Exercise 9.2 (Revised) - Chapter 11 - Mensuration - Ncert Solutions class 8 - Maths

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Chapter 9 Mensuration - NCERT Solutions Class 8 Maths | Free PDF Download

Ex 9.2 Question 1.

There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Solution.

(a) Length of cuboidal box (l) $=60 \mathrm{~cm}$
Breadth of cuboidal box $(b)=40 \mathrm{~cm}$
Height of cuboidal box $(h)=50 \mathrm{~cm}$
$\therefore$ Total surface area of cuboidal box $=2(l b+b h+h l)$
$=2(60 \times 40+40 \times 50+50 \times 60) \mathrm{cm}^2$
$=2(2400+2000+3000) \mathrm{cm}^2$
$=2 \times 7400 \mathrm{~cm}^2$
$
=14800 \mathrm{~cm}^2
$
(b) Length of the cube is $50 \mathrm{~cm}$
$\therefore$ Total surface area of cuboidal box $=6(\text { side })^2$
$=6(50)^2 \mathrm{~cm}^2$
$=6(2500) \mathrm{cm}^2$
$=15000 \mathrm{~cm}^2$

Thus, the cuboidal box (a) requires the lesser amount of materal.
Ex 9.2 Question 2.

A suitcase with measures $80 \mathrm{~cm} \times 48 \mathrm{~cm} \times 24 \mathrm{~cm}$ is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width $96 \mathrm{~cm}$ is required to cover 100 such suitcases?

Solution.

Given: Length of suitcase box $(l)=80 \mathrm{~cm}$, Breadth of suitcase box $(b)=48 \mathrm{~cm}$

And Height of cuboidal box $(h)=24 \mathrm{~cm}$
$
\begin{aligned}
& \therefore \text { Total surface area of suitcase box }=2(l b+b h+h l) \\
& =2(80 \times 48+48 \times 24+24 \times 80) \mathrm{cm}^2 \\
& =2(3840+1152+1920) \\
& =2 \times 6912=13824 \mathrm{~cm}^2
\end{aligned}
$

Area of Tarpaulin cloth $=$ Surface area of suitcase
$
\begin{aligned}
& \Rightarrow I \times b=13824 \\
& \Rightarrow I \times 96=13824 \\
& \Rightarrow I=\frac{13824}{96} \\
& =144 \mathrm{~cm}
\end{aligned}
$

$
\begin{aligned}
& \text { Required tarpaulin for } 100 \text { suitcases }=(144 \times 100) \mathrm{cm} \\
& =14400 \mathrm{~cm} \\
& =144 \mathrm{~m}\left[1 \mathrm{~cm}=\frac{1}{100} \mathrm{~m}\right]
\end{aligned}
$

Thus, $144 \mathrm{~m}$ tarpaulin cloth required to cover 100 suitcases.
Ex 9.2 Question 3.

Find the side of a cube whose surface area id $600 \mathrm{~cm}^2$.

Solution.

Here Surface area of cube $=600 \mathrm{~cm}^2$
$
\begin{aligned}
& \Rightarrow 6 l^2=600 \mathrm{~cm}^2 \\
& \Rightarrow l^2=100 \mathrm{~cm}^2 \\
& \Rightarrow l=\sqrt{1} 00 \mathrm{~cm} \\
& \Rightarrow l=10 \mathrm{~cm}
\end{aligned}
$

Hence the side of cube is $10 \mathrm{~cm}$
Ex 9.2 Question 4.

Rukshar painted the outside of the cabinet of measure $1 \mathrm{~m} \times 2 \mathrm{~m} \times 1.5 \mathrm{~m}$. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution.

Length of cabinet $(l)=2 \mathrm{~m}$
Breadth of cabinet $(b)=1 \mathrm{~m}$
Height of cabinet $(h)=1.5 \mathrm{~m}$
$
\begin{aligned}
& \therefore \text { Surface area of cabinet = (Area of Base of cabinet (Cuboid) + Area of four walls) } \\
& =l b+2(l+b) h \\
& =\{2 \times 1+2(1+2) 1.5\} \mathrm{m}^2 \\
& =2+2(3) 1.5 \mathrm{~m}^2 \\
& =2+6(1.5) \mathrm{m}^2 \\
& =(2+9.0) \mathrm{m}^2 \\
& =11 \mathrm{~m}^2
\end{aligned}
$

Hence required surface area of cabinet is $11 \mathrm{~m}^2$.
Ex 9.2 Question 5.

Daniel is paining the walls and ceiling of a cuboidal hall with length, breadth and height of $15 \mathrm{~m}, 10 \mathrm{~m}$ and $7 \mathrm{~m}$ respectively. From each can of paint $100 \mathrm{~m}^2$ of area is

painted. How many cans of paint will she need to paint the room?
Sol. Length of wall $(l)=15 \mathrm{~m}$
Breadth of wall $(b)=10 \mathrm{~m}$
Height of wall $(h)=7 \mathrm{~m}$
$\therefore$ Total Surface area of classroom=(Area of Base of ceiling (Cuboid) + Area of four walls)
$=l b+2(l+b) h$
$=(15 \times 10+2(10+15)(7)) \mathrm{m}^2$
$=(150+2(25)(7)) \mathrm{m}^2$
$=(150+350) \mathrm{m}^2$
$
=500 \mathrm{~m}^2
$

Area of one can is $100 \mathrm{~m}^2$
Now Required number of cans $=\frac{\text { Area of hall }}{\text { Area of one can }}=\frac{500}{100}=5$ cans
Hence 5 cans are required to paint the room.
Ex 9.2 Question 6.

