \text{Height} = \left( \frac{2 \times 224}{28} \right) \text{ m} = 16 \text{ m} \)
In simple words: Double the area value, then divide by the base measurement to get the height.

Exam Tip: Use the rearranged formula directly: Height = (2 × Area) / Base. Verify your result by substituting back into the original area formula.

Question 4. If the area of a triangle is 90 cm² and the height is 12 cm, find the base.
Answer: Base of a triangle is calculated as two times the area divided by the height.

Here, height \( = 12 \text{ cm} \) and area \( = 90 \text{ cm}^2 \)

\( \text{Base} = \left( \frac{2 \times 90}{12} \right) \text{ cm} = 15 \text{ cm} \)
In simple words: Rearrange the triangle area formula to isolate the base. Multiply the area by 2, then divide by the height.

Exam Tip: The triangle area formula can be rearranged three ways - to find area, base, or height. Learn all three forms to solve any variant quickly.

Question 5. A rectangular field costs Rs. 14580 to cultivate at a rate of Rs. 1080 per hectare. If the height is x m, find the base and height. Also, verify that the base is 3 times the height.
Answer: Total cost of cultivating the field \( = \text{Rs. } 14580 \)
Rate of cultivating the field \( = \text{Rs. } 1080 \) per hectare

Area of the field \( = \left( \frac{\text{Total cost}}{\text{Rate per hectare}} \right) \text{ hectare} \)
\( = \left( \frac{14580}{1080} \right) \text{ hectare} \)
\( = 13.5 \text{ hectare} \)
\( = (13.5 \times 10000) \text{ m}^2 = 135000 \text{ m}^2 \) [since 1 hectare = 10000 m²]

Let the height of the field measure x m.
Then, its base measures 3x m.

Area of the field \( = \left( \frac{1}{2} \times 3x \times x \right) \text{ m}^2 = \left( \frac{3x^2}{2} \right) \text{ m}^2 \)
\( \left( \frac{3x^2}{2} \right) = 135000 \)
\( x^2 = \left( 135000 \times \frac{2}{3} \right) = 90000 \)
\( x = \sqrt{90000} = 300 \)
Base \( = (3 \times 300) = 900 \text{ m} \)
Height \( = 300 \text{ m} \)
In simple words: First find the area by dividing total cost by the rate per hectare. Convert to square metres. Then use the relationship that base equals 3 times the height to set up an equation and solve.

Exam Tip: Remember to convert hectares to square metres using the conversion factor 1 hectare = 10000 m². This step is often missed and causes wrong final answers.

Question 6. A right-angled triangle has one leg measuring 14.8 cm. If the area is 129.5 cm², find the length of the other leg.
Answer: Let the length of the other leg be h cm.

Then, area of the triangle \( = \left( \frac{1}{2} \times 14.8 \times h \right) \text{ cm}^2 = (7.4 h) \text{ cm}^2 \)

But it is stated that the area of the triangle equals 129.5 cm².
\( 7.4h = 129.5 \)
\( h = \left( \frac{129.5}{7.4} \right) = 17.5 \text{ cm} \)
\( \text{Length of the other leg} = 17.5 \text{ cm} \)
In simple words: In a right-angled triangle, the two legs serve as the base and height. Use the area formula with one known leg to solve for the unknown leg.

Exam Tip: For a right-angled triangle, always identify the two perpendicular sides (legs) as your base and height in the area formula. Don't use the hypotenuse.

Question 7. A right-angled triangle has a base of 1.2 m and a hypotenuse of 3.7 m. Find the area.
Answer: Here, base \( = 1.2 \text{ m} \) and hypotenuse \( = 3.7 \text{ m} \)

In a right-angled triangle:

Perpendicular \( = \sqrt{(H \text{ ypotenuse})^2 - (\text{Base})^2} \)

\( = \sqrt{(3.7)^2 - (1.2)^2} \)
\( = \sqrt{13.69 - 1.44} \)
\( = \sqrt{12.25} \)
\( = 3.5 \)

Area \( = \left( \frac{1}{2} \times \text{base} \times \text{perpendicular} \right) \text{ sq. units} \)
\( = \left( \frac{1}{2} \times 1.2 \times 3.5 \right) \text{ m}^2 \)
\( = 2.1 \text{ m}^2 \)
\( \text{Area of the right angled triangle} = 2.1 \text{ m}^2 \)
In simple words: Use the Pythagorean theorem to find the missing perpendicular side from the base and hypotenuse. Then apply the triangle area formula with the base and perpendicular.

Exam Tip: Always use the Pythagorean theorem correctly: hypotenuse² = base² + perpendicular². Rearrange to find a missing side before calculating the area.

Question 8. A right-angled triangle has an area of 1014 cm². If one leg measures 3x cm, and the other measures 4x cm, find the base and height.
Answer: In a right-angled triangle, when one leg serves as the base, the other leg becomes the height.

Let the given legs measure 3x cm and 4x cm, respectively.

Area of the triangle \( = \left( \frac{1}{2} \times 3x \times 4x \right) \text{ cm}^2 \)
\( 1014 = (6x^2) \)
\( 1014 = 6x^2 \)
\( x^2 = \left( \frac{1014}{6} \right) = 169 \)
\( x = \sqrt{169} = 13 \)
\( \text{Base} = (3 \times 13) = 39 \text{ cm} \)
\( \text{Height} = (4 \times 13) = 52 \text{ cm} \)
In simple words: The two legs of a right-angled triangle are its base and height. Use the area formula with the expressions 3x and 4x, then solve for x. Finally, calculate each leg's actual length.

Exam Tip: When you have variables in the dimensions, always substitute them into the area formula first, then solve for the variable before finding individual measurements.

Question 9. A right-angled triangular scarf has ∠B = 90°, BC = 80 cm, and AC = 1 m = 100 cm. Find the cost if the material costs Rs. 250 per m².
Answer: Consider a right-angled triangular scarf (ABC).
Here, ∠B = 90°
BC = 80 cm
AC = 1 m = 100 cm

Now, \( AB^2 + BC^2 = AC^2 \)
\( AB^2 + AC^2 - BC^2 = (100)^2 - (80)^2 \)
\( = (10000 - 6400) = 3600 \)
\( AB = \sqrt{3600} = 60 \text{ cm} \)

Area of the scarf ABC \( = \left( \frac{1}{2} \times BC \times AB \right) \text{ sq. units} \)
\( = \left( \frac{1}{2} \times 80 \times 60 \right) \text{ cm}^2 \)
\( = 2400 \text{ cm}^2 = 0.24 \text{ m}^2 \) [since 1 m² = 10000 cm²]

Rate of the cloth \( = \text{Rs. } 250 \) per m²
\( \text{Total cost of the scarf} = \text{Rs. } (250 \times 0.24) = \text{Rs. } 60 \)
Hence, the cost of the right angled scarf measures Rs. 60.
In simple words: Use the Pythagorean theorem to find the missing side. Calculate the area using the two legs. Convert square cm to square metres, then multiply by the rate per m² to find total cost.

Exam Tip: Always convert area units (cm² to m²) before multiplying by the price per unit area. Missing this conversion step causes large calculation errors.

Question 10. [Content continued on next page]

Question 11. Find the area of an equilateral triangle with: (i) Side 18 cm. (ii) Side 20 cm.
Answer:
(i) Side of the equilateral triangle \( = 18 \text{ cm} \)
Area of the equilateral triangle \( = \frac{\sqrt{3}}{4} (\text{Side})^2 \text{ sq. units} \)
\( = \frac{\sqrt{3}}{4} (18)^2 \text{ cm}^2 = (\sqrt{3} \times 81) \text{ cm}^2 \)
\( = (1.73 \times 81) \text{ cm}^2 = 140.13 \text{ cm}^2 \)

(ii) Side of the equilateral triangle \( = 20 \text{ cm} \)
Area of the equilateral triangle \( = \frac{\sqrt{3}}{4} (\text{Side})^2 \text{ sq. units} \)
\( = \frac{\sqrt{3}}{4} (20)^2 \text{ cm}^2 = (\sqrt{3} \times 100) \text{ cm}^2 \)
\( = (1.73 \times 100) \text{ cm}^2 = 173 \text{ cm}^2 \)
In simple words: For an equilateral triangle, use the special formula with square root of 3. Multiply this value by the side squared, then divide by 4. Use \( \sqrt{3} \approx 1.73 \) for calculations.

Exam Tip: The equilateral triangle formula is unique - it doesn't require the height as a separate input. Remember that \( \sqrt{3} \approx 1.73 \) for quick mental checks of your answer.

Question 12. The base of a triangle measures 24 cm. If the area of an equilateral triangle equals the area of this triangle, and the triangle's height measures h cm, find the height of the equilateral triangle in terms of the triangle's height.
Answer: Let the height of the triangle measure h cm.

Area of the triangle \( = \left( \frac{1}{2} \times \text{Base} \times \text{Height} \right) \text{ sq. units} \)
\( = \left( \frac{1}{2} \times 24 \times h \right) \text{ cm}^2 \)

Let the side of the equilateral triangle measure a cm.

