RBSE Solutions Class 9 Maths Chapter 11 Area of Plane Figures More Ques

Multiple Choice Questions

Question 1. The length of side of an equilateral triangle is 8 cm, then the area of triangle is:
(A) 16√3 sq.cm
(B) 8√3 sq.cm
(C) 64√3 sq.cm
(D) 4√3 sq.cm
Answer: (A) 16√3 sq.cm
In simple words: To find the area of a triangle where all sides are equal (equilateral), you use a special formula. For a side length of 8 cm, the area calculates to 16 times the square root of 3 square centimeters.

🎯 Exam Tip: Remember the formula for the area of an equilateral triangle: \( \frac{\sqrt{3}}{4} \times (\text{side})^2 \). This is a common formula to apply.

Question 2. The sides of a triangle are 40 cm, 70 cm and 90 cm, then the area of triangle is:
(A) 600√5 sq. cm
(B) 500√6 sq. cm
(C) 482√5 sq. cm
(D) 60√5 sq. cm
Answer: (A) 600√5 sq. cm
In simple words: When a triangle has sides of different lengths (40 cm, 70 cm, and 90 cm), you can use Heron's formula to find its area. After calculating the semi-perimeter and applying the formula, the area comes out to 600 times the square root of 5 square centimeters.

🎯 Exam Tip: For triangles with three different side lengths, Heron's formula is essential. First, calculate the semi-perimeter (s), then use \( \sqrt{s(s-a)(s-b)(s-c)} \).

Question 3. The equal sides of an isosceles triangle is 6 cm and another side is 8 cm then its area will be:
(A) 8√5 sq. cm
(B) [Option text missing]
Answer: (A) 8√5 sq. cm
In simple words: For a triangle with two equal sides (6 cm) and one different side (8 cm), you can find the area by using the base and height formula. If you drop a perpendicular from the top vertex to the base, it divides the base into two equal halves. The area is 8 times the square root of 5 square centimeters.

🎯 Exam Tip: In an isosceles triangle, drawing an altitude to the unequal side creates two congruent right-angled triangles. Use the Pythagorean theorem to find the height, then apply the area formula \( \frac{1}{2} \times \text{base} \times \text{height} \).

Question 4. The perimeter of an equilateral triangle is 60 cm then its area will be:
(A) 400√3 sq.cm
(B) 100√3 sq.cm
(C) 50√3 sq. cm
(D) 200√3 sq. cm
Answer: (B) 100√3 sq.cm
In simple words: An equilateral triangle has three equal sides. If its total boundary (perimeter) is 60 cm, each side must be 20 cm. Using the area formula for an equilateral triangle, the area comes out to 100 times the square root of 3 square centimeters.

🎯 Exam Tip: Always find the side length first if the perimeter of an equilateral triangle is given. Side length = Perimeter / 3.

Question 5. The area of a right angled triangle is 36 sq. cm and its base 9 cm then the length of perpendicular will be:
(A) 8 cm
(B) 4 cm
(C) 16 cm
(D) 32 cm
Answer: (A) 8 cm
In simple words: For a right-angled triangle, the area is found by multiplying half of its base by its height (the perpendicular side). If the area is 36 sq. cm and the base is 9 cm, the perpendicular length must be 8 cm to get that area.

🎯 Exam Tip: The area of a right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). If any two values (area, base, or height) are known, you can easily find the third.

Question 6. The side of any square is 10 cm then its perimeter is:
(A) 20 cm
(B) 10 cm
(C) 40 cm
(D) 30 cm
Answer: (C) 40 cm
In simple words: A square has four sides of equal length. To find its perimeter, you just add up the lengths of all four sides. If one side is 10 cm, then the total perimeter will be 40 cm.

🎯 Exam Tip: The perimeter of a square is simply 4 times its side length. This is a very basic but important concept.

Question 7. The diagonals of a rhombus are 8 cm and 6 cm then its area will be:
(A) 48 sq. cm
(B) 24 sq. cm
(C) 12 sq. cm
(D) [Option text missing]
Answer: (B) 24 sq. cm
In simple words: A rhombus has two diagonals. To find its area, you multiply the lengths of the two diagonals and then divide by two. With diagonals of 8 cm and 6 cm, the area is 24 square centimeters.

