Samacheer Kalvi Class 9 Maths Solutions Chapter 7 Mensuration More Ques

I. Choose the Correct Answer

Question 1. If the sides of a triangles are 5 cm, 6 cm and 7 cm then the area is ........
(a) 18 \( \text{cm}^2 \)
(b) 6 \( \sqrt{2} \text{ cm}^2 \)
(c) 6 \( \sqrt{6} \text{ cm}^2 \)
(d) 0 \( \sqrt{3} \text{ cm} \)
Answer: (c) 6 \( \sqrt{6} \text{ cm}^2 \)
In simple words: We find the area of the triangle using Heron's formula, which involves calculating the semi-perimeter first. The calculation shows that the area is 6 multiplied by the square root of 6, in square centimeters.

🎯 Exam Tip: Remember Heron's formula for finding the area of a triangle when all three side lengths are known: \( A = \sqrt{s(s-a)(s-b)(s-c)} \), where s is the semi-perimeter.

Question 2. The perimeter of an equilateral triangle is 60 cm then the area is .........
(a) 60 \( \sqrt{3} \text{ cm}^2 \)
(b) 20 \( \sqrt{3} \text{ cm}^2 \)
(c) 50 \( \sqrt{3} \text{ cm}^2 \)
(d) 100 \( \sqrt{3} \text{ cm}^2 \)
Answer: (d) 100 \( \sqrt{3} \text{ cm}^2 \)
In simple words: An equilateral triangle has all sides equal. If its perimeter is 60 cm, each side is 20 cm. Using the formula for the area of an equilateral triangle, which is \( \frac{\sqrt{3}}{4} \times \text{side}^2 \), we get 100 root 3 square centimeters.

🎯 Exam Tip: Always recall the specific properties of equilateral triangles (all sides and angles equal) and their direct area formula to solve such problems quickly.

Question 3. The total surface area of the cuboid with dimension 20 cm x 30 cm x 15 cm is ........
(a) 2700 \( \text{cm}^2 \)
(b) 1500 \( \text{cm}^2 \)
(c) 2500 \( \text{cm}^2 \)
(d) 3000 \( \text{cm}^2 \)
Answer: (a) 2700 \( \text{cm}^2 \)
In simple words: To find the total surface area of a cuboid, you calculate the area of all six faces and add them up. For a cuboid with length 20 cm, breadth 30 cm, and height 15 cm, the total surface area comes out to be 2700 square centimeters.

🎯 Exam Tip: Remember the formula for the total surface area of a cuboid: \( \text{TSA} = 2(lb + bh + hl) \), where l, b, and h are length, breadth, and height respectively. Be careful with units.

Question 4. The number of bricks each measuring 70 cm x 80 cm x 40 cm that will be required to build a wall whose dimensions are 7 m x 8 m x 4 m is ........
(a) 4000
(b) 3000
(c) 2000
(d) 1000
Answer: (d) 1000
In simple words: To find out how many bricks are needed, first make sure all measurements are in the same unit. Convert the wall's dimensions from meters to centimeters. Then, divide the total volume of the wall by the volume of one brick. This gives us 1000 bricks.

🎯 Exam Tip: Always convert all units to be consistent (e.g., all to cm or all to m) before performing any calculations to avoid errors in volume and quantity problems.

Question 5. The volume of a cube is 4913 \( \text{m}^3 \) then the length of its side is ........
(a) 13 m
(b) 17 m
(c) 34 m
(d) 27 m
Answer: (b) 17 m
In simple words: The volume of a cube is found by multiplying its side length by itself three times (side\(^3\)). To find the side length when you know the volume, you need to calculate the cube root of the volume. The cube root of 4913 is 17.

🎯 Exam Tip: Familiarize yourself with common cubes and cube roots (up to at least 20) to quickly solve problems involving cube volumes and side lengths.