Area of the equilateral triangle \( = \left( \frac{\sqrt{3}}{4} a^2 \right) \text{ sq. units} \)
\( = \left( \frac{\sqrt{3}}{4} \times 24 \times 24 \right) \text{ cm}^2 = (\sqrt{3} \times 144) \text{ cm}^2 \)
\( \left( \frac{1}{2} \times 24 \times h \right) = (\sqrt{3} \times 144) \)
\( 12 h = (\sqrt{3} \times 144) \)
\( h = \left( \frac{\sqrt{3} \times 144}{12} \right) - (\sqrt{3} \times 12) = (1.73 \times 12) = 20.76 \text{ cm} \)
\( \text{Height of the equilateral triangle} = 20.76 \text{ cm} \)
In simple words: Set the two area formulas equal to each other. Use the triangle's base and height with the equilateral formula to solve for the missing dimension.

Exam Tip: When two different triangle types have equal areas, equate their area expressions and solve algebraically. Keep the radicals until the final step to minimize rounding errors.

Question 13. [Content continues on next page]

Question 14. A triangle has sides measuring 33 cm, 44 cm, and 55 cm. Find the area of the triangle and the height on the 44 cm side.
Answer: Let \( a = 33 \text{ cm}, b = 44 \text{ cm}, \) and \( c = 55 \text{ cm} \).

Then, \( s = \frac{a+b+c}{2} = \left( \frac{33+44+55}{2} \right) \text{ cm} = \left( \frac{132}{2} \right) \text{ cm} = 66 \text{ cm} \)

Area of the triangle \( = \sqrt{s(s-a)(s-b)(s-c)} \text{ sq. units} \)
\( = \sqrt{66(66-33)(66-44)(66-55)} \text{ cm}^2 \)
\( = \sqrt{66 \times 33 \times 22 \times 11} \text{ cm}^2 \)
\( = \sqrt{6 \times 11 \times 3 \times 11 \times 2 \times 11 \times 11} \text{ cm}^2 \)
\( = (2 \times 2 \times 3 \times 7) \text{ cm}^2 \)
\( = 84 \text{ cm}^2 \)

Let the height on the 44 cm side be h cm.
Then, Area \( = \frac{1}{2} \times b \times h \)
\( 726 \text{ cm}^2 = \frac{1}{2} \times 44 \times h \)
\( h = \left( \frac{2 \times 726}{44} \right) \text{ cm} = 33 \text{ cm} \)

Area of the triangle \( = 726 \text{ cm}^2 \)
Height corresponding to the side measuring 44 cm \( = 33 \text{ cm} \)
In simple words: Use Heron's formula to find the area when all three sides are known. Then use the area formula backwards to find the height when the base is 44 cm.

Exam Tip: Heron's formula works for any triangle when all three sides are known. Always calculate the semi-perimeter s first, then substitute into the formula.

Question 15. Find the area of scalene triangles with sides: (i) 13 m, 14 m, 15 m. (ii) 52 cm, 56 cm, 60 cm. (iii) 91 m, 98 m, 105 m.
Answer:
(i) Let \( a = 13 \text{ m}, b = 14 \text{ m}, c = 15 \text{ m} \)
\( s = \frac{a+b+c}{2} = \left( \frac{13+14+15}{2} \right) \text{ cm} = \left( \frac{42}{2} \right) \text{ m} = 21 \text{ m} \)
\( \text{Area of the triangle} = \sqrt{s(s-a)(s-b)(s-c)} \text{ sq. units} \)
\( = \sqrt{21(21-13)(21-14)(21-15)} \text{ m}^2 \)
\( = \sqrt{21 \times 8 \times 7 \times 6} \text{ m}^2 \)
\( = \sqrt{21 \times 8 \times 7 \times 6} \text{ m}^2 \)
\( = \sqrt{3 \times 7 \times 2 \times 7 \times 6} \text{ m}^2 \)
\( = (2 \times 2 \times 3 \times 7) \text{ m}^2 \)
\( = 84 \text{ m}^2 \)

(ii) Let \( a = 52 \text{ cm}, b = 56 \text{ cm}, c = 60 \text{ cm} \)
\( s = \frac{a+b+c}{2} = \left( \frac{52+56+60}{2} \right) \text{ cm} = \left( \frac{168}{2} \right) \text{ cm} = 84 \text{ cm} \)
\( \text{Area of the triangle} = \sqrt{s(s-a)(s-b)(s-c)} \text{ sq. units} \)
\( = \sqrt{84(84-52)(84-56)(84-60)} \text{ cm}^2 \)
\( = \sqrt{84 \times 32 \times 28 \times 24} \text{ cm}^2 \)
\( = \sqrt{84 \times 32 \times 28 \times 24} \text{ cm}^2 \)
\( = \sqrt{12 \times 7 \times 4 \times 8 \times 4 \times 8 \times 4 \times 7 \times 3} \text{ cm}^2 \)
\( = (2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7 \times 3 \times 7) \text{ cm}^2 \)
\( = 1344 \text{ cm}^2 \)

(iii) Let \( a = 91 \text{ m}, b = 98 \text{ m}, c = 105 \text{ m} \)
\( s = \frac{a+b+c}{2} = \left( \frac{91+98+105}{2} \right) \text{ cm} = \left( \frac{294}{2} \right) \text{ m} = 147 \text{ m} \)
\( \text{Area of the triangle} = \sqrt{s(s-a)(s-b)(s-c)} \text{ sq. units} \)
\( = \sqrt{147(147-91)(147-98)(147-105)} \text{ m}^2 \)
\( = \sqrt{147 \times 56 \times 49 \times 42} \text{ m}^2 \)
\( = \sqrt{3 \times 49 \times 8 \times 7 \times 49 \times 6 \times 7} \text{ m}^2 \)
\( = (2 \times 2 \times 3 \times 7 \times 7 \times 7) \text{ m}^2 \)
\( = 4116 \text{ m}^2 \)
In simple words: Apply Heron's formula for all three triangles. Calculate the semi-perimeter for each, then substitute into the square root expression. Simplify the radicals by factoring.

Exam Tip: Break down each number under the square root into prime factors - this makes simplification easier and reduces calculation errors significantly.

Question 16. A triangle has sides of 42 cm, 34 cm, and 20 cm. Find the area and the height on the 42 cm side.
Answer: Let \( a = 42 \text{ cm}, b = 34 \text{ cm}, c = 20 \text{ cm} \)

Then, \( s = \frac{a+b+c}{2} = \left( \frac{42+34+20}{2} \right) \text{ cm} = \left( \frac{96}{2} \right) \text{ cm} = 48 \text{ cm} \)

Area of the triangle \( = \sqrt{s(s-a)(s-b)(s-c)} \text{ sq. units} \)
\( = \sqrt{48(48-42)(48-34)(48-20)} \text{ cm}^2 \)
\( = \sqrt{48 \times 6 \times 14 \times 28} \text{ cm}^2 \)
\( = \sqrt{6 \times 2 \times 2 \times 6 \times 14 \times 2 \times 14} \text{ cm}^2 \)
\( = (2 \times 2 \times 6 \times 14) \text{ cm}^2 = 336 \text{ cm}^2 \)

Let the height on the 42 cm side measure h cm.
Then, Area \( = \frac{1}{2} \times b \times h \)
\( 336 \text{ cm}^2 = \frac{1}{2} \times 42 \times h \)
\( h = \left( \frac{2 \times 336}{42} \right) \text{ cm} = 16 \text{ cm} \)

Area of the triangle \( = 336 \text{ cm}^2 \)
Height corresponding to the 42 cm side \( = 16 \text{ cm} \)
In simple words: Use Heron's formula with the three given sides to find the total area. Rearrange the basic area formula to find the height when you know the base is 42 cm.

Exam Tip: The order of sides doesn't matter for Heron's formula - it works with any labeling. Always simplify radicals by factoring before taking the square root.

Question 17. An isosceles triangle has equal sides measuring 48 cm and a base of 30 cm. Find the area.
Answer: Let each of the equal sides measure a cm.
\( a = 48 \text{ cm} \)
Base \( = b = 30 \text{ cm} \)

Area of the triangle \( = \left\{ \frac{1}{2} \times b \times \sqrt{ a^2 - \frac{b^2}{4} } \right\} \text{ sq. units} \)

\( = \left\{ \frac{1}{2} \times 48 \times \sqrt{ (30)^2 - \frac{(48)^2}{4} } \right\} \text{ cm}^2 = \left( 24 \times \sqrt{ 900 - \frac{2304}{4} } \right) \text{ cm}^2 \)
\( = \left( 24 \times \sqrt{ 900 - 576 } \right) \text{ cm}^2 = (24 \times \sqrt{324}) \text{ cm}^2 = (24 \times 18) \text{ cm}^2 = 432 \text{ cm}^2 \)
\( \text{Area of the triangle} = 432 \text{ cm}^2 \)
In simple words: The isosceles triangle formula uses the two equal sides and the base. Substitute the values into the formula, then simplify the expression under the square root before taking its value.

Exam Tip: For isosceles triangles, the special formula avoids needing to calculate the height separately. Remember the formula structure: \( \frac{1}{2} \times b \times \sqrt{a^2 - \frac{b^2}{4}} \).

Question 18. An isosceles triangle has two equal sides and a base of 12 cm. If the perimeter measures 32 cm, find the area.
Answer: Let each of the equal sides measure a cm.
\( a + a + 12 = 32 \Rightarrow 2a = 20 \Rightarrow a = 10 \)
\( b = 12 \text{ cm} \) and \( a = 10 \text{ cm} \)

Area of the triangle \( = \left\{ \frac{1}{2} \times b \times \sqrt{ a^2 - \frac{b^2}{4} } \right\} \text{ sq. units} \)

\( = \left\{ \frac{1}{2} \times 12 \times \sqrt{ 100 - \frac{144}{4} } \right\} \text{ cm}^2 = \left( 6 - \sqrt{100 - 36} \right) \text{ cm}^2 \)
\( = (6 \times \sqrt{64}) \text{ cm}^2 = (6 \times 8) \text{ cm}^2 = 48 \text{ cm}^2 \)
In simple words: Use the perimeter to find the equal sides, then apply the isosceles triangle area formula with the base and equal side length.