🎯 Exam Tip: The area of a rhombus is \( \frac{1}{2} \times d_1 \times d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of its diagonals.

Question 8. [Question text missing]
(A) 40 sq. cm
(B) 80 sq. cm
(C) 120 sq. cm
(D) 160 sq. cm
Answer: (D) 160 sq. cm
In simple words: This question asks for an area, but the full question text is not available. The given options suggest it's a multiple-choice question. The correct answer is 160 square centimeters.

🎯 Exam Tip: When dealing with area problems, always double-check the units and ensure they are consistent throughout your calculation (e.g., all in cm or all in m).

Question 9. Find the length of the side of an equilateral triangle whose area is 9√3 sq. cm.
Answer:
Area of an equilateral triangle \( = \frac{\sqrt{3}}{4} (\text{side})^2 \)
\( \implies 9\sqrt{3} = \frac{\sqrt{3}}{4} (\text{side})^2 \)
\( \implies (\text{side})^2 = \frac{9\sqrt{3} \times 4}{\sqrt{3}} \)
\( \implies (\text{side})^2 = 36 \)
\( \implies \text{side} = \sqrt{36} \)
\( \implies \text{side} = 6 \text{ cm} \)
In simple words: We are given the area of an equilateral triangle and need to find its side length. We use the area formula for an equilateral triangle, rearrange it to solve for the side, and find that the side length is 6 cm. This shows how area can be used to find other dimensions.

🎯 Exam Tip: When working backward from area to side length, remember to take the square root at the final step. Always write down the formula first.

Question 10. Write the formula to find the area of the cyclic quadrilateral.
Answer:
Area of cyclic quadrilateral \( = \sqrt{ (s-a)(s-b)(s-c)(s-d) } \)
In simple words: The formula for the area of a cyclic quadrilateral (a four-sided shape whose vertices all lie on a circle) is called Brahmagupta's formula. Here, 's' is the semi-perimeter (half the sum of all sides) and 'a, b, c, d' are the lengths of the four sides.

🎯 Exam Tip: Make sure to correctly calculate the semi-perimeter 's' before plugging values into Brahmagupta's formula. This formula is similar to Heron's formula but for four sides.

Question 11. If the area of a square is 144 Are, then write its perimeter.
Answer:
Area of square \( = 144 \text{ Are} \)
We know that \( 1 \text{ Are} = 100 \text{ m}^2 \).
\( \implies (\text{side})^2 = 144 \times 100 \text{ m}^2 \)
\( \implies (\text{side})^2 = 14400 \text{ m}^2 \)
\( \implies \text{side} = \sqrt{14400} \text{ m} \)
\( \implies \text{side} = 120 \text{ m} \)
Perimeter of square \( = 4 \times \text{one side} \)
\( \implies \text{Perimeter} = 4 \times 120 \text{ m} \)
\( \implies \text{Perimeter} = 480 \text{ m} \)
In simple words: An 'Are' is a unit of area, equal to 100 square meters. First, we convert the given area of the square from 'Are' to square meters, then find the side length by taking the square root. Once we have the side length, we multiply it by 4 to get the perimeter.

🎯 Exam Tip: Always pay attention to unit conversions (e.g., Are to square meters) before starting calculations. A common mistake is to forget the conversion.

Question 12. If the area and base of a parallelogram are 174.60 sq. m and 18 m respectively, write its height.
Answer: The solution for this question is not provided in the source material. However, we can calculate it as follows:
Area of a parallelogram = base × height.
Given Area = \( 174.60 \text{ m}^2 \)
Given Base = \( 18 \text{ m} \)
\( \implies 174.60 = 18 \times \text{height} \)
\( \implies \text{height} = \frac{174.60}{18} \)
\( \implies \text{height} = 9.7 \text{ m} \)
In simple words: We know that the area of a parallelogram is found by multiplying its base by its height. So, if we divide the given area by the given base, we can find the height of the parallelogram.

🎯 Exam Tip: The height of a parallelogram is the perpendicular distance between its parallel bases. Ensure you use the correct height, not the slant side.

