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

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Example 1:

The area of a trapezium shaped field is $480 \mathrm{~m}^2$, the distance between two parallel sides is $15 \mathrm{~m}$ and one of the parallel side is $20 \mathrm{~m}$. Find the other parallel side.
Solution:

One of the parallel sides of the trapezium is $a=20 \mathrm{~m}$, let another parallel side be $b$, height $h=15 \mathrm{~m}$.
The given area of trapezium $=480 \mathrm{~m}^2$.
Area of a trapezium $=\frac{1}{2} h(a+b)$
So $\quad 480=\frac{1}{2} \times 15 \times(20+b) \quad$ or $\quad \frac{480 \times 2}{15}=20+b$
or $64=20+b$ or $b=44 \mathrm{~m}$

Hence the other parallel side of the trapezium is $44 \mathrm{~m}$.

Example 2:

The area of a rhombus is $240 \mathrm{~cm}^2$ and one of the diagonals is $16 \mathrm{~cm}$.
Find the other diagonal.
Solution:

Let length of one diagonal $d_1=16 \mathrm{~cm}$
and $\quad$ length of the other diagonal $=d_2$
Area of the rhombus $=\frac{1}{2} d_1 \cdot d_2=240$

So,
$
\frac{1}{2} 16 \cdot d_2=240
$

Therefore,
$
d_2=30 \mathrm{~cm}
$

Hence the length of the second diagonal is $30 \mathrm{~cm}$.
Example 3:

There is a hexagon MNOPQR of side $5 \mathrm{~cm}$ (Fig 9.6). Aman and Ridhima divided it in two different ways (Fig 9.7).
Find the area of this hexagon using both ways.

Solution:

Aman's method:
Since it is a hexagon so NQ divides the hexagon into two congruent trapeziums. You can verify it by paper folding (Fig 9.8 ).
Now area of trapezium MNQR $=4 \times \frac{(11+5)}{2}=2 \times 16=32 \mathrm{~cm}^2$.

Now area of trapezium MNQR $=4 \times \frac{(11+5)}{2}=2 \times 16=32 \mathrm{~cm}^2$.
So the area of hexagon $\mathrm{MNOPQR}=2 \times 32=64 \mathrm{~cm}^2$.
Ridhima's method:
$\Delta \mathrm{MNO}$ and $\triangle \mathrm{RPQ}$ are congruent triangles with altitude $3 \mathrm{~cm}$ (Fig 9.9).

You can verify this by cutting off these two triangles and placing them on one another.
$
\text { Area of } \triangle \mathrm{MNO}=\frac{1}{2} \times 8 \times 3=12 \mathrm{~cm}^2=\text { Area of } \triangle \mathrm{RPQ}
$

Area of rectangle MOPR $=8 \times 5=40 \mathrm{~cm}^2$.
Now, area of hexagon $\mathrm{MNOPQR}=40+12+12=64 \mathrm{~cm}^2$.

Example 4:

An aquarium is in the form of a cuboid whose external measures are $80 \mathrm{~cm} \times 30 \mathrm{~cm} \times 40 \mathrm{~cm}$. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed?
Solution:

The length of the aquarium $=l=80 \mathrm{~cm}$
Width of the aquarium $=b=30 \mathrm{~cm}$

$
\begin{aligned}
\text { Height of the aquarium }= & h=40 \mathrm{~cm} \\
\text { Area of the base } & =l \times b=80 \times 30=2400 \mathrm{~cm}^2 \\
\text { Area of the side face } & =b \times h=30 \times 40=1200 \mathrm{~cm}^2 \\
\text { Area of the back face }= & l \times h=80 \times 40=3200 \mathrm{~cm}^2 \\
\text { Required area }= & \text { Area of the base }+ \text { area of the back face } \\
& +(2 \times \text { area of a side face }) \\
= & 2400+3200+(2 \times 1200)=8000 \mathrm{~cm}^2
\end{aligned}
$

Hence the area of the coloured paper required is $8000 \mathrm{~cm}^2$.

Example 5:

The internal measures of a cuboidal room are $12 \mathrm{~m} \times 8 \mathrm{~m} \times 4 \mathrm{~m}$. Find the total cost of whitewashing all four walls of a room, if the cost of white washing is ₹ 5 per $\mathrm{m}^2$. What will be the cost of white washing if the ceiling of the room is also whitewashed.
Solution:

Let the length of the room $=l=12 \mathrm{~m}$
Width of the room $=b=8 \mathrm{~m}$
Height of the room $=h=4 \mathrm{~m}$
Area of the four walls of the room $=$ Perimeter of the base $\times$ Height of the room
$
\begin{aligned}
& =2(l+b) \times h=2(12+8) \times 4 \\
& =2 \times 20 \times 4=160 \mathrm{~m}^2 .
\end{aligned}
$