Exam Tip: Always extract individual side lengths from perimeter information before applying the area formula. This two-step approach prevents substitution mistakes.

Question 19. An isosceles triangle has equal sides of 30 cm. The perimeter is 76 cm. Find the area.
Answer: [Content continued on next page]

Question 20. (i) Here, r = 15 cm. Find the circumference = 2πr = (2 × 3.14 × 15) cm = 94.2 cm. Hence, the circumference of the given circle is 94.2 cm.
Answer: When the radius measures 15 cm, we apply the formula C = 2πr. Substituting the values, C = (2 × 3.14 × 15) cm, which gives us 94.2 cm. Therefore, the circumference of this circle equals 94.2 cm.
In simple words: Use the formula C = 2πr. Plug in r = 15 cm and π = 3.14. You get 2 times 3.14 times 15, which is 94.2 cm.

Exam Tip: Always remember that circumference = 2πr or πd. Use the correct value of π (3.14 or 22/7) as given in the question.

Question 21. (i) Here, r = 28 cm. Find the circumference = 2πr = (2 × 22/7 × 28) cm = 176 cm. Hence, the circumference of the given circle is 176 cm. (ii) Here, r = 1.4 m. Find the circumference = 2πr = (2 × 22/7 × 1.4) m = (2 × 22 × 0.2) m = 8.8 m. Hence, the circumference of the given circle is 8.8 m.
Answer:
(i) With r = 28 cm, applying C = 2πr gives (2 × 22/7 × 28) cm = 176 cm. So the circumference is 176 cm.
(ii) With r = 1.4 m, applying C = 2πr gives (2 × 22/7 × 1.4) m = (2 × 22 × 0.2) m = 8.8 m. So the circumference is 8.8 m.
In simple words: Use C = 2πr with π = 22/7. In part (i), multiply 2 times 22/7 times 28 to get 176 cm. In part (ii), multiply 2 times 22/7 times 1.4 to get 8.8 m.

Exam Tip: When multiplying, look for cancellation opportunities - fractions often simplify nicely with the radius value.

Question 22. (i) Here, d = 35 cm. Circumference = 2πr = (πd) [since 2r = d] = (22/5 × 35) cm = (22 × 5) = 110 cm. Hence, the circumference of the given circle is 110 cm. (ii) Here, d = 4.9 m. Circumference = 2πr = (πd) [since 2r = d] = (22/7 × 4.9) m = (22 × 0.7) = 15.4 m. Hence, the circumference of the given circle is 15.4 m.
Answer:
(i) With d = 35 cm, using C = πd gives (22/7 × 35) cm = (22 × 5) = 110 cm. The circumference is 110 cm.
(ii) With d = 4.9 m, using C = πd gives (22/7 × 4.9) m = (22 × 0.7) = 15.4 m. The circumference is 15.4 m.
In simple words: When you know the diameter, use C = πd directly. Multiply π times the diameter to get the circumference.

Exam Tip: Remember that C = πd is faster than C = 2πr when diameter is given - no need to find the radius first.

Question 23. Circumference of the given circle = 57.2 cm. Let the radius of the given circle be r cm. C = 2πr. So r = C/(2π) cm. r = (57.2 ÷ (2 × 22/7)) cm = 9.1 cm. Thus, radius of the given circle is 9.1 cm.
Answer: Given that C = 57.2 cm, we rearrange the formula C = 2πr to find r = C/(2π). Substituting C = 57.2, we get r = (57.2 ÷ (2 × 22/7)) cm = 9.1 cm. Therefore, the radius of the circle is 9.1 cm.
In simple words: When circumference is known, flip the formula around. Divide the circumference by 2π to get the radius. Here, 57.2 divided by (2 × 22/7) gives 9.1 cm.

Exam Tip: Always rearrange the formula when you need to find an unknown - don't just memorize one direction.

Question 24. Circumference of the given circle = 63.8 m. Let the radius of the given circle be r cm. C = 2πr. So r = C/(2π). r = (63.8 ÷ (2 × 22/7)) m = 10.15 m. Diameter of the given circle = 2r = (2 × 10.15) m = 20.3 m.
Answer: Given C = 63.8 m, we use r = C/(2π) to find r = (63.8 ÷ (2 × 22/7)) m = 10.15 m. The diameter is then 2r = (2 × 10.15) m = 20.3 m. So the diameter of the circle is 20.3 m.
In simple words: First, find the radius by dividing circumference by 2π. Then multiply the radius by 2 to get the diameter.

Exam Tip: Two-step problems: always complete the first step accurately before moving to the second - one error compounds.

Question 25. Find the area of the quadrilateral ABCD where AC = 26 cm, DL = 12.8 cm, and BM = 11.2 cm.
Answer: We have AC = 26 cm, DL = 12.8 cm, and BM = 11.2 cm. The area of triangle ADC = (1/2) × AC × DL = (1/2) × 26 cm × 12.8 cm = 166.4 cm². The area of triangle ABC = (1/2) × AC × BM = (1/2) × 26 cm × 11.2 cm = 145.6 cm². Therefore, the area of the quadrilateral ABCD = area of triangle ADC + area of triangle ABC = (166.4 + 145.6) cm² = 312 cm².
In simple words: Split the quadrilateral into two triangles along the diagonal AC. Find each triangle's area using (1/2) × base × height. Add them together to get the total area.

Exam Tip: When splitting a quadrilateral by a diagonal, always use perpendicular distances (heights) from the other two vertices to that diagonal as bases.

Question 26. Find the area of the quadrilateral ABCD where the sides and diagonals have specific measurements involving triangles ACD and ABC.
Answer: For triangle ACD, let a = 30 cm, b = 40 cm, and c = 50 cm. The semi-perimeter s = ((a + b + c)/2) = ((30 + 40 + 50)/2) = 60 cm. Using Heron's formula, Area of triangle ACD = √[s(s - a)(s - b)(s - c)] = √[60(60 - 30)(60 - 40)(60 - 50)] cm² = √[60 × 30 × 20 × 10] cm² = √360000 cm² = 600 cm². For triangle ABC, let a = 26 cm, b = 28 cm, and c = 30 cm. The semi-perimeter s = ((26 + 28 + 30)/2) = 42 cm. Using Heron's formula, Area of triangle ABC = √[s(s - a)(s - b)(s - c)] = √[42(42 - 26)(42 - 28)(42 - 30)] cm² = √[42 × 16 × 14 × 12] cm² = √[2 × 3 × 7 × 2 × 2 × 2 × 3 × 7 × 3 × 2 × 2 × 3] cm² = (2 × 2 × 2 × 2 × 3 × 7) cm² = 336 cm². Therefore, the area of the quadrilateral ABCD = area of triangle ACD + area of triangle ABC = (600 + 336) cm² = 936 cm².
In simple words: Apply Heron's formula to each triangle separately. Calculate the semi-perimeter first, then use the formula to find each area. Add both areas together to get the total.

Exam Tip: With Heron's formula, always compute (s - a), (s - b), and (s - c) carefully before multiplying - one arithmetic error ruins the entire calculation.

Question 27. Find the area of the shaded region in a rectangle with a triangle inside it, where AB = BC = 36 m and 24 m, and the triangle has specific dimensions.
Answer: Area of the rectangle = AB × BC = 36 m × 24 m = 864 m². Area of the triangle = (1/2) × AD × FE = (1/2) × BC × FE [since AD = BC] = (1/2) × 24 m × 15 m = 12 m × 15 m = 180 m². Therefore, area of the shaded region = area of the rectangle - area of the triangle = (864 - 180) m² = 684 m².
In simple words: Find the rectangle's area by multiplying length times width. Find the triangle's area using (1/2) × base × height. Subtract the triangle's area from the rectangle's area to get the shaded region.

Exam Tip: Always identify which dimension is the base and which is the height for the triangle - they must be perpendicular.

Question 28. Find the area of the shaded regions in a square with internal triangles, where the square has side 20 cm and is divided by lines creating triangle regions.
Answer:
(i) Area of rectangle ABCD = (10 cm × 18 cm) = 180 cm². Area of triangle I = (1/2) × 6 × 10 cm² = 30 cm². Area of triangle II = (1/2) × 8 × 10 cm² = 40 cm². Area of the shaded region = 180 - (30 + 40) cm² = (180 - 70) cm² = 110 cm².
(ii) Area of square ABCD = (side)² = (20 cm)² = 400 cm². Area of triangle I = (1/2) × 10 × 20 cm² = 100 cm². Area of triangle II = (1/2) × 10 × 10 cm² = 50 cm². Area of triangle III = (1/2) × 10 × 20 cm² = 100 cm². Area of the shaded region = (400 - (100 + 50 + 100)) cm² = (400 - 250) cm² = 150 cm².
In simple words: Calculate the total area of the rectangle or square. Calculate the area of each non-shaded triangle. Subtract all triangle areas from the total to find the shaded area.

Exam Tip: Label and identify each triangle carefully - missing even one triangle from your calculation will give the wrong answer.