Question 13. [Question text incomplete] of perpendiculars upon the diagonals is 12 cm.
Answer:
Area of a quadrilateral \( = \frac{1}{2} (\text{sum of offsets}) \times \text{diagonal} \)
\( = \frac{1}{2} \times 12 \times 6 \)
\( = 36 \text{ cm}^2 \)
In simple words: This calculation finds the area of a specific type of quadrilateral, probably one where the diagonal and perpendiculars (offsets) from other vertices to that diagonal are given. We multiply half of the sum of the perpendiculars (12 cm in this case) by the length of the diagonal (which seems to be 6 cm here) to get the area.

🎯 Exam Tip: For quadrilaterals where a diagonal and perpendiculars to it from the other two vertices are given, this formula is key. Ensure you correctly identify the 'sum of offsets' and the 'diagonal'.

Question 14. Sides of a triangle are in the ratio 25 : 17 : 12 and its perimeter is 540 m. Find its area.
Answer:
Let sides of the triangle be \( 25x, 17x \text{ and } 12x \).
According to the question, perimeter \( = 540 \text{ m} \)
\( \implies 25x + 17x + 12x = 540 \text{ m} \)
\( \implies 54x = 540 \text{ m} \)
\( \implies x = \frac{540}{54} \)
\( \implies x = 10 \text{ m} \)
So, the actual sides are:
\( 25 \times 10 = 250 \text{ m} \)
\( 17 \times 10 = 170 \text{ m} \)
\( 12 \times 10 = 120 \text{ m} \)
Now, calculate the semi-perimeter (s):
\( s = \frac{250 + 170 + 120}{2} \)
\( s = \frac{540}{2} \)
\( s = 270 \text{ m} \)
Using Heron's formula for the area of the triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{270(270-250)(270-170)(270-120)} \)
\( = \sqrt{270 \times 20 \times 100 \times 150} \)
\( = \sqrt{ (3 \times 3 \times 3 \times 10) \times (2 \times 10) \times (10 \times 10) \times (3 \times 5 \times 10) } \)
\( = \sqrt{ 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 10 \times 10 \times 10 \times 10 } \)
\( = \sqrt{ (2 \times 3 \times 3 \times 5 \times 10 \times 10)^2 } \)
\( = 2 \times 3 \times 3 \times 5 \times 10 \times 10 \)
\( = 9000 \text{ m}^2 \)
Hence, the area of the triangle is \( 9000 \text{ m}^2 \).
In simple words: First, we use the given ratio of the sides and the perimeter to find the actual lengths of the three sides of the triangle. Then, we calculate the semi-perimeter. Finally, we apply Heron's formula to find the area of the triangle using its three side lengths.

🎯 Exam Tip: Always use a variable (like 'x') to represent the ratio when finding actual side lengths from a given perimeter. Remember Heron's formula for area when all three sides are known.

Question 15. Find the base of an isosceles triangle whose area is 12 sq. cm and length of one of the equal side is 5 cm.
Answer:
Let the equal sides be \( a = 5 \text{ cm} \). Let the base be \( b \).
Area of an isosceles triangle \( = 12 \text{ sq. cm} \)
The formula for the area of an isosceles triangle is: \( \frac{1}{4} b \sqrt{4a^2 - b^2} \)
\( \implies 12 = \frac{1}{4} b \sqrt{4(5^2) - b^2} \)
\( \implies 48 = b \sqrt{100 - b^2} \)
Squaring both sides:
\( (48)^2 = (b \sqrt{100 - b^2})^2 \)
\( 2304 = b^2 (100 - b^2) \)
\( 2304 = 100b^2 - b^4 \)
Rearranging the equation to form a quadratic in \( b^2 \):
\( b^4 - 100b^2 + 2304 = 0 \)
Let \( y = b^2 \). Then \( y^2 - 100y + 2304 = 0 \)
We need to find two numbers that multiply to 2304 and add to -100. These are -36 and -64.
\( y^2 - 64y - 36y + 2304 = 0 \)
\( y(y - 64) - 36(y - 64) = 0 \)
\( (y - 64)(y - 36) = 0 \)
So, \( y - 64 = 0 \) or \( y - 36 = 0 \)
\( y = 64 \) or \( y = 36 \)
Since \( y = b^2 \):
\( b^2 = 64 \implies b = \sqrt{64} \implies b = \pm 8 \)
\( b^2 = 36 \implies b = \sqrt{36} \implies b = \pm 6 \)
Since length cannot be negative, possible base lengths are 8 cm or 6 cm.
Both are valid. If base is 8, height is \( \sqrt{5^2 - 4^2} = \sqrt{25-16} = \sqrt{9} = 3 \). Area \( = \frac{1}{2} \times 8 \times 3 = 12 \).
If base is 6, height is \( \sqrt{5^2 - 3^2} = \sqrt{25-9} = \sqrt{16} = 4 \). Area \( = \frac{1}{2} \times 6 \times 4 = 12 \).
Hence, the base can be either 8 cm or 6 cm.
In simple words: We start with the formula for the area of an isosceles triangle, using the given area and the length of the equal sides. This leads to a complex equation for the base. By solving this equation, we find two possible values for the base: 8 cm or 6 cm. Both values give the correct area for the triangle.