II. Answer the Following Questions

Question 6. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non parallel sides are 14 m and 13 m. Find the area of the field.
Answer: Let the parallel sides of the trapezium be \( a = 25 \text{ m} \) and \( b = 10 \text{ m} \). The non-parallel sides are \( c = 13 \text{ m} \) and \( d = 14 \text{ m} \). We can draw a line BE parallel to AD from B to DC to form a parallelogram ABED and a triangle BCE. This makes ABED a parallelogram, so \( AD = BE = 13 \text{ m} \) and \( AB = DE = 10 \text{ m} \). Then, \( EC = DC - DE = 25 \text{ m} - 10 \text{ m} = 15 \text{ m} \). Now we find the area of triangle BCE using Heron's formula. The sides of \( \triangle BCE \) are \( 13 \text{ m} \), \( 14 \text{ m} \), and \( 15 \text{ m} \). First, find the semi-perimeter \( s \): \( s = \frac{a+b+c}{2} = \frac{13+15+14}{2} = \frac{42}{2} = 21 \text{ m} \) Next, calculate \( s-a \), \( s-b \), \( s-c \): \( s-a = 21-13 = 8 \text{ m} \) \( s-b = 21-15 = 6 \text{ m} \) \( s-c = 21-14 = 7 \text{ m} \) Area of \( \triangle BCE = \sqrt{s(s-a)(s-b)(s-c)} \) \( = \sqrt{21 \times 8 \times 6 \times 7} \) \( = \sqrt{(3 \times 7) \times (2^3) \times (2 \times 3) \times 7} \) \( = \sqrt{2^4 \times 3^2 \times 7^2} \) \( = 2^2 \times 3 \times 7 \) \( = 4 \times 3 \times 7 = 84 \text{ m}^2 \) Now, we find the height \( h \) of \( \triangle BCE \) (which is also the height of the trapezium). Area of \( \triangle BCE = \frac{1}{2} \times \text{base} \times \text{height} \) \( 84 = \frac{1}{2} \times EC \times h \) \( 84 = \frac{1}{2} \times 15 \times h \) \( h = \frac{84 \times 2}{15} = \frac{168}{15} = \frac{56}{5} = 11.2 \text{ m} \) Area of parallelogram ABED = base \( \times \) height \( = AB \times h = 10 \text{ m} \times 11.2 \text{ m} = 112 \text{ m}^2 \) Total area of the field = Area of \( \triangle BCE \) + Area of parallelogram ABED \( = 84 \text{ m}^2 + 112 \text{ m}^2 = 196 \text{ m}^2 \) Alternatively, using the direct formula for the area of a trapezium: Area of the field = \( \frac{1}{2} \times h \times (a+b) \) \( = \frac{1}{2} \times 11.2 \times (25+10) \) \( = \frac{1}{2} \times 11.2 \times 35 \) \( = 5.6 \times 35 = 196 \text{ m}^2 \)

🎯 Exam Tip: When dealing with complex shapes like trapeziums, try to break them down into simpler shapes like triangles and rectangles/parallelograms. Ensure consistent units throughout your calculations.

Question 7. Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm and \( \angle B = 90^\circ \).
Answer: We need to find the area of quadrilateral ABCD. We can split it into two triangles by the diagonal AC. So, Area(ABCD) = Area( \( \triangle ABC \) ) + Area( \( \triangle ACD \) ). Given \( AB = 8 \text{ cm} \), \( BC = 6 \text{ cm} \), \( CD = 8 \text{ cm} \), \( DA = 10 \text{ cm} \), \( AC = 10 \text{ cm} \), and \( \angle B = 90^\circ \). **For \( \triangle ABC \):** Since \( \angle B = 90^\circ \), \( \triangle ABC \) is a right-angled triangle. Area of \( \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} \) \( = \frac{1}{2} \times AB \times BC \) \( = \frac{1}{2} \times 8 \text{ cm} \times 6 \text{ cm} \) \( = 4 \times 6 \text{ cm}^2 = 24 \text{ cm}^2 \) **For \( \triangle ACD \):** The sides are \( a = CD = 8 \text{ cm} \), \( b = DA = 10 \text{ cm} \), and \( c = AC = 10 \text{ cm} \). We use Heron's formula for this triangle. First, find the semi-perimeter \( s \): \( s = \frac{a+b+c}{2} = \frac{8+10+10}{2} = \frac{28}{2} = 14 \text{ cm} \) Next, calculate \( s-a \), \( s-b \), \( s-c \): \( s-a = 14-8 = 6 \text{ cm} \) \( s-b = 14-10 = 4 \text{ cm} \) \( s-c = 14-10 = 4 \text{ cm} \) Area of \( \triangle ACD = \sqrt{s(s-a)(s-b)(s-c)} \) \( = \sqrt{14 \times 6 \times 4 \times 4} \) \( = \sqrt{(2 \times 7) \times (2 \times 3) \times 4 \times 4} \) \( = \sqrt{2^2 \times 4^2 \times 21} \) \( = 2 \times 4 \times \sqrt{21} \text{ cm}^2 \) \( = 8 \sqrt{21} \text{ cm}^2 \) Using \( \sqrt{21} \approx 4.58 \): \( = 8 \times 4.58 \text{ cm}^2 = 36.64 \text{ cm}^2 \) **Total Area of Quadrilateral ABCD:** Area(ABCD) = Area( \( \triangle ABC \) ) + Area( \( \triangle ACD \) ) \( = 24 \text{ cm}^2 + 36.64 \text{ cm}^2 \) \( = 60.64 \text{ cm}^2 \)

🎯 Exam Tip: When finding the area of complex polygons, a common strategy is to break them into simpler, known shapes (like triangles or rectangles) whose areas can be calculated easily. Always check for right angles first!