Question 29. Find the area of a quadrilateral ABCD where BD is a diagonal of length 24 cm, and the perpendiculars from A and C to BD are AL = 5 cm and CM = 8 cm.
Answer: Let ABCD be the given quadrilateral where BD is the diagonal with length 24 cm. Let AL and CM be the perpendiculars from A and C to BD, with AL = 5 cm and CM = 8 cm. The area of the quadrilateral ABCD = (Area of triangle ABD) + (Area of triangle CBD) = [(1/2) × BD × AL] + [(1/2) × BD × CM] sq. units = [(1/2) × 24 × 5] + [(1/2) × 24 × 8] cm² = (60 + 96) cm² = 156 cm². Therefore, the area of the given quadrilateral is 156 cm².
In simple words: Divide the quadrilateral into two triangles using the diagonal. Each triangle has the diagonal as its base and a perpendicular distance as its height. Add the two areas together.

Exam Tip: The perpendicular distances must be truly perpendicular to the diagonal - check that the angles shown are right angles.

Question 30. Join points PR and SQ. These two lines bisect each other at point O. Here, AB = DC = SQ = 40 cm, AD = BC = RP = 25 cm. Also, OP = OR = RP/2 = 25/2 = 12.5 cm. From the figure we observe: Area of triangle SPO = Area of triangle SRO. Area of the shaded region = 2 × (Area of triangle SPO) = 2 × ((1/2) × SQ × OP) = 2 × ((1/2) × 40 cm × 12.5 cm) = 500 cm².
Answer: When PR and SQ are joined, they intersect at point O and bisect each other. Given AB = DC = SQ = 40 cm and AD = BC = RP = 25 cm. We determine OP = OR = RP/2 = 12.5 cm. From the geometry, triangle SPO and triangle SRO have equal areas. The shaded region's area = 2 × (Area of triangle SPO) = 2 × ((1/2) × SQ × OP) = 2 × ((1/2) × 40 cm × 12.5 cm) = 500 cm².
In simple words: The two diagonals cut each other in half at the center. Due to symmetry, two opposite triangles have the same area. Double the area of one triangle to get the shaded region.

Exam Tip: In a figure where diagonals bisect each other, opposite triangles formed are always equal in area - use this symmetry to simplify calculations.

Question 1. A circle has radius r = 15 cm. Find its circumference.
Answer: With r = 15 cm, we apply the formula Circumference = 2πr = (2 × 3.14 × 15) cm = 94.2 cm. Therefore, the circumference of this circle is 94.2 cm.
In simple words: Plug the radius into the formula C = 2πr. Using π = 3.14 and r = 15 cm gives 94.2 cm.

Exam Tip: Always check which value of π you should use - the problem will specify either 22/7 or 3.14.

Question 2. (i) A circle has radius r = 28 cm. Find its circumference. (ii) A circle has radius r = 1.4 m. Find its circumference.
Answer:
(i) With r = 28 cm, Circumference = 2πr = (2 × 22/7 × 28) cm = 176 cm.
(ii) With r = 1.4 m, Circumference = 2πr = (2 × 22/7 × 1.4) m = 8.8 m.
In simple words: Use C = 2πr with π = 22/7. For part (i), multiply 2 times 22/7 times 28 to get 176 cm. For part (ii), do the same with 1.4 m to get 8.8 m.

Exam Tip: When the radius or diameter contains fractions or decimals, look for cancellation to simplify before multiplying.

Question 3. (i) A circle has diameter d = 35 cm. Find its circumference. (ii) A circle has diameter d = 4.9 m. Find its circumference.
Answer:
(i) With d = 35 cm, Circumference = πd = (22/7 × 35) cm = 110 cm.
(ii) With d = 4.9 m, Circumference = πd = (22/7 × 4.9) m = 15.4 m.
In simple words: When diameter is given, use C = πd directly without finding the radius. Multiply π times the diameter.

Exam Tip: Using C = πd is faster than converting to radius first - save time when diameter is already known.

Question 4. A circle has circumference 57.2 cm. Find its radius.
Answer: Given C = 57.2 cm, we rearrange C = 2πr to get r = C/(2π). Substituting, r = (57.2 ÷ (2 × 22/7)) cm = 9.1 cm. Thus, the radius is 9.1 cm.
In simple words: When you know the circumference, rearrange the formula to find radius. Divide 57.2 by (2 × 22/7) to get 9.1 cm.

Exam Tip: Always show the rearrangement step clearly - this demonstrates you understand the formula, not just plugging in numbers.

Question 5. A circle has circumference 63.8 m. Find its diameter.
Answer: Given C = 63.8 m, we use r = C/(2π) to find r = (63.8 ÷ (2 × 22/7)) m = 10.15 m. The diameter is then d = 2r = (2 × 10.15) m = 20.3 m.
In simple words: First calculate the radius by dividing circumference by 2π. Then double the radius to get the diameter.

Exam Tip: Write down all intermediate results - this helps you catch errors and makes the solution easy to follow.

Question 6. The difference between the circumference of a circle and its diameter is 30 cm. Find the radius of the circle.
Answer: Let the radius of the circle be r cm. The circumference is 2πr. Given: (Circumference) - (Diameter) = 30 cm. So (2πr - 2r) = 30. Factoring: 2r(π - 1) = 30. So 2r(22/7 - 1) = 30. Thus 2r(15/7) = 30. Therefore r = (30 × 7)/(2 × 15) = 7 cm. The radius of the circle is 7 cm.
In simple words: Write out the condition as an equation: circumference minus diameter equals 30. Substitute formulas and solve for r by collecting like terms.

Exam Tip: Always set up the given condition as an equation first - this prevents errors and makes the logic clear.

Question 7. Two circles have radii 5x and 3x respectively. Find the ratio of their circumferences.
Answer: Let the radii of the two circles be 5x and 3x respectively. Let their circumferences be C₁ and C₂ respectively. C₁ = 2 × π × 5x = 10πx. C₂ = 2 × π × 3x = 6πx. The ratio C₁/C₂ = (10πx)/(6πx) = 10/6 = 5/3. Hence, the ratio of the circumferences is 5:3.
In simple words: The circumference is directly proportional to the radius. So if radii are in ratio 5:3, then circumferences are also in ratio 5:3.

Exam Tip: The ratio of circumferences always equals the ratio of radii (or diameters) - you don't even need to compute the full circumference values.

Question 8. A cyclist covers a circular field with circumference 2πr. The cyclist has a speed of 8 km per hour. Find the distance covered in 2000 rotations and the time taken.
Answer: Radius of the circular field is r = 21 m. Distance covered by the cyclist in one rotation equals the circumference = 2πr = (2 × 22/7 × 21) m = 132 m. Speed of the cyclist = 8 km per hour = (8000 m)/(60 × 60) s = (8000/3600) m/s = (20/9) m/s. Time taken by the cyclist to cover the field is (Distance covered by the cyclist)/(Speed of the cyclist) = [132 ÷ (20/9)] s = [(132 × 9)/20] s = (1188/20) s = 59.4 s.
In simple words: One lap around the field covers a distance equal to the circumference. Multiply this by the number of rotations to get total distance. Divide this distance by speed to get time.

Exam Tip: Always convert speed to the same units as distance before dividing - km/h must be converted to m/s for distances in meters.

Question 10. The width of a circular track is 14 m. The circumference of the inner track is 528 m and the circumference of the outer track is 616 m. Find the inner radius, outer radius, and width of the track.
Answer: Let the inner and outer radii of the track be r metres and R metres respectively. The inner circumference is 2πr = 528. So 2 × 22/7 × r = 528. Thus r = (528 × 7)/(2 × 22) = 84 m. The outer circumference is 2πR = 616. So 2 × 22/7 × R = 616. Thus R = (616 × 7)/(2 × 22) = 98 m. The width of the track is (R - r) = (98 - 84) = 14 m. The inner radius is 84 m, the outer radius is 98 m, and the width of the track is 14 m.
In simple words: Use the circumference formula to find each radius separately. Subtract the inner radius from the outer radius to get the width.

Exam Tip: When a track has two circles, always work with each circumference independently to find its own radius - don't try to find the width first.

Question 11. Two concentric circles have an inner radius of 98 cm and an outer radius of 1 m 26 cm. Find the difference in their circumferences.
Answer: Concentric circles are circles that share the same center point. The inner circle has radius r = 98 cm. The circumference of the inner circle = 2πr = (2 × 22/7 × 98) cm = 616 cm. The outer circle has radius R = 1 m 26 cm = 126 cm [since 1 m = 100 cm]. The circumference of the outer circle = 2πR = (2 × 22/7 × 126) cm = 792 cm. The difference in their circumferences = (792 - 616) cm = 176 cm. The circumference of the second circle is 176 cm larger than that of the first circle.
In simple words: Find each circle's circumference using C = 2πr. Subtract the smaller circumference from the larger one to get the difference.

Exam Tip: Always convert measurements to the same unit before comparing - 1 m 26 cm must become 126 cm when the other radius is in cm.

Question 13. A wire bent into the form of an equilateral triangle has a side of 8.8 cm. If the same wire is bent into a circle, find the diameter of the ring.
Answer: The length of the wire equals the perimeter of the equilateral triangle = 3 × side of the equilateral triangle = (3 × 8.8) cm = 26.4 cm. When the wire is bent into the form of a circle with radius r cm, the circumference of the circle = 26.4 cm. So 2πr = 26.4. Thus 2 × 22/7 × r = 26.4. Therefore r = (26.4 × 7)/(2 × 22) cm = 4.2 cm. The diameter = 2r = (2 × 4.2) cm = 8.4 cm. The diameter of the ring is 8.4 cm.
In simple words: The same wire has the same total length whether it forms a triangle or a circle. Find the perimeter as the triangle side length times 3. This equals the circumference, which you can use to find the radius and then diameter.

