Exercise 7.2 - Chapter 7 Mensuration 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.2$
Question $1 .$
Find the total surface area and the lateral surface area of a cuboid whose dimensions are
(i) length $=20 \mathrm{~cm}$,
breadth $=15 \mathrm{~cm}$,
height $=8 \mathrm{~cm}$
Solution:

(i) (a) Total surface Area of cuboid $=2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})$ sq. units
$=2(20 \times 15+15 \times 8+8 \times 20)$
$=2(300+120+160)=2 \times 580$
(b) Lateral surface area of a cuboid $=2 \mathrm{~h}(1+\mathrm{b})$ sq. units
$=2 \times 8(20+15)=16 \times 35=560 \mathrm{~cm}^{2}$

Question 2.
The dimensions of a cuboidal box are $6 \mathrm{~m} \times 400 \mathrm{~cm} \times 1.5 \mathrm{~m}$. Find the cost of painting its entire outer surface at the rate of $₹ 22$ per $\mathrm{m}^{2}$
Solution:
$1 \times b \times h=6 \mathrm{~m} \times 400 \mathrm{~cm} \times 1.5 \mathrm{~m}$
$\mathrm{l}=6 \mathrm{~m}$,
$\mathrm{b}=4 \mathrm{~m}$,
$\mathrm{h}=1.5 \mathrm{~m}$

$\therefore$ Total surface area of the cuboid $=$ Outer surface area
$=2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})=2((6 \times 4)+(4 \times 1.5)+(1.5 \times 6))$
$=2(24+6+9)=2(39) \mathrm{m}^{2}$
Cost of painting $1 \mathrm{~m}^{2}=₹ 22$
Cost of painting $78 \mathrm{~m}^{2}=78 \times 22=₹ 1716$

Question $3 .$
The dimensions of a hall is $10 \mathrm{~m} \times 9 \mathrm{~m} \times 8 \mathrm{~m}$. Find the cost of white washing the walls and ceiling at the rate of $₹ 8.50$ per $\mathrm{m}^{2}$.
Solution:

Dimensions of a hall $10 \mathrm{~m} \times 9 \mathrm{~m} \times 8 \mathrm{~m}$
$\begin{aligned}
&\mathrm{l}=10 \mathrm{~m} \\
&\mathrm{~b}=9 \mathrm{~m} \\
&\mathrm{~h}=8 \mathrm{~m}
\end{aligned}$
White washing to be done for the area of the surface
$\begin{aligned}
&=2(\mathrm{lh}+\mathrm{bh})+\mathrm{lb} \\
&=2(10 \times 8+9 \times 8)+10 \times 9 \\
&=2(80+72)+90=2 \times 152+90 \\
&=304+90=394 \mathrm{~m}^{2}
\end{aligned}$
Cost of white washing per $\mathrm{m}^{2}=₹ 8.50$
Cost of white washing $394 \mathrm{~m}^{2}=394 \times 8.50$
Total cost = ₹ 3349

Question $4 .$
Find the TSA and LSA of the cube whose side is
(i) $8 \mathrm{~m}$
(ii) $21 \mathrm{~cm}$
(iii) $7.5 \mathrm{~cm}$
Solution:

(i) side of a cube $=8 \mathrm{~m}$
TSA of the cube $=6 a^{2}=6 \times 64=384 \mathrm{~m}^{2}$
$\mathrm{LSA}$ of the cube $=4 \mathrm{a}^{2}=4 \times 64=256 \mathrm{~m}^{2}$
(ii) side a $21 \mathrm{~cm}$
$\mathrm{TSA}=6 \mathrm{a}^{2}=6 \times 21 \times 21=2646 \mathrm{~cm}^{2}$.
$\mathrm{LSA}=4 \mathrm{a}^{2}=4 \times 21 \times 21=1764 \mathrm{~cm}^{2}$.
(iii) side $\mathrm{a}=7.5 \mathrm{~cm}$
TSA $=6 \mathrm{a}^{2}=6 \times 7.5 \times 7.5 \mathrm{~cm}^{2}=337.5 \mathrm{~cm}^{2}$
LSA $=4 \mathrm{a}^{2}=4 \times 7.5 \times 7.5 \mathrm{~cm}^{2}=225 \mathrm{~cm}^{2}$

Question $5 .$
If the total surface area of a cube is $2400 \mathrm{~cm}^{2}$ then, find its lateral surface area.
Solution:
$\begin{aligned}
6 a^{2} &=2400 \mathrm{~cm}^{2} \\
\Rightarrow a^{2} &=\frac{2400}{6}
\end{aligned}$
$\therefore 4 a^{2}=\frac{4 \times 2400}{6}=1600 \mathrm{~cm}^{2} .$
 

Question 6.
A cubical container of side $6.5 \mathrm{~m}$ 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 $₹ 24$ per $\mathrm{m}^{2}$.
Solution:
$a=6.5 \mathrm{~m}$
$6 \mathrm{a}^{2}=6 \times 6.5 \times 6.5=253.5 \mathrm{~m}^{2}$
Area to be painted $=253.5 \mathrm{~m}^{2}$
Cost of painting $1 \mathrm{~m}^{2}=₹ 24$
$\therefore$ Cost of painting $253.5 \mathrm{~m}^{2}=253.5 \times 24=₹ 6084$
 

Question 7.
Three identical cubes of side $4 \mathrm{~cm}$ are joined end to end. Find the total surface area and lateral surface area of the new resulting cuboid.
Solution:
$\mathrm{a}=4 \mathrm{~cm}$
TSA of the cuboid $=2(\mathrm{lb}+\mathrm{bh}+\mathrm{hl})$

$\mathrm{l}=12 \mathrm{~cm}$
$\mathrm{~b}=4 \mathrm{~cm}$
$\mathrm{~h}=4 \mathrm{~cm}$
$\therefore \mathrm{TSA}=2(12 \times 4+4 \times 4+4 \times 12)$
$=2(48+16+48)=2 \times 112=224 \mathrm{~cm}^{2}$
$\mathrm{LSA}=2 \mathrm{~h}(1+\mathrm{b})=2 \times 4(12+4)=8 \times 16=128 \mathrm{~cm}^{2}$