🎯 Exam Tip: For problems involving area and unknown side lengths, sometimes a quadratic equation needs to be solved. Always check if both positive solutions make geometric sense for lengths.

Question 16. The perimeter of a triangle is 40 cm and two of its sides are 8 cm and 15 cm. Find the area of the triangle and length of the perpendicular of the largest side from the opposite vertex.
Answer:
Perimeter of \( \triangle ABC = 40 \text{ cm} \) (given)
We know that \( AB + BC + CA = \text{Perimeter} \)
\( \implies 8 + 15 + CA = 40 \text{ cm} \)
\( \implies 23 + CA = 40 \text{ cm} \)
\( \implies CA = 40 - 23 \)
\( \implies CA = 17 \text{ cm} \)
So, the sides of the triangle are \( a=8 \text{ cm}, b=15 \text{ cm}, c=17 \text{ cm} \).
Calculate the semi-perimeter (s):
\( s = \frac{8+15+17}{2} \)
\( s = \frac{40}{2} \)
\( s = 20 \text{ cm} \)
Using Heron's formula for the area of the triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{20(20-8)(20-15)(20-17)} \)
\( = \sqrt{20 \times 12 \times 5 \times 3} \)
\( = \sqrt{(4 \times 5) \times (3 \times 4) \times 5 \times 3} \)
\( = \sqrt{2^2 \times 5 \times 3 \times 2^2 \times 5 \times 3} \)
\( = \sqrt{2^4 \times 3^2 \times 5^2} \)
\( = 2^2 \times 3 \times 5 \)
\( = 4 \times 3 \times 5 \)
\( = 60 \text{ cm}^2 \)
The area of \( \triangle ABC \) is \( 60 \text{ cm}^2 \).
To find the length of the perpendicular from the opposite vertex to the largest side:
The largest side is \( CA = 17 \text{ cm} \). Let the perpendicular be \( h \).
Area of triangle \( = \frac{1}{2} \times \text{base} \times \text{height} \)
\( \implies 60 = \frac{1}{2} \times CA \times h \)
\( \implies 60 = \frac{1}{2} \times 17 \times h \)
\( \implies 120 = 17 \times h \)
\( \implies h = \frac{120}{17} \)
\( \implies h = 7\frac{1}{17} \text{ cm} \)
Hence, the area of \( \triangle ABC = 60 \text{ cm}^2 \) and the length of the perpendicular on the largest side from the opposite vertex is \( 7\frac{1}{17} \text{ cm} \).
In simple words: First, we find the length of the third side using the given perimeter and two sides. Then, we use Heron's formula to calculate the total area of the triangle. Finally, using the area and the longest side as the base, we calculate the length of the altitude (perpendicular) to that side.

🎯 Exam Tip: For perpendicular length, remember that any side can be a base. Choose the largest side as the base to find the perpendicular to it. Also, check if the triangle is a right-angled triangle by Pythagorean theorem (8, 15, 17 is a Pythagorean triplet). If so, one side is the altitude directly.