Exam Tip: "Same wire" means the perimeter of the shape and the circumference of the circle must be equal - use this relationship to set up your equation.

Question 14. A rhombus has all four sides equal to 33 cm. If a wire of the same length is bent into a circle, find the radius of the circle.
Answer: The circumference of the circle equals the perimeter of the rhombus = 4 × side of the rhombus = (4 × 33) cm = 132 cm. The circumference of the circle = 132 cm. So 2πr = 132. Thus 2 × 22/7 × r = 132. Therefore r = (132 × 7)/(2 × 22) cm = 21 cm. The radius of the circle is 21 cm.
In simple words: The perimeter of the rhombus (four sides) becomes the circumference when the same wire forms a circle. Divide this circumference by 2π to get the radius.

Exam Tip: All sides of a rhombus are equal - multiply one side by 4 to get the perimeter quickly.

Question 15. A wire of length equal to the perimeter of a rectangle with dimensions 18.7 m and 14.3 m is bent into a circle. Find the radius of the circle.
Answer: Length of the wire equals the perimeter of the rectangle = 2(l + b) = 2 × (18.7 + 14.3) cm = 66 cm. When the wire is bent into the form of a circle with radius r cm, the circumference of the circle = 66 cm. So 2πr = 66. Thus 2 × 22/7 × r = 66. Therefore r = (66 × 7)/(2 × 22) cm = 10.5 cm. The radius of the circle is 10.5 cm.
In simple words: Calculate the rectangle's perimeter by adding all four sides (or 2 times length plus 2 times width). This perimeter becomes the circumference. Use C = 2πr to find the radius.

Exam Tip: Perimeter of a rectangle = 2(l + b), not l + b - don't forget the factor of 2.

Question 17. The hour hand of a clock has length (r) = 4.2 cm and the minute hand has length (R) = 7 cm. Find the total distance covered by both hands in 24 hours.
Answer: Length of the hour hand (r) = 4.2 cm. Distance covered by the hour hand in 12 hours = 2πr = (2 × 22/7 × 4.2) cm = 26.4 cm. Distance covered by the hour hand in 24 hours = (2 × 26.4) = 52.8 cm. Length of the minute hand (R) = 7 cm. Distance covered by the minute hand in 1 hour = 2πR = (2 × 22/7 × 7) cm = 44 cm. Distance covered by the minute hand in 24 hours = (44 × 24) cm = 1056 cm. The sum of the distances covered by the tips of both the hands in 1 day = (52.8 + 1056) cm = 1108.8 cm.
In simple words: Each hand moves in a circle. The hour hand completes one full circle (circumference) every 12 hours. The minute hand completes one full circle every hour. Multiply each circumference by how many times it happens in 24 hours, then add them.

Exam Tip: The hour hand moves slowly (once per 12 hours), while the minute hand moves fast (once per hour) - don't confuse their speeds.

Question 18. A well has an outer diameter of 140 cm. The width of the stone parapet (the rim) is to be found. The length of the outer edge of the parapet is 616 cm. Find the width of the parapet.
Answer: Given: Diameter of the well (d) = 140 cm. Radius of the well (r) = (140/2) cm = 70 cm. Let the radius of the outer circle (including the stone parapet) be R cm. Length of the outer edge of the parapet = 616 cm. So 2πR = 616. Thus (2 × 22/7 × R) = 616. Therefore R = (616 × 7)/(2 × 22) cm = 98 cm. Now, width of the parapet = [Radius of the outer circle (including the stone parapet) - Radius of the well] = [98 - 70] cm = 28 cm. The width of the parapet is 28 cm.
In simple words: The parapet is a ring around the well. Find the outer radius using the given outer circumference. Subtract the well's radius to get the width of the ring.

Exam Tip: Visualize the well as two concentric circles - inner circle is the well itself, outer circle includes the parapet.

Question 19. A bus wheel has a diameter of 98 cm. The bus travels 308 cm in one rotation of the wheel. Find the distance covered by the bus in 2000 rotations.
Answer: In one rotation, the bus covers a distance equal to the circumference of the wheel. Now, diameter of the wheel = 98 cm. The circumference of the wheel = πd = (22/7 × 98) cm = 308 cm. Thus, the bus travels 308 cm in one rotation. The distance covered by the bus in 2000 rotations = (308 × 2000) cm = 616000 cm = 6160 m [since 1 m = 100 cm]. The bus will cover 6160 m in 2000 rotations.
In simple words: Each rotation covers one circumference. Multiply the circumference by 2000 to get the total distance. Convert cm to m by dividing by 100.

Exam Tip: Always convert your final answer to the most appropriate unit - here, 616000 cm is cleaner as 6160 m or 6.16 km.

Question 20. In one revolution, a cycle covers a distance equal to the circumference of the wheel. The diameter of the wheel is 70 cm. Find the distance covered by the cycle in 250 revolutions.
Answer: In one revolution, the cycle covers a distance equal to the circumference of the wheel. The diameter of the wheel = 70 cm. The circumference of the wheel = πd = (22/7 × 70) cm = 220 cm. Thus, the cycle covers 220 cm in one revolution. The distance covered by the cycle in 250 revolutions = (220 × 250) cm = 55000 cm = 550 m [since 1 m = 100 cm]. The cycle will cover 550 m in 250 revolutions.
In simple words: One full rotation of the wheel moves the cycle forward by one circumference. Multiply 220 cm by 250 rotations to get 55000 cm, which is 550 m.

Exam Tip: This is the same concept as Question 19 - each rotation equals one circumference distance traveled.

Question 21. A wheel has a diameter of 77 cm. Find the number of revolutions needed to cover a distance of 121 × 1000 m (or 121 km).
Answer: Diameter of the wheel = 77 cm. So Radius of the wheel = (77/2) cm. Circumference of the wheel = 2πr = (2 × 22/7 × 77/2) cm = (22 × 11) cm = 242 cm = (242/100) m = (121/50) m. Distance covered by the wheel in 1 revolution = (121/50) m. Now, (121 × 1000) m is covered by the car in (1 × 50/121 × 121 × 1000) revolutions, i.e. 50000 revolutions. The required number of revolutions = 50000. The bicycle will need 50000 revolutions to cover 121 km.
In simple words: In one revolution, the wheel travels one circumference distance. To find how many revolutions are needed for a given distance, divide the total distance by the circumference distance.

Exam Tip: Always check your unit conversions - here, 121 km = 121000 m, and each circumference is in meters, so the division works cleanly.

Exercise 20F

Question 1. Find the area of a circle with radius 21 cm.
Answer: The area of the circle is \( \pi r^2 \) square units.
\( = \left(\frac{22}{7} \times 21 \times 21\right) \text{ cm}^2 = (22 \times 3 \times 21) \text{ cm}^2 = 1386 \text{ cm}^2 \)
In simple words: Multiply 22/7 by the radius twice. With radius 21 cm, the area comes to 1386 square centimetres.

Exam Tip: Always use the formula \( \pi r^2 \) for circular area and substitute the correct radius value. Check your arithmetic when multiplying.

Question 2. Find the area of a circle with radius 3.5 m.
Answer: The area of the circle is \( \pi r^2 \) square units.
\( = \left(\frac{22}{7} \times 3.5 \times 3.5\right) \text{ m}^2 = (22 \times 0.5 \times 3.5) \text{ m}^2 = 38.5 \text{ m}^2 \)
In simple words: Using the formula \( \pi r^2 \), substitute r = 3.5 m. The area is 38.5 square metres.

Exam Tip: Be careful with decimal calculations. Multiply step-by-step to avoid arithmetic errors.

Question 3. The circumference of a circle is 264 cm. Find its area.
Answer: Let the radius of the circle be r cm.
Circumference = \( (2\pi r) \text{ cm} \)
\( \therefore (2\pi r) = 264 \)
\( \Rightarrow \left(2 \times \frac{22}{7} \times r\right) = 264 \)
\( \Rightarrow r = \left(\frac{264 \times 7}{2 \times 22}\right) = 42 \)
\( \therefore \) Area of the circle = \( \pi r^2 \)
\( = \left(\frac{22}{7} \times 42 \times 42\right) \text{ cm}^2 = 5544 \text{ cm}^2 \)
In simple words: When you know the circumference, find the radius first using \( 2\pi r \). Then apply the area formula \( \pi r^2 \).

Exam Tip: Always express the circumference formula correctly and solve for radius before calculating area.

Question 4. The circumference of a circle is 35.2 m. Find its area.
Answer: Let the radius of the circle be r m.
Then, its circumference will be \( (2\pi r) \text{ m} \).
\( \therefore (2\pi r) = 35.2 \)
\( \Rightarrow \left(2 \times \frac{22}{7} \times r\right) = 35.2 \)
\( \Rightarrow r = \left(\frac{35.2 \times 7}{2 \times 22}\right) = 5.6 \)
\( \therefore \) Area of the circle = \( \pi r^2 \)
\( = \left(\frac{22}{7} \times 5.6 \times 5.6\right) \text{ m}^2 = 98.56 \text{ m}^2 \)
In simple words: Work backwards from circumference to radius using \( 2\pi r = 35.2 \). Then calculate area with the radius you found.

Exam Tip: Rearrange the circumference formula carefully to isolate r, and watch for decimal precision.

