Samacheer Kalvi Class 9 Maths Solutions Chapter 7 Mensuration Exercise 7.4

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.4

Question 1. The semi-perimeter of a triangle having sides 15 cm, 20 cm and 25 cm is ........
(a) 60 cm
(b) 45 cm
(c) 30 cm
(d) 15 cm
Answer: (c) 30 cm
The semi-perimeter is found by adding all three sides of the triangle and then dividing the sum by two. It is half of the total boundary length of the triangle.
Hint:
Sides of the triangle are \( a = 15 \) cm, \( b = 20 \) cm, \( c = 25 \) cm.
Semi-perimeter \( s = \frac{a+b+c}{2} \)
\( s = \frac{15+20+25}{2} \)
\( s = \frac{60}{2} \)
\( s = 30 \) cm
In simple words: To find the semi-perimeter, add all three side lengths of the triangle and then divide that total by two.

🎯 Exam Tip: Remember that semi-perimeter is half the perimeter. This concept is crucial for applying Heron's formula to find the area of a triangle when only its sides are known.

Question 2. If the sides of a triangle are 3 cm, 4 cm and 5 cm, then the area is .........
(a) 3 cm²
(b) 6 cm²
(c) 9 cm²
(d) 12 cm²
Answer: (b) 6 cm²
To find the area of this triangle, we use Heron's formula after calculating the semi-perimeter. This triangle is also a right-angled triangle, which is a common trick in such questions.
Hint:
Sides of the triangle are \( a = 3 \) cm, \( b = 4 \) cm, \( c = 5 \) cm.
First, find the semi-perimeter \( s \):
\( s = \frac{a+b+c}{2} \)
\( s = \frac{3+4+5}{2} \)
\( s = \frac{12}{2} \)
\( s = 6 \) cm
Now, use Heron's formula for the area of the triangle:
Area \( = \sqrt{s(s-a)(s-b)(s-c)} \)
Area \( = \sqrt{6(6-3)(6-4)(6-5)} \)
Area \( = \sqrt{6 \times 3 \times 2 \times 1} \)
Area \( = \sqrt{36} \)
Area \( = 6 \) cm²
In simple words: First, find half of the total length of all sides added together. Then, use that number in Heron's formula to calculate the triangle's area.

🎯 Exam Tip: Always check if the triangle is a right-angled triangle (e.g., using the Pythagorean theorem \( a^2 + b^2 = c^2 \)) as its area can be calculated more simply using \( \frac{1}{2} \times \text{base} \times \text{height} \).

Question 3. The perimeter of an equilateral triangle is 30 cm. The area is ........
(a) 10 √3 cm²
(b) 12 √3 cm²
(c) 15 √3 cm²
(d) 25 √3 cm²
Answer: (d) 25 √3 cm²
Since all sides of an equilateral triangle are equal, we can find the length of one side from its perimeter. Then, we use the specific formula for the area of an equilateral triangle.
Hint:
Perimeter of an equilateral triangle \( = 30 \) cm
For an equilateral triangle, all three sides are equal. Let the side length be \( a \).
So, \( 3a = 30 \) cm
\( a = \frac{30}{3} \)
\( a = 10 \) cm
The formula for the area of an equilateral triangle is:
Area \( = \frac{\sqrt{3}}{4} a^2 \) sq.units
Area \( = \frac{\sqrt{3}}{4} \times 10 \times 10 \)
Area \( = \frac{\sqrt{3}}{4} \times 100 \)
Area \( = 25 \sqrt{3} \) cm²
In simple words: First, find the length of one side of the triangle by dividing the perimeter by three. Then, use the special formula for equilateral triangles to find its area.

🎯 Exam Tip: Memorize the formula for the area of an equilateral triangle \( (\frac{\sqrt{3}}{4} a^2) \) as it saves time compared to using Heron's formula.