Describe how the two figures below are alike and how they are different. Which box has larger lateral surface area?

Solution.

Diameter of cylinder $=7 \mathrm{~cm}$
$\therefore$ Radius of cylinder $(r)=\frac{7}{2} \mathrm{~cm}$
Height of cylinder $(h)=7 \mathrm{~cm}$
Lateral surface area of cylinder $=2 \pi r h$

$
\begin{aligned}
& =2 \times \frac{22}{7} \times \frac{7}{2} \times 7 \\
& =154 \mathrm{~cm}^2
\end{aligned}
$

Now lateral surface area of cube $=4(\text { Side })^2=4(7)^2 \mathrm{~cm}^2$
$
\begin{aligned}
& =(4 \times 49) \mathrm{cm}^2 \\
& =196 \mathrm{~cm}^2
\end{aligned}
$

Hence the cube has larger lateral surface area.
Ex 9.2 Question 7.

A closed cylindrical tank of radius $7 \mathrm{~m}$ and height $3 \mathrm{~m}$ is made from a sheet of metal. How much sheet of metal is required?

Solution

$
\begin{aligned}
& \text {Radius of cylindrical tank }(r)=7 \mathrm{~m} \\
& \text { Height of cylindrical tank }(h)=3 \mathrm{~m} \\
& \text { Total surface area of cylindrical tank }=\text { (Curved surface area }+ \text { Area of upper end } \\
& \text { (circle) }+ \text { Area of Lower (circle) end) } \\
& =\left(2 \pi r h+\pi r^2+\pi r^2\right) \\
& =\left(2 \pi r h+2 \pi r^2\right) \\
& =2 \pi r(h+r) \\
& =2 \times \frac{22}{7} \times 7(3+7) \mathrm{m}^2 \\
& =44 \times 10 \mathrm{~m}^2 \\
& =440 \mathrm{~m}^2
\end{aligned}
$

Hence $440 \mathrm{~m}^2$ metal sheet is required.

Ex 9.2 Question 8.

The lateral surface area of a hollow cylinder is $4224 \mathrm{~cm}^2$. It is cut along its height and formed a rectangular sheet of width $33 \mathrm{~cm}$. Find the perimeter of rectangular sheet?

Solution.

Lateral surface area of hollow cylinder $=4224 \mathrm{~cm}^2$
Height of hollow cylinder $=33 \mathrm{~cm}$
Curved surface area of hollow cylinder $=2 \pi h$
$
\begin{aligned}
& \Rightarrow 4224=2 \times \frac{22}{7} \times r \times 33 \\
& \Rightarrow r=\frac{4224 \times 7}{2 \times 22 \times 33} \\
& =\frac{64 \times 7}{22} \mathrm{~cm}
\end{aligned}
$

Now Length of rectangular sheet $=2 \pi \gamma$
$
\begin{aligned}
& \Rightarrow l=2 \times \frac{22}{7} \times \frac{64 \times 7}{22} \\
& =128 \mathrm{~cm}
\end{aligned}
$

Perimeter of rectangular sheet $=2(l+b)$
$
\begin{aligned}
& =2(128+33) \\
& =2 \times 161 \\
& =322 \mathrm{~cm}
\end{aligned}
$

Hence perimeter of rectangular sheet is $322 \mathrm{~cm}$.
Ex 9.2 Question 9.

A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is $84 \mathrm{~cm}$ and length $1 \mathrm{~m}$.

Solution.

Diameter of road roller $=84 \mathrm{~cm}$
$\therefore$ Radius of road roller $(r)=\frac{d}{2}=\frac{84}{2}$
$
=42 \mathrm{~cm}
$

Length of road roller $(h)=1 \mathrm{~m}=100 \mathrm{~cm}$

Curved surface area of road roller $=2 \pi r h$
$
\begin{aligned}
& =2 \frac{22}{7} \times 42 \times 100 \\
& =26400 \mathrm{~cm}^2
\end{aligned}
$
$\therefore$ Area covered by road roller in 750 revolutions $=26400 \times 750 \mathrm{~cm}^2$
$
\begin{aligned}
& =1,98,00,000 \mathrm{~cm}^2 \\
& =1980 \mathrm{~m}^2\left[\because 1 \mathrm{~m}^2=10,000 \mathrm{~cm}^2\right]
\end{aligned}
$

Thus, the area of the road is $1980 \mathrm{~m}^2$.
Ex 9.2 Question 10.

A company packages its milk powder in cylindrical container whose base has a diameter of $14 \mathrm{~cm}$ and height $20 \mathrm{~cm}$. Company places a label around the surface of the container (as shown in figure). If the label is placed $2 \mathrm{~cm}$ from top and bottom, what is the area of the label?

Solution .

Diameter of cylindrical container $=14 \mathrm{~cm}$

$\therefore$ Radius of cylindrical container $(r)=\frac{d}{2}=\frac{14}{2}=7 \mathrm{~cm}$
Height of cylindrical container $=20 \mathrm{~cm}$
Height of the label $(h)=(20-2-2)$
$
=16 \mathrm{~cm}
$

Curved surface area of label $=2 \pi \mu h$
$
\begin{aligned}
& =2 \times \frac{22}{7} \times 7 \times 16 \\
& =704 \mathrm{~cm}^2
\end{aligned}
$

Hence the area of the label of $704 \mathrm{~cm}^2$.