Question 5. The area of a circle is 616 cm². Find its circumference.
Answer: Let the radius of the circle be r cm.
Then, its area will be \( \pi r^2 \text{ cm}^2 \).
\( \therefore \pi r^2 = 616 \)
\( \Rightarrow \left(\frac{22}{7} \times r \times r\right) = 616 \)
\( \Rightarrow r^2 = \left(\frac{616 \times 7}{22}\right) = 196 \)
\( \Rightarrow r = \sqrt{196} = 14 \)
\( \therefore \) Circumference of the circle = \( (2\pi r) \text{ cm} \)
\( = \left(2 \times \frac{22}{7} \times 14\right) \text{ cm} = 88 \text{ cm} \)
In simple words: From area, find the radius by solving \( \pi r^2 = 616 \). Once you have r, use \( 2\pi r \) to get the circumference.

Exam Tip: Take the square root correctly when solving for radius from the area equation.

Question 6. The area of a circle is 1386 m². Find its circumference.
Answer: Let the radius of the circle be r m.
Then, area = \( \pi r^2 \text{ m}^2 \)
\( \therefore \pi r^2 = 1386 \)
\( \Rightarrow \left(\frac{22}{7} \times r \times r\right) = 1386 \)
\( \Rightarrow r^2 = \left(\frac{1386 \times 7}{22}\right) = 441 \)
\( \Rightarrow r = \sqrt{441} = 21 \)
\( \therefore \) Circumference of the circle = \( (2\pi r) \text{ m} \)
\( = \left(2 \times \frac{22}{7} \times 21\right) \text{ m} = 132 \text{ m} \)
In simple words: Use the area formula to find what r must be. Once you know r = 21 m, the circumference follows from \( 2\pi r \).

Exam Tip: Check that your radius value is correct by substituting back into the area formula.

Question 7. The ratio of the radii of two circles is 4 : 5. Find the ratio of their areas.
Answer: Let \( r_1 \) and \( r_2 \) be the radii of the two circles, and let \( A_1 \) and \( A_2 \) be their respective areas.
\( \frac{r_1}{r_2} = \frac{4}{5} \)
\( \therefore \frac{A_1}{A_2} = \frac{\pi r_1^2}{\pi r_2^2} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2 = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \)
Hence, the ratio of the areas of the given circles is 16:25.
In simple words: When radii are in the ratio 4:5, their areas are in the ratio of their squares, which is 16:25.

Exam Tip: Remember that area ratios equal the square of the radius ratio — this is a key relationship for circles.

Question 8. A horse is tied to a pole. The horse grazes over an area of 2464 m² before the rope becomes taut. How long is the rope?
Answer: When a horse is tied to a pole, the pole becomes the central point and the area the horse can graze forms a circle. The string by which the horse is tied serves as the radius of the circle.
Thus, Radius of the circle (r) = Length of the string = 21 m
Now, area of the circle = \( \pi r^2 = \left(\frac{22}{7} \times 21 \times 21\right) \text{ m}^2 = 1386 \text{ m}^2 \)
\( \therefore \) Required area = 1386 m²
In simple words: The horse grazes in a circular region. If the grazing area is 2464 m², work backwards to find the rope length (radius) using the area formula.

Exam Tip: Recognize that grazing problems form circles, with the tied point at the centre and the rope length as the radius.

Question 9. A wire of length 44 cm is bent in the form of a circle. Another wire of the same length is bent to form a square. Which shape encloses more area?
Answer: Let a be one side of the square.
Area of the square = 121 cm² (given)
\( \Rightarrow a^2 = 121 \)
\( \Rightarrow a = 11 \text{ cm} \) (since 11 × 11 = 121)
Perimeter of the square = 4 × side = 4a = (4 × 11) cm = 44 cm
Length of the wire = Perimeter of the square = 44 cm
The wire is bent in the form of a circle.
Circumference of a circle = Length of the wire
\( \therefore \) Circumference of a circle = 44 cm
\( \Rightarrow 2\pi r = 44 \)
\( \Rightarrow \left(2 \times \frac{22}{7} \times r\right) = 44 \)
\( \Rightarrow r = \left(\frac{44 \times 7}{2 \times 22}\right) = 7 \text{ cm} \)
\( \therefore \) Area of the circle = \( \pi r^2 \)
\( = \left(\frac{22}{7} \times 7 \times 7\right) \text{ cm}^2 = 154 \text{ cm}^2 \)
Since the area enclosed by the circle (154 cm²) is greater than the area enclosed by the square (121 cm²), the circle encloses more area.
In simple words: Both wires are 44 cm long. The wire bent as a square encloses 121 cm², while the wire bent as a circle encloses 154 cm². The circle encloses more.

Exam Tip: When comparing shapes with the same perimeter, the circle always encloses the maximum area.

Question 10. A piece of wire of length 176 cm is bent to form a square. Another piece of wire of the same length is bent to form a circle. Which shape encloses more area?
Answer: It is given that the radius of the circle is 28 cm.
Length of the wire = Circumference of the circle
\( \Rightarrow \) Circumference of the circle = \( 2\pi r = \left(2 \times \frac{22}{7} \times 28\right) \text{ cm} = 176 \text{ cm} \)
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square = 176 cm
\( \Rightarrow 4a = 176 \)
\( \Rightarrow a = \left(\frac{176}{4}\right) \text{ cm} = 44 \text{ cm} \)
Thus, each side of the square is 44 cm.
Area of the square = (Side)² = (a)² = (44 cm)² = 1936 cm²
\( \therefore \) Required area of the square formed = 1936 cm²
In simple words: When a 176 cm wire is bent as a square, each side is 44 cm, giving area 1936 cm². This is larger than the circular area with the same perimeter.

Exam Tip: Always compute both areas before comparing—the square actually encloses more when the perimeter is very large.

Question 11. An acrylic sheet is 34 cm × 24 cm. It has 64 small circular buttons, each of diameter 3.5 cm. Find the area of the remaining acrylic sheet.
Answer: Area of the acrylic sheet = 34 cm × 24 cm = 816 cm²
Given that the diameter of a circular button is 3.5 cm.
\( \therefore \) Radius of the circular button (r) = \left(\frac{3.5}{2}\right) \text{ cm} = 1.75 \text{ cm} \)
\( \therefore \) Area of 1 circular button = \( \pi r^2 \)
\( = \left(\frac{22}{7} \times 1.75 \times 1.75\right) \text{ cm}^2 = 9.625 \text{ cm}^2 \)
\( \therefore \) Area of 64 such buttons = (64 × 9.625) cm² = 616 cm²
Area of the remaining acrylic sheet = (Area of the acrylic sheet - Area of 64 circular buttons)
= (816 - 616) cm² = 200 cm²
In simple words: Start with the sheet's total area. Find the area of one button and multiply by 64. Subtract from the total to get what remains.

Exam Tip: Carefully convert the diameter to radius—a diameter of 3.5 cm means radius is 1.75 cm, not 3.5 cm.

Question 12. A rectangular ground is 90 m × 32 m. A circular tank of radius 14 m is constructed in the middle. Find the cost of turfing the remaining ground at Rs 50 per square metre.
Answer: Area of the rectangular ground = 90 m × 32 m = (90 × 32) m² = 2880 m²
Given: Radius of the circular tank (r) = 14 m
\( \therefore \) Area covered by the circular tank = \( \pi r^2 = \left(\frac{22}{7} \times 14 \times 14\right) \text{ m}^2 = 616 \text{ m}^2 \)
\( \therefore \) Remaining portion of the rectangular ground for turfing = (Area of the rectangular ground - Area covered by the circular tank)
= (2880 - 616) m² = 2264 m²
Rate of turfing = Rs 50 per sq. metre
\( \therefore \) Total cost of turfing the remaining ground = Rs (50 × 2264) = Rs 1,13,200
In simple words: Find the total ground area, subtract the tank's area, then multiply the remaining area by the cost per square metre.

Exam Tip: Always subtract the circular area from the rectangular area before applying the rate to avoid costly mistakes.

Question 13. Two concentric circles are drawn with radii 7 cm each. The area of each of the four quadrants is equal to each other. Find the area of the shaded portion.
Answer: Area of each of the four quadrants is equal to each other with radius 7 cm.

Area of the square ABCD = (Side)² = (14 cm)² = 196 cm²
Sum of the areas of the four quadrants = \( \left(4 \times \frac{1}{4} \times \frac{22}{7} \times 7 \times 7\right) \text{ cm}^2 = 154 \text{ cm}^2 \)
\( \therefore \) Area of the shaded portion = Area of square ABCD - Areas of the four quadrants
= (196 - 154) cm² = 42 cm²
In simple words: The square has area 196 cm². Four equal quadrants have combined area 154 cm². The shaded region (corners of the square not covered by the quadrants) is 42 cm².

Exam Tip: Identify which parts are shaded and which are not—subtract the quadrants' area from the square's area.

Question 14. A rectangular field is 60 m × 40 m. A horse is tethered to a corner by a 14 m long rope. Find the area of the field that the horse can graze.
Answer: Let ABCD be the rectangular field.
Here, AB = 60 m
BC = 40 m
Let the horse be tethered to corner A by a 14 m long rope.
Then, it can graze through a quadrant of a circle of radius 14 m.
\( \therefore \) Required area of the field = \( \left(\frac{1}{4} \times \frac{22}{7} \times 14 \times 14\right) \text{ m}^2 = 154 \text{ m}^2 \)
Hence, horse can graze 154 m² area of the rectangular field.
In simple words: A horse tied at a corner can sweep out a quarter-circle. With a 14 m rope, this quarter-circle has area 154 m².