Question 4. The lateral surface area of a cube of side 12 cm is ........
(a) 144 cm²
(b) 196 cm²
(c) 576 cm²
(d) 664 cm²
Answer: (c) 576 cm²
The lateral surface area of a cube includes the area of its four side faces, excluding the top and bottom faces. Since all faces are squares, it's simply four times the area of one face.
Hint:
Side of a cube \( (a) = 12 \) cm
Lateral Surface Area (L.S.A.) of a cube \( = 4a^2 \) sq.units
L.S.A. \( = 4 \times 12 \times 12 \) cm²
L.S.A. \( = 4 \times 144 \) cm²
L.S.A. \( = 576 \) cm²
In simple words: The lateral surface area of a cube is the area of its four side walls. You find it by multiplying 4 by the side length squared.

🎯 Exam Tip: Distinguish between lateral surface area (four sides) and total surface area (all six sides) of a cube to avoid common errors.

Question 5. If the lateral surface area of a cube is 600 cm², then the total surface area is .........
(a) 150 cm²
(b) 400 cm²
(c) 900 cm²
(d) 1350 cm²
Answer: (c) 900 cm²
Given the lateral surface area, we can first find the area of one face, which helps us calculate the total surface area easily. This is because a cube has 6 equal faces, and lateral area is 4 of these.
Hint:
Lateral Surface Area (L.S.A.) of a cube \( = 600 \) cm²
We know that L.S.A. \( = 4a^2 \), where \( a \) is the side length.
So, \( 4a^2 = 600 \)
\( a^2 = \frac{600}{4} \)
\( a^2 = 150 \)
Now, the Total Surface Area (T.S.A.) of a cube \( = 6a^2 \) sq.units
T.S.A. \( = 6 \times 150 \) cm²
T.S.A. \( = 900 \) cm²
In simple words: If you know the area of the four side walls of a cube, you can find the area of one wall. Then, multiply that by six to get the total area of all its surfaces.

🎯 Exam Tip: Remember that \( a^2 \) represents the area of one face of the cube. Lateral surface area is \( 4a^2 \) and total surface area is \( 6a^2 \).

Question 6. The total surface area of a cuboid with dimension 10 cm × 6 cm × 5 cm is .........
(a) 280 cm²
(b) 300 cm²
(c) 360 cm²
(d) 600 cm²
Answer: (a) 280 cm²
A cuboid has six rectangular faces. The total surface area is the sum of the areas of all these faces. We use the formula that accounts for the pairs of identical faces.
Hint:
Dimensions of the cuboid: length \( (l) = 10 \) cm, breadth \( (b) = 6 \) cm, height \( (h) = 5 \) cm.
Total Surface Area (T.S.A.) of a cuboid \( = 2(lb + bh + lh) \) sq.units
T.S.A. \( = 2( (10 \times 6) + (6 \times 5) + (10 \times 5) ) \) cm²
T.S.A. \( = 2(60 + 30 + 50) \) cm²
T.S.A. \( = 2(140) \) cm²
T.S.A. \( = 280 \) cm²
In simple words: To find the total surface area of a box-shape (cuboid), you add up the areas of its top, bottom, and four side faces.

🎯 Exam Tip: Ensure you correctly substitute the length, breadth, and height into the formula and perform the multiplication and addition steps carefully.

Question 7. If the ratio of the sides of two cubes are 2 : 3, then ratio of their surface areas will be .........
(a) 4:6
(b) 4:9
(c) 6:9
(d) 16:36
Answer: (b) 4:9
The surface area of a cube is proportional to the square of its side length. Therefore, if the sides are in a certain ratio, their surface areas will be in the square of that ratio. This is a general rule for similar 3D shapes.
Hint:
Let the side lengths of the two cubes be \( a_1 \) and \( a_2 \).
Given, \( a_1 : a_2 = 2 : 3 \)
The surface area of a cube \( = 6a^2 \).
Ratio of their surface areas \( = 6a_1^2 : 6a_2^2 \)
This simplifies to \( a_1^2 : a_2^2 \)
Substitute the ratio of the sides:
Ratio \( = 2^2 : 3^2 \)
Ratio \( = 4 : 9 \)
In simple words: If you have two cubes and you know how their side lengths compare, then their surface areas will compare by squaring those numbers.

🎯 Exam Tip: For any two similar 3D shapes, the ratio of their surface areas is the square of the ratio of their corresponding linear dimensions (like sides, radii, or heights).