Question 17. The perimeter of a rhombus is 146 cm and length of one of its diagonal is 55 cm. Find the other diagonal and area of the rhombus.
Answer:
Perimeter of a rhombus \( = 146 \text{ cm} \)
A rhombus has 4 equal sides. Let the side length be \( a \).
\( \implies 4a = 146 \text{ cm} \)
\( \implies a = \frac{146}{4} \)
\( \implies a = 36.5 \text{ cm} \)
Let the diagonals be \( d_1 \) and \( d_2 \). We are given one diagonal, say \( d_1 = 55 \text{ cm} \).
In a rhombus, the diagonals bisect each other at right angles. This means we can form a right-angled triangle with half of each diagonal and one side of the rhombus.
The formula relating side and diagonals is: \( 4a^2 = d_1^2 + d_2^2 \)
\( \implies 4(36.5)^2 = (55)^2 + d_2^2 \)
\( \implies 4(1332.25) = 3025 + d_2^2 \)
\( \implies 5329 = 3025 + d_2^2 \)
\( \implies d_2^2 = 5329 - 3025 \)
\( \implies d_2^2 = 2304 \)
\( \implies d_2 = \sqrt{2304} \)
\( \implies d_2 = 48 \text{ cm} \)
The length of the other diagonal is 48 cm.
Now, find the area of the rhombus:
Area \( = \frac{1}{2} \times d_1 \times d_2 \)
\( = \frac{1}{2} \times 55 \times 48 \)
\( = 55 \times 24 \)
\( = 1320 \text{ sq. cm} \)
In simple words: First, we find the side length of the rhombus by dividing its perimeter by four. Then, we use the relationship between the side and diagonals of a rhombus to find the length of the second diagonal. Finally, we calculate the area of the rhombus using the lengths of both diagonals.

🎯 Exam Tip: Remember the two key properties of a rhombus: all four sides are equal, and its diagonals bisect each other at right angles. This allows you to use the Pythagorean theorem or the formula \( 4a^2 = d_1^2 + d_2^2 \).

Question 18. A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Answer:
Let the rhombus be ABCD.
Side length \( a = 30 \text{ m} \). Longer diagonal \( d_1 = 48 \text{ m} \).
To find the area of the rhombus, we first need to find the other diagonal \( d_2 \).
Using the formula: \( 4a^2 = d_1^2 + d_2^2 \)
\( \implies 4(30)^2 = (48)^2 + d_2^2 \)
\( \implies 4(900) = 2304 + d_2^2 \)
\( \implies 3600 = 2304 + d_2^2 \)
\( \implies d_2^2 = 3600 - 2304 \)
\( \implies d_2^2 = 1296 \)
\( \implies d_2 = \sqrt{1296} \)
\( \implies d_2 = 36 \text{ m} \)
Now, calculate the total area of the rhombus field:
Area \( = \frac{1}{2} \times d_1 \times d_2 \)
\( = \frac{1}{2} \times 48 \times 36 \)
\( = 24 \times 36 \)
\( = 864 \text{ m}^2 \)
The total area for 18 cows is \( 864 \text{ m}^2 \).
Area of grass field each cow will get \( = \frac{\text{Total Area}}{\text{Number of cows}} \)
\( = \frac{864}{18} \)
\( = 48 \text{ m}^2 \)
So, each cow will get \( 48 \text{ m}^2 \) of grass field.
In simple words: We first find the length of the shorter diagonal of the rhombus using its side length and the longer diagonal. Then, we calculate the total area of the rhombus field. Finally, we divide the total area by the number of cows to find out how much grass area each cow gets.

🎯 Exam Tip: Always break down complex problems into smaller, manageable steps. Here, finding the second diagonal is crucial before calculating the area and then dividing it among the cows.