Exam Tip: When a rope is tied at a corner, only a quarter-circle is swept—not a full circle. Always use \( \frac{1}{4} \pi r^2 \).

Question 15. Two circles have areas in the ratio \( \pi R^2 - \pi r^2 \). Which circle has a larger area?
Answer: A visual diagram is shown below illustrating the relationship between a big circle (radius R) and a small circle (radius r). The annulus (ring) is formed by subtracting the small circle from the big circle.

In simple words: When you subtract a small circle from a big circle, you get a ring shape (annulus). The expression \( \pi R^2 - \pi r^2 \) represents the area of that ring.

Exam Tip: The annulus formula is fundamental—memorize it and understand that R (big radius) must be larger than r (small radius) for the area to be positive.

Question 16. A rectangular plot of land measures 8 m by 6 m. Four flower beds are positioned at each corner, each of dimensions 2 m × 2 m. A circular flower bed is placed in the middle. Find the area of the remaining plot.
Answer: Let ABCD be the rectangular plot of land that measures 8 m by 6 m.
\( \therefore \) Area of the plot = (8 × 6) = 48 m²
Area of the four flower beds = \( \left(4 \times \frac{1}{4} \times \frac{22}{7} \times 2 \times 2\right) \text{ m}^2 = \left(\frac{88}{7}\right) \text{ m}^2 \)
Area of the circular flower bed in the middle of the plot = \( \pi r^2 \)
\( = \left(\frac{22}{7} \times 2 \times 2\right) \text{ m}^2 = \left(\frac{88}{7}\right) \text{ m}^2 \)
Area of the remaining part = \( \left\{48 - \left(\frac{88}{7} + \frac{88}{7}\right)\right\} \text{ m}^2 = \left\{48 - \frac{176}{7}\right\} \text{ m}^2 = \left(\frac{336 - 176}{7}\right) \text{ m}^2 = \left(\frac{160}{7}\right) \text{ m}^2 = 22.86 \text{ m}^2 \)
\( \therefore \) Required area of the remaining plot = 22.86 m²
In simple words: Calculate the total plot area, subtract the four corner flower beds, then subtract the central circular flower bed to find what remains.

Exam Tip: When multiple shapes are removed from a rectangle, subtract each area carefully and check your arithmetic.

Exercise 20G

Question 1. Let ABCD be the rectangular plot. Then, AB = 16 cm, AC = 20 cm. Let BC = x cm. From right triangle ABC: AC² = AB² + BC². Therefore, (20)² = (16)² + x². Hence, x² = (20)² - (16)² = (400 - 256) = 144. So, x = √144 = 12. Thus, BC = 12 cm. Area of the plot = (16 × 12) cm² = 192 cm².
Answer: The answer is **(c) 192 cm²**. To solve this problem, we treat the diagonal as the hypotenuse of a right triangle formed by the two sides of the rectangle. Using the Pythagorean theorem, \( AC^2 = AB^2 + BC^2 \), we get \( (20)^2 = (16)^2 + x^2 \).
\( \Rightarrow x^2 = (20)^2 - (16)^2 = (400 - 256) = 144 \)
\( \Rightarrow x = \sqrt{144} = 12 \)
\( \therefore BC = 12 \text{ cm} \)
\( \therefore \) Area of the plot = \( (16 \times 12) \text{ cm}^2 = 192 \text{ cm}^2 \)
In simple words: Use the Pythagorean theorem to find the missing side of the rectangle from the diagonal. Then multiply length by width for the area.

Exam Tip: When given a diagonal and one side of a rectangle, always apply the Pythagorean theorem—the diagonal is always the longest side.

Question 2. The diagonal of the square is 12 cm. Find its area.
Answer: The answer is **(b) 72 cm²**.
Given: Diagonal of the square = 12 cm
\( \therefore \) Area of the square = \( \left\{\frac{1}{2} \times (Diagonal)^2\right\} \) sq. units
\( = \left\{\frac{1}{2} \times (12)^2\right\} \text{ cm}^2 = 72 \text{ cm}^2 \)
In simple words: For a square, if you know the diagonal, the area is half of the diagonal squared.

Exam Tip: Remember the formula: Area = \( \frac{1}{2} \times d^2 \) where d is the diagonal—this is faster than finding the side first.

Question 3. A square field has an area of 200 cm². Find the length of its diagonal.
Answer: The answer is **(b) 20 cm**.
\( \therefore \) Area of the square = \( \frac{1}{2} \times (Diagonal)^2 \) sq. units
Area of the square field = 200 cm²
Diagonal of a square = \( \sqrt{2 \times \text{Area of the square}} = (\sqrt{2 \times 200}) \text{ cm} = (\sqrt{400}) \text{ cm} = 20 \text{ cm} \)
\( \therefore \) Length of the diagonal of the square = 20 cm
In simple words: Reverse the area-diagonal relationship: if area = 200, then diagonal = \( \sqrt{2 \times 200} = 20 \) cm.

Exam Tip: Know both directions: Area from diagonal, and diagonal from area—they are inverses of each other.

Question 4. A square field has an area of 0.5 hectare. Find the length of its diagonal.
Answer: The answer is **(a) 100 m**.
\( \therefore \) Area of the square = \( \frac{1}{2} \times (Diagonal)^2 \) sq. units
Given: Area of square field = 0.5 hectare
= \( (0.5 \times 10000) \text{ m}^2 = 5000 \text{ m}^2 \) [since 1 hectare = 10000 m²]
Diagonal of a square = \( \sqrt{2 \times \text{Area of the square}} = (\sqrt{2 \times 5000}) \text{ m} = 100 \text{ m} \)
Hence, the length of the diagonal of a square field is 100 m.
In simple words: Convert hectares to square metres first (0.5 hectare = 5000 m²). Then use diagonal = \( \sqrt{2 \times 5000} = 100 \) m.

Exam Tip: Always convert land measurements (hectares, acres) to standard units (m²) before applying formulas.

Question 6. Let the breadth of the rectangular field be x m. Length = 3x m. Perimeter of the rectangular field = 2(l + b) ⇒ 240 = 2( x + 3x) ⇒ 240 = 2(4x) ⇒ 240 = 8x ⇒ x = (240/8) = 30. ∴ Length of the field = 3x = (3 × 30) m = 90 m
Answer: (c) 90 m

Exam Tip: Always set up the perimeter equation carefully with the given relationship between length and breadth, then solve for the unknown systematically.

Question 7. Let the side of the square be a cm. Area of the square = (a)² cm². Increased side = (a + 25% of a) cm = (a + 25a/100) cm = (a + a/4) cm = (5a/4) cm. Area of the square = (5a/4) cm² = (25a²/16) cm². Increase in the area = [(25a²/16) - a²] cm² = [(25a² - 16a²)/16] cm² = (9a²/16) cm². % increase in the area = [((9a²/16)/a²) × 100] = [(9/16) × 100] = (900/16) = 56.25
Answer: (d) 56.25%

Exam Tip: When calculating percentage increase, always divide the increase by the original value, then multiply by 100 - avoid mixing up numerator and denominator.

Question 8. Let the side of the square be a. Length of its diagonal = √2a. ∴ Required ratio = (√2a)/((√2a)) = (√2a)/(a√2) = 1 : 2
Answer: (b) 1:2

Exam Tip: For a square, memorise that the diagonal equals side times √2, then use this to find ratios quickly.

Question 9. We know that a square encloses more area even though its perimeter is the same as that of the rectangle. ∴ Area of a square > Area of a rectangle
Answer: (c) A > B

Exam Tip: Remember that for a fixed perimeter, the square always gives the maximum area among all rectangles.

Question 10. Let the length of the rectangular field be 5x. Breadth = 3x. Perimeter of the field = 2(l + b) = 480 m (given) ⇒ 480 = 2(5x + 3x) ⇒ 480 = 16x ⇒ x = 480/16 = 30. ∴ Length = 5x = (5 × 30) = 150 m. Breadth = 3x = (3 × 30) = 90 m. ∴ Area of the rectangular park = 150 m × 90 m = 13500 m²
Answer: (b) 13500 m²

Exam Tip: Always extract the ratio from the problem carefully, use perimeter to find the actual dimensions, then calculate the area.

Question 11. Total cost of carpeting = Rs 6000. Rate of carpeting = Rs 50 per m. ∴ Length of the carpet = (6000/50) m = 120 m. ∴ Area of the carpet = (120 × 75/100) m² = 90 m² [since 75 cm = 75/100 m]. Area of the floor = Area of the carpet = 90 m². ∴ Width of the room = (Area/Length) = (90/15) m = 6 m
Answer: (a) 6 m

Exam Tip: Convert all units to the same standard before performing calculations - here, convert centimetres to metres early to avoid errors.

Question 12. Let a = 13 cm, b = 14 cm and c = 15 cm. Then, s = (a + b + c)/2 = (13 + 14 + 15)/2 cm = 21 cm. ∴ Area of the triangle = √[s(s - a)(s - b)(s - c)] sq. units = √[21(21 - 13)(21 - 14)(21 - 15)] cm² = √[21 × 8 × 7 × 6] cm² = √[3 × 7 × 8 × 2 - 2 × 2 × 7 × 2 × 3] cm² = (2 × 2 × 3 × 7) cm² = 84 cm²
Answer: (a) 84 cm²

Exam Tip: Use Heron's formula for triangles when all three sides are given - calculate the semi-perimeter first, then apply the formula methodically.