Question 8. The volume of a cuboid is 660 cm³ and the area of the base is 33 cm². Its height is .........
(a) 10 cm
(b) 12 cm
(c) 20 cm
(d) 22 cm
Answer: (c) 20 cm
The volume of a cuboid can be found by multiplying the area of its base by its height. Knowing any two of these values allows us to calculate the third.
Hint:
Volume of a cuboid \( = 660 \) cm³
Area of the base \( = 33 \) cm²
We know that Volume \( = \text{Area of base} \times \text{height} \)
\( l \times b \times h = 660 \)
Since \( \text{Area of base} = l \times b \), we have:
\( 33 \times h = 660 \)
To find the height \( h \), divide the volume by the base area:
\( h = \frac{660}{33} \)
\( h = 20 \) cm
In simple words: Imagine a box. If you know how much space it takes up (volume) and the size of its bottom (base area), you can find its height by dividing the volume by the base area.

🎯 Exam Tip: Remember the fundamental relationship: Volume = Base Area × Height. This formula is applicable to many prisms and cylinders.

Question 9. The capacity of a water tank of dimensions 10 m × 5 m × 1.5 m is .........
(a) 75 litres
(b) 750 litres
(c) 7500 litres
(d) 75000 litres
Answer: (d) 75000 litres
First, we calculate the volume of the tank in cubic meters using its dimensions. Then, we convert this volume from cubic meters to litres, knowing that one cubic meter holds 1000 litres.
Hint:
Dimensions of the water tank: length \( (l) = 10 \) m, breadth \( (b) = 5 \) m, height \( (h) = 1.5 \) m.
Capacity of a tank (Volume) \( = l \times b \times h \)
Volume \( = (10 \times 5 \times 1.5) \) m³
Volume \( = 50 \times 1.5 \) m³
Volume \( = 75 \) m³
To convert cubic meters to litres, use the conversion factor: \( 1 \text{ m}^3 = 1000 \text{ litres} \).
Capacity in litres \( = 75 \times 1000 \text{ litres} \)
Capacity \( = 75000 \) litres
In simple words: First, find out how much space the tank takes up in cubic meters. Then, multiply that number by 1000 to change it into litres.

🎯 Exam Tip: Always remember the conversion factor \( 1 \text{ m}^3 = 1000 \text{ litres} \) when dealing with capacities of large containers.

Question 10. The number of bricks each measuring 50 cm x 30 cm × 20 cm that will be required to build a wall whose dimensions are 5 m x 3 m x 2 m is .........
(a) 1000
(b) 2000
(d) 5000
Answer: (a) 1000
To find the number of bricks, we need to calculate the volume of the wall and the volume of a single brick. It's important that both volumes are calculated using the same units before division. Then, divide the total volume of the wall by the volume of one brick.
Hint:
Dimensions of one brick: \( 50 \) cm \( \times 30 \) cm \( \times 20 \) cm.
Volume of one brick \( = 50 \times 30 \times 20 = 30000 \) cm³

Dimensions of the wall: length \( = 5 \) m, breadth \( = 3 \) m, height \( = 2 \) m.
Convert wall dimensions to centimeters (since brick dimensions are in cm):
Length \( (l) = 5 \) m \( = 5 \times 100 = 500 \) cm
Breadth \( (b) = 3 \) m \( = 3 \times 100 = 300 \) cm
Height \( (h) = 2 \) m \( = 2 \times 100 = 200 \) cm
Volume of the wall \( = l \times b \times h \)
Volume of the wall \( = 500 \times 300 \times 200 = 30,000,000 \) cm³

Number of bricks \( = \frac{\text{Volume of the wall}}{\text{Volume of one brick}} \)
Number of bricks \( = \frac{30,000,000}{30,000} \)
Number of bricks \( = 1000 \)
In simple words: Find how much space the whole wall takes up, and how much space one brick takes up. Make sure both are measured in the same units (like all in centimeters). Then, divide the wall's space by the brick's space to find how many bricks are needed.

🎯 Exam Tip: Always ensure all dimensions are in the same units (e.g., all in cm or all in m) before calculating volumes, otherwise your answer will be incorrect.