Question 8. The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per \( \text{m}^2 \).
Answer: Given the dimensions of the room: Length (l) = 5 m Breadth (b) = 4 m Height (h) = 3 m Rate of whitewashing = Rs 7.50 per \( \text{m}^2 \) The area to be whitewashed includes the four walls and the ceiling. Area for whitewashing = Lateral Surface Area of four walls + Area of the ceiling Lateral Surface Area of four walls \( = 2(l+b)h \) Area of the ceiling \( = l \times b \) So, Total Area \( = 2(l+b)h + lb \) Substitute the given values: Total Area \( = 2(5+4) \times 3 + (5 \times 4) \) \( = 2(9) \times 3 + 20 \) \( = 18 \times 3 + 20 \) \( = 54 + 20 \) \( = 74 \text{ m}^2 \) Now, calculate the total cost: Cost of whitewashing \( = \text{Total Area} \times \text{Rate per } \text{m}^2 \) \( = 74 \times \text{Rs } 7.50 \) \( = \text{Rs } 555 \) Therefore, the cost of whitewashing the walls and the ceiling is Rs 555. The whitewashing typically covers the visible surfaces, and not the floor.
In simple words: First, calculate the area of the four walls and the ceiling that need whitewashing. The formula for this is \( 2 \times (\text{length} + \text{breadth}) \times \text{height} + (\text{length} \times \text{breadth}) \). Then, multiply this total area by the cost per square meter to find the total money needed.

🎯 Exam Tip: For problems involving painting or whitewashing, always identify which surfaces are to be covered. Usually, it's walls and the ceiling, but not the floor, unless specified. Be careful with unit conversions if present.

Question 9. How many hollow blocks of size 30 cm x 15 cm x 20 cm are needed to construct a wall 60 m in length 0.3 m in breadth and 2 m in height.
Answer: First, we need to ensure all measurements are in the same units. We will convert all dimensions to centimeters. Dimensions of the hollow block: Length \( = 30 \text{ cm} \) Breadth \( = 15 \text{ cm} \) Height \( = 20 \text{ cm} \) Volume of one hollow block \( = l \times b \times h = 30 \times 15 \times 20 = 9000 \text{ cm}^3 \) Dimensions of the wall: Length \( = 60 \text{ m} = 60 \times 100 \text{ cm} = 6000 \text{ cm} \) Breadth \( = 0.3 \text{ m} = 0.3 \times 100 \text{ cm} = 30 \text{ cm} \) Height \( = 2 \text{ m} = 2 \times 100 \text{ cm} = 200 \text{ cm} \) Volume of the wall \( = L \times B \times H = 6000 \times 30 \times 200 = 36,000,000 \text{ cm}^3 \) Number of hollow blocks required \( = \frac{\text{Volume of the wall}}{\text{Volume of one hollow block}} \) \( = \frac{36,000,000}{9000} \) \( = 4000 \) So, 4000 hollow blocks are needed to construct the wall. This calculation assumes the blocks are solid and fit perfectly without mortar space.
In simple words: To find the number of blocks, first change all measurements to the same unit, like centimeters. Then, calculate the total space (volume) of the wall and the space of one block. Divide the wall's volume by the block's volume to see how many blocks fit.

🎯 Exam Tip: Always pay close attention to units! Mixed units (meters and centimeters) are a common source of error in these problems. Convert all measurements to a single consistent unit before calculating volumes.

Question 10. Find the number of cubes of side 3 cm that can be cut from a cuboid of dimensions 10 cm x 9 cm x 6 cm.
Answer: We need to find how many small cubes can be made from a larger cuboid. Side of one cube \( = 3 \text{ cm} \) Volume of one cube \( = \text{side}^3 = 3 \times 3 \times 3 = 27 \text{ cm}^3 \) Dimensions of the cuboid: Length (l) \( = 10 \text{ cm} \) Breadth (b) \( = 9 \text{ cm} \) Height (h) \( = 6 \text{ cm} \) Volume of the cuboid \( = l \times b \times h = 10 \times 9 \times 6 = 540 \text{ cm}^3 \) Number of cubes \( = \frac{\text{Volume of the cuboid}}{\text{Volume of one cube}} \) \( = \frac{540}{27} \) \( = 20 \) Therefore, 20 cubes of side 3 cm can be cut from the given cuboid. This is a common method for understanding volume packing.
In simple words: To find how many small cubes fit inside a big cuboid, first calculate the volume of the small cube. Then, calculate the volume of the big cuboid. Finally, divide the cuboid's volume by the cube's volume to get the total count.

🎯 Exam Tip: This problem involves dividing volumes. Ensure you correctly calculate the volume of both the cuboid and the cube before performing the division. If dimensions aren't perfectly divisible, the number of cubes would be based on whole numbers only, but here they are exact.