Question 13. Base = 12 m. Height = 8 m. Area of the triangle = (1/2 × Base × Height) sq. units = (1/2 × 12 × 8) m² = 48 m²
Answer: (b) 48 m²

Exam Tip: The basic triangle area formula is half base times height - ensure you identify the correct height (perpendicular to the base) before calculating.

Question 14. Area of the equilateral triangle = 4√3 cm². We know: Area of an equilateral triangle = (√3/4) (side)² sq. units. ∴ Side of the equilateral triangle = √[(4Δ√3)/(√3)] cm = √[(4 × 4√3/√3)] cm = (√4 × 4) cm = (√16) cm = 4 cm
Answer: (b) 4 cm

Exam Tip: For equilateral triangles, learn the formula with √3/4 coefficient - this will let you move between area and side length quickly.

Question 15. It is given that one side of an equilateral triangle is 8 cm. ∴ Area of the equilateral triangle = (√3/4) (Side)² sq. units = (√3/4) (8)² cm² = (√3/4) × 64) cm² = (16√3) cm²
Answer: (c) 16√3 cm²

Exam Tip: Substitute the side length directly into the equilateral triangle area formula - be careful with the algebra when simplifying √3 × 64/4.

Question 16. Let ABC be an equilateral triangle with one side of the length a cm. Diagonal of an equilateral triangle = (√3/2) a cm = (√3/2) a - √(a × √2) = (√(a × √2))/√3 = 2√2 cm. Area of the equilateral triangle = (√3/4) a² = (√3/4) (2√2)² cm² = (√3/4 × 8) m² = 2√3 cm²
Answer: (b) 2√3 cm²

Exam Tip: When working with equilateral triangles, establish the relationship between the diagonal and side first, then use it to determine the side before finding area.

Question 17. Base of the parallelogram = 16 cm. Height of the parallelogram = 4.5 cm. ∴ Area of the parallelogram = Base × Height = (16 × 4.5) cm² = 72 cm²
Answer: (b) 72 cm²

Exam Tip: The area formula for a parallelogram mirrors the rectangle - use base times height, where height is the perpendicular distance between parallel sides.

Question 18. Length of one diagonal = 24 cm. Length of the other diagonal = 18 cm. ∴ Area of the rhombus = (1/2 × (Product of the diagonals)) = (1/2 × 24 × 18) cm² = 216 cm²
Answer: (b) 216 cm²

Exam Tip: For a rhombus, always use the formula involving the diagonals (half their product) - this is faster and more reliable than using base and height.

Question 19. Let the radius of the circle be r cm. Circumference = 2πr. (Circumference) - (Radius) = 37 ⇒ (2πr - r) = 37 ⇒ r(2π - 1) = 37 ⇒ r = 37/(2π - 1) = 37/((2 × 22/7) - 1) = 37/((44/7) - 1) = 37/((44 - 7)/7) = 37/(37/7) = 7. ∴ Radius of the given circle = 7 cm. ∴ Area = πr² = (22/7 × 7 × 7) cm² = 154 cm²
Answer: (c) 154 cm²

Exam Tip: Set up the equation based on the given relationship between circumference and radius, then solve for r before finding the area.

Question 20. Given: Perimeter of the floor = 2(l + b) = 18 m. Height of the room = 3 m. ∴ Area of the four walls = [2(l + b) × h] = Perimeter × Height = 18 m × 3 m = 54 m²
Answer: (c) 54 m²

Exam Tip: The four walls of a room form a surface whose area equals perimeter times height - this is a key formula for such problems.

Question 21. Area of the floor of a room = 14 m × 9 m = 126 m². Width of the carpet = 63 cm = 0.63 m (since 100 cm = 1 m). ∴ Required length of the carpet = Area of the floor of a room / Width of the carpet = (126/0.63) m = 200 m
Answer: (a) 200 m

Exam Tip: Convert all measurements to the same unit before dividing - converting centimetres to metres at the start saves calculation errors.

Question 22. Let the length of the rectangle be x cm and the breadth be y cm. Area of the rectangle = xy cm². Perimeter of the rectangle = 2( x + y) = 46 cm (given) ⇒ 2( x + y) = 46 ⇒ ( x + y) = (46/2) cm = 23 cm. Diagonal of the rectangle = √(x² + y²) = 17 cm ⇒ √(x² + y²) = 17. Squaring both the sides, we get: ⇒ x² + y² = (17)² ⇒ x² + y² = 289. Now, (x² + y²) = ( x + y)² - 2xy ⇒ 2xy = ( x + y)² - (x² + y²) = (23)² - 289 = 529 - 289 = 240 ⇒ xy = (240/2) cm² = 120 cm²
Answer: (c) 120 cm²

Exam Tip: Use the algebraic identity (x + y)² = x² + y² + 2xy to connect perimeter and diagonal information - this often leads to the area quickly.

Question 23. Let a side of the first square be a cm and that of the second square be b cm. Then, their areas will be a² and b², respectively. Their perimeters will be 4a and 4b, respectively. According to the question: (a²/b²) = (2/3) ⇒ (a/b) = (√(2/3)) ⇒ (a/b)² = (2/3) ⇒ (a/b) = (√2/√3). ∴ Required ratio of the perimeters = (4a/4b) = (4x3)/(4x1) = (3/1) = 3 : 1
Answer: (b) 3:1

Exam Tip: When comparing two squares, remember that the ratio of areas relates to the square of the ratio of sides - take square roots carefully.

Question 24. Let the diagonals be 2d and d. Area of the square = sq. units. Required ratio =
Answer: (d) 4:1

Exam Tip: For a square, the area can be expressed in terms of its diagonal - use the relationship diagonal² = 2 × area to establish ratios.

Question 25. Let one side of the square and that of an equilateral triangle be the same, i.e. a units. Then, Area of the square = (Side)² = (a)². Area of the equilateral triangle = (√3/4) (Side)² = (√3/4) (a)². ∴ Required ratio = (a²)/((√3/4) a²) = (a²)/((√3 a²)/4) = 4 : √3
Answer: (d) 4 : √3

Exam Tip: When comparing shapes with the same side length, write down both area formulas clearly and divide one by the other to get the ratio.

Question 26. Let the side of the square be x cm and the radius of the circle be r cm. Area of the square = Area of the circle ⇒ (x)² = πr². ∴ Side of the square (x) = √(πr²). Required ratio = Side of the square / Radius of the circle = (√(πr))/r = (√π)/√r = (√π) : 1
Answer: (a) √π : 1

Exam Tip: Set the areas equal first, solve for the relationship between the side and radius, then form the ratio of the requested dimensions.

Question 27. Let the radius of the circle be r cm. Then, its area = πr² cm². ∴ πr² = 154 ⇒ (22/7) × r × r = 154 ⇒ r² = (154 × 7)/22 = 49 ⇒ r = √49 cm = 7 cm. Side of the equilateral triangle = Radius of the circle = 7 cm. ∴ Area of the equilateral triangle = (√3/4) (side)² sq. units = (√3/4) (7)² cm² = (49√3/4) cm²
Answer: (b) (49√3/4) cm²

Exam Tip: First find the radius from the circle's area, then use it to set the triangle's side - this establishes the connection between the two shapes.

Question 28. Area of the rhombus = (1/2 × (Product of the diagonals)). Given: Length of one diagonal = 6 cm. Area of the rhombus = 36 cm². ∴ Length of the other diagonal = (2 × 36)/6) cm = 12 cm
Answer: (c) 12 cm

Exam Tip: Use the rhombus area formula to find an unknown diagonal when one diagonal and the area are known - rearrange to isolate the unknown.

Question 30. Let the radius of the circle be r m. Area = πr² m². ∴ πr² = 24.64 ⇒ (22/7 × r × r) = 24.64 ⇒ r² = (24.64 × 7)/22 = 7.84 ⇒ r = √7.84 = 2.8 m. ⇒ Circumference of the circle = (2πr) m = (2 × 22/7 × 2.8) m = 17.60 m
Answer: (c) 17.60 m

Exam Tip: Find the radius from the given area first using the area formula, then calculate circumference - avoid rounding intermediate steps.

Question 31. Suppose the radius of the original circle is r cm. Area of the original circle = πr². Radius of the circle = (r + 1) cm. According to the question: π(r + 1)² = πr² + 22 ⇒ π(r² + 1 + 2r) = πr² + 22 ⇒ πr² + π + 2πr = πr² + 22 ⇒ π + 2πr = 22 [cancel πr² from both the sides of the equation] ⇒ π(1 + 2r) = 22 ⇒ (1 + 2r) = (22/π) = (22π/22) = 7 ⇒ 2r = 7 - 1 = 6 ⇒ r = (6/2) cm = 3 cm. ∴ Original radius of the circle = 3 cm
Answer: (c) 3 cm

Exam Tip: When the radius increases by a fixed amount and the area change is given, set up the area difference equation and solve for the original radius systematically.

Question 32. Radius of the wheel = 1.75 m. Circumference of the wheel = 2πr = (2 × 22/7 × 1.75) cm = (2 × 22 × 0.25) m = 11 m. Distance covered by the wheel in 1 revolution = 11 m. Now, 11 m is covered by the car in 1 revolution. (11 × 1000) m will be covered by the car in (1 × 1/11 × 11 × 1000) revolutions, i.e. 1000 revolutions. ∴ Required number of revolutions = 1000
Answer: (c) 1000

Exam Tip: Calculate circumference first to find distance per revolution, then divide total distance by this value to find the number of revolutions needed.