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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 7 Mensuration Ex 7.2

Question 1.
Find the Total Surface Area and the Lateral Surface Area of a cuboid whose dimensions are length = 20 cm, breadth = 15 cm, height = 8 cm.
Solution:
Here l = 20 cm, b = 15 cm, h = 8 cm
L.S.A. of the cuboid = 2(1 + b)h sq.m
= 2(20 + 15) × 8
= 2 × 35 × 8
= 560 sq.m
Total surface area of the cuboid = 2(lb + bh + lh) sq.units
= 2(20 × 15 + 15 × 8 + 8 × 20) sq. cm
= 2(300 + 120 + 160) sq. cm
= 2 × 580 sq. cm
= 1160 sq. cm

Question 2.
The dimensions of a cuboidal box are 6 m x 400 cm x 1.5 m. Find the cost of painting its entire outer surface at the rate of Rs 22 per m².
Solution:
Length of the cuboid box (l) = 6 m
Breadth of the cuboid box (b) = 400 cm = 4m
Height of the cuboid box (h) = 1.5 m
T.S.A. of the cuboid = 2(lb + bh + lh) sq.units
= 2(6 × 4 + 4 × 1.5 + 1.5 × 6) sq.units
= 2(24 + 6 + 9)
= 2 × 39 sq.m
= 78 sq.m
Cost of painting for one sq.m = Rs 22
Total cost of painting = Rs 78 × 22
= Rs 1716

Question 3.
The dimensions of a hall is 10 m × 9 m × 8 m. Find the cost of white washing the walls and ceiling at the rate of Rs 8.50 per m².
Solution:
Length of the hall (l) = 10 m
Breath of the hall (b) = 9 m
Height of the hall (h) = 8 m
Area to be white wash = L.S.A. + Ceiling of the building
= 2(l + b)h + lb sq.units
= 2(10 + 9)8 + 10 × 9 sq.m
= 2 × 19 × 8 + 10 × 9 sq. m
= (304 + 90) sq.m
= 394 sq.m
Cost of white washing one sq.m = Rs 8.50
Cost of white washing for 394 sq.m = Rs 394 × 8.50
= Rs 3349
Total cost of white washing = Rs 3349

Question 4.
Find the TSA and LSA of the cube whose side is
(i) 8 m
(ii) 21 cm
(iii) 7.5 cm
Solution:
(i) 8m
Side of a cube (a) = 8m
T.S.A. of the cube = 6a² sq.units
= 6 × 8 × 8 sq. m
= 384 sq.m
L.S.A. of the cube = 4a² sq.units
= 4 × 8 × 8 sq.m
= 256 sq.m

(ii) 21 cm
Solution:
Side of a cube (a) = 21 cm
T.S.A. of the cube = 6a² sq. units
= 6 × 21 × 21 cm²
= 2646 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 21 × 21 sq.cm
= 4 × 441 cm²
= 1764 cm²

(iii) 7.5 cm
Solution:
Side of a cube (a) = 7.5 cm
T.S.A. of the cube = 6a² sq.units
= 6 × 7.5 × 7.5 cm²
= 337.5 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 7.5 × 7.5 sq.cm
= 225 cm²

Question 5.
If the total surface area of a cube is 2400 cm² then, find its lateral surface area.
Solution:
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = $$\frac{2400}{6}$$
= 400 cm²
L.S.A. of the cube = 4a² sq.units
= 4 × 400 cm²
= 1600 cm²
(OR)
T.S.A. of the cube = 2400 cm²
6a² = 2400
a² = $$\frac{2400}{6}$$
= 400
a = $$\sqrt{400}$$
= 20 cm
Side of a cube (a) = 20 cm
L.S.A. of the cube = 4a² sq.units
= 4 × 20 × 20 cm²
= 1600 cm²

Question 6.
A cubical container of side 6.5 m is to be painted on the entire outer surface. Find the area to be painted and the total cost of painting it at the rate of Rs 24 per m².
Solution:
Side of a cube (a) = 6.5 m
Total surface area of the cube = 6a² sq.units
= 6 × 6.5 × 6.5 sq.m
= 253.50 sq.m
Cost of painting for 1 sq.m = Rs 24
Cost of painting for 253.5 sq.m = 253.5 × 24
= Rs 6084
∴ Area to be painted = 253.50 m²
Total cost of painting = Rs 6084

Question 7.
Three identical cubes of side 4 cm are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid.
Solution:
Joint the three identical cubes we get a new cuboid
Length of the cuboid (l) = (4 + 4 + 4) cm
l = 12 cm
Breadth of the cuboid (b) = 4 cm
Height of the cuboid (h) = 4 cm
Total surface area of the new cuboid = 2(lb + bh + lh) sq.units
= 2(12 × 4 + 4 × 4 + 4 × 12)
= 2(48 + 16 + 48) cm
= 2(112) cm²
= 224 cm²
Lateral surface area of the new cuboid = 2(l + b)h sq.units
= 2(12 + 4)4 cm²
= 2 × 16 × 4 cm²
= 128 cm²
∴ T.S.A of the new cuboid = 224 cm²
L.S.A of the new cuboid = 128 cm²