Question 19. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
Answer:
The sides of one triangular piece of cloth are \( a=20 \text{ cm}, b=50 \text{ cm}, c=50 \text{ cm} \).
First, find the semi-perimeter (s) of one triangular piece:
\( s = \frac{20+50+50}{2} \)
\( s = \frac{120}{2} \)
\( s = 60 \text{ cm} \)
Now, use Heron's formula to find the area of one triangular piece:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
\( = \sqrt{60(60-20)(60-50)(60-50)} \)
\( = \sqrt{60 \times 40 \times 10 \times 10} \)
\( = \sqrt{(6 \times 10) \times (4 \times 10) \times 10 \times 10} \)
\( = \sqrt{2 \times 3 \times 10 \times 2 \times 2 \times 10 \times 10 \times 10} \)
\( = \sqrt{2^4 \times 3 \times 10^4} \)
\( = \sqrt{16 \times 3 \times 10000} \)
\( = 4 \times 100 \times \sqrt{3} \)
\( = 400\sqrt{6} \text{ cm}^2 \) (The source calculation leads to 200√6, let's re-verify) \( \sqrt{60 \times 40 \times 10 \times 10} = \sqrt{2400 \times 100} = 100 \sqrt{24} = 100 \sqrt{4 \times 6} = 100 \times 2 \sqrt{6} = 200\sqrt{6} \text{ cm}^2 \)
So, the area of one triangular piece is \( 200\sqrt{6} \text{ cm}^2 \).
The umbrella is made of 10 such triangular pieces of two different colours.
This means there are 5 pieces of each colour.
Area of cloth of each colour required \( = 5 \times \text{Area of one triangular piece} \)
\( = 5 \times 200\sqrt{6} \text{ cm}^2 \)
\( = 1000\sqrt{6} \text{ cm}^2 \)
In simple words: First, we find the area of one triangular piece of cloth using its side lengths and Heron's formula. Since the umbrella uses 10 pieces of two different colours, there are 5 pieces of each colour. We then multiply the area of one piece by 5 to find the total cloth needed for each colour.

🎯 Exam Tip: When using Heron's formula, be careful with factorization and simplification of the square root. Always double-check your calculations, especially with larger numbers.

Question 20. A trapezium with its parallel sides in the ratio 16 : 5, is cut from a rectangle whose sides are 63 metre and 5 metre respectively. The area of the trapezium is \( \frac{4}{15} \) of the rectangle. Find the length of the parallel sides of the trapezium.
Answer:
Let the parallel sides of the trapezium be \( AB = 16x \) and \( CD = 5x \).
The trapezium is cut from a rectangle whose sides are 63 m and 5 m.
Area of the rectangle \( = \text{length} \times \text{breadth} = 63 \text{ m} \times 5 \text{ m} = 315 \text{ m}^2 \).
The height of the trapezium is the shorter side of the rectangle, so height \( h = 5 \text{ m} \).
Area of the trapezium \( = \frac{4}{15} \) of the area of the rectangle.
\( \implies \text{Area of trapezium} = \frac{4}{15} \times 315 \text{ m}^2 \)
\( \implies \text{Area of trapezium} = 4 \times 21 \text{ m}^2 \)
\( \implies \text{Area of trapezium} = 84 \text{ m}^2 \)
Now, using the formula for the area of a trapezium:
Area \( = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height} \)
\( \implies 84 = \frac{1}{2} \times (16x + 5x) \times 5 \)
\( \implies 84 = \frac{1}{2} \times (21x) \times 5 \)
\( \implies 84 = \frac{105x}{2} \)
\( \implies 168 = 105x \)
\( \implies x = \frac{168}{105} \)
\( \implies x = \frac{56}{35} = \frac{8}{5} = 1.6 \)
Now, find the lengths of the parallel sides:
\( AB = 16x = 16 \times 1.6 = 25.6 \text{ m} \)
\( CD = 5x = 5 \times 1.6 = 8 \text{ m} \)
Hence, the parallel sides are 25.6 m and 8 m respectively.
In simple words: First, we find the area of the rectangle from which the trapezium is cut. Then, we calculate the area of the trapezium using the given fraction of the rectangle's area. We then set up an equation using the trapezium's area formula and the given ratio of its parallel sides to find the value of 'x'. Finally, we use 'x' to find the actual lengths of the parallel sides.

🎯 Exam Tip: Understand how the trapezium's dimensions relate to the rectangle it's cut from. The height of the trapezium is usually the height of the rectangle, unless specified otherwise.

Question 21. A rectangular field is of length 99 metre and area 4356 sq. m. A road 4.5 m wide has been constructed centrally in the field parallel to its length and breadth. Find the total number of square blocks of side 1.5 m to cover the road.
Answer:
Area of the rectangular field \( = 4356 \text{ m}^2 \)
Length of the field \( = 99 \text{ m} \)
Breadth of the field \( = \frac{\text{Area}}{\text{Length}} = \frac{4356}{99} = 44 \text{ m} \)
Width of the road \( = 4.5 \text{ m} \)
Area of the road parallel to the length \( = \text{Length of field} \times \text{Road width} \)
\( = 99 \text{ m} \times 4.5 \text{ m} = 445.5 \text{ m}^2 \)
Area of the road parallel to the breadth \( = \text{Breadth of field} \times \text{Road width} \)
\( = 44 \text{ m} \times 4.5 \text{ m} = 198 \text{ m}^2 \)
The common square portion (where the roads intersect) will be counted twice if we simply add the two road areas. We need to subtract its area once.
Area of the common square portion \( = \text{Road width} \times \text{Road width} \)
\( = 4.5 \text{ m} \times 4.5 \text{ m} = 20.25 \text{ m}^2 \)
Total area of the road \( = (\text{Area of horizontal road}) + (\text{Area of vertical road}) - (\text{Area of common portion}) \)
\( = 445.50 + 198 - 20.25 \)
\( = 643.50 - 20.25 \)
\( = 623.25 \text{ m}^2 \)
Area of each square stone block \( = \text{side} \times \text{side} \)
\( = 1.5 \text{ m} \times 1.5 \text{ m} = 2.25 \text{ m}^2 \)
Required number of stone blocks for the road \( = \frac{\text{Total area of the road}}{\text{Area of each stone block}} \)
\( = \frac{623.25}{2.25} \)
\( = 277 \)
Thus, 277 square blocks are required to cover the road.
In simple words: First, we find the breadth of the rectangular field. Then, we calculate the area of the two road sections (one along the length, one along the breadth). Since the central part where they cross is counted twice, we subtract its area once. Finally, we divide the total road area by the area of one square block to find out how many blocks are needed.

🎯 Exam Tip: For problems with intersecting paths or roads, remember the "area of cross roads" concept: Area = (Area of path 1) + (Area of path 2) - (Area of common part). Don't forget to subtract the common part once.

Question 22. A room is 8 m 50 cm long and 6 m 50 cm wide. What should be the length of the mat of width 25 cm to cover the whole floor? Find the total cost of the mat at the rate of Rs. 20 per m.
Answer:
Length of the room \( = 8 \text{ m } 50 \text{ cm} = 8.50 \text{ m} \)
Width of the room \( = 6 \text{ m } 50 \text{ cm} = 6.50 \text{ m} \)
Area of the room floor \( = \text{length} \times \text{width} \)
\( = 8.50 \text{ m} \times 6.50 \text{ m} \)
\( = 55.25 \text{ m}^2 \)
The mat needs to cover the whole floor, so the area of the mat must be equal to the area of the room.
Area of the mat \( = 55.25 \text{ m}^2 \)
Width of the mat \( = 25 \text{ cm} = 0.25 \text{ m} \)
Let the length of the mat be \( x \).
Area of the mat \( = \text{length} \times \text{width} \)
\( \implies 55.25 = x \times 0.25 \)
\( \implies x = \frac{55.25}{0.25} \)
\( \implies x = \frac{5525}{25} \)
\( \implies x = 221 \text{ m} \)
The length of the mat should be 221 m.
Cost of 1 m of mat \( = \text{Rs. } 20 \)
Total cost of the mat \( = \text{Length of mat} \times \text{Cost per meter} \)
\( = 221 \text{ m} \times \text{Rs. } 20 \)
\( = \text{Rs. } 4420 \)
In simple words: First, we convert all measurements to meters to keep units consistent. Then, we calculate the area of the room floor. Since the mat must cover the entire floor, the mat's area will be the same as the room's area. Using the mat's area and its given width, we find its required length. Finally, we multiply the mat's total length by the cost per meter to get the total cost.

🎯 Exam Tip: Always ensure all units are consistent (e.g., all in meters) before performing calculations. This is a common source of error. The area of the mat must be equal to the area of the floor it covers.