Cost of white washing per $\mathrm{m}^2=₹ 5$
Hence the total cost of white washing four walls of the room $=₹(160 \times 5)=₹ 800$
Area of ceiling is $12 \times 8=96 \mathrm{~m}^2$
Cost of white washing the ceiling $=₹(96 \times 5)=₹ 480$
So the total cost of white washing $=₹(800+480)=₹ 1280$

Example 6:

In a building there are 24 cylindrical pillars. The radius of each pillar is $28 \mathrm{~cm}$ and height is $4 \mathrm{~m}$. Find the total cost of painting the curved surface area of all pillars at the rate of $₹ 8$ per $\mathrm{m}^2$.

Solution:

Radius of cylindrical pillar, $r=28 \mathrm{~cm}=0.28 \mathrm{~m}$
$
\text { height, } h=4 \mathrm{~m}
$

curved surface area of a cylinder $=2 \pi r h$
curved surface area of a pillar $=2 \times \frac{22}{7} \times 0.28 \times 4=7.04 \mathrm{~m}^2$
curved surface area of 24 such pillar $=7.04 \times 24=168.96 \mathrm{~m}^2$
cost of painting an area of $1 \mathrm{~m}^2=₹ 8$

Therefore, cost of painting $1689.6 \mathrm{~m}^2=168.96 \times 8=₹ 1351.68$
Example 7:

Find the height of a cylinder whose radius is $7 \mathrm{~cm}$ and the total surface area is $968 \mathrm{~cm}^2$.

Solution:

Let height of the cylinder $=h$, radius $=r=7 \mathrm{~cm}$
Total surface area $=2 \pi r(h+r)$

i.e.,
$
\begin{aligned}
2 \times \frac{22}{7} \times 7 \times(7+h) & =968 \\
h & =15 \mathrm{~cm}
\end{aligned}
$

Hence, the height of the cylinder is $15 \mathrm{~cm}$.

Example 8:

Find the height of a cuboid whose volume is $275 \mathrm{~cm}^3$ and base area is $25 \mathrm{~cm}^2$.
Solution:
$
\begin{aligned}
\text { Volume of a cuboid } & =\text { Base area } \times \text { Height } \\
\text { Hence height of the cuboid } & =\frac{\text { Volume of cuboid }}{\text { Base area }} \\
& =\frac{275}{25}=11 \mathrm{~cm}
\end{aligned}
$

Height of the cuboid is $11 \mathrm{~cm}$.
Example 9:

Agodown is in the form of a cuboid of measures $60 \mathrm{~m} \times 40 \mathrm{~m} \times 30 \mathrm{~m}$. How many cuboidal boxes can be stored in it if the volume of one box is $0.8 \mathrm{~m}^3$ ?
Solution:

$\quad$ Volume of one box $=0.8 \mathrm{~m}^3$
Volume of godown $=60 \times 40 \times 30=72000 \mathrm{~m}^3$
$
\begin{aligned}
\text { Number of boxes that can be stored in the godown } & =\frac{\text { Volume of the godown }}{\text { Volume of one box }} \\
& =\frac{60 \times 40 \times 30}{0.8}=90,000
\end{aligned}
$

Hence the number of cuboidal boxes that can be stored in the godown is 90,000 .
Example 10:

Arectangular paper of width $14 \mathrm{~cm}$ is rolled along its width and a cylinder of radius $20 \mathrm{~cm}$ is formed. Find the volume of the cylinder (Fig 9.31). (Take $\frac{22}{7}$ for $\pi$ )
Solution:

Acylinder is formed by rolling a rectangle about its width. Hence the width of the paper becomes height and radius of the cylinder is $20 \mathrm{~cm}$.

$\begin{aligned}
\text { Height of the cylinder } & =h=14 \mathrm{~cm} \\
\text { Radius } & =r=20 \mathrm{~cm}
\end{aligned}$

$
\begin{aligned}
\text { Volume of the cylinder } & =\mathrm{V}=\pi r^2 h \\
& =\frac{22}{7} \times 20 \times 20 \times 14=17600 \mathrm{~cm}^3
\end{aligned}
$

Hence, the volume of the cylinder is $17600 \mathrm{~cm}^3$.
Example 11:

A rectangular piece of paper $11 \mathrm{~cm} \times 4 \mathrm{~cm}$ is folded without overlapping to make a cylinder of height $4 \mathrm{~cm}$. Find the volume of the cylinder.

Solution:

Length of the paper becomes the perimeter of the base of the cylinder and width becomes height.
Let radius of the cylinder $=r$ and height $=h$
Perimeter of the base of the cylinder $=2 \pi r=11$
or
$
2 \times \frac{22}{7} \times r=11
$

Therefore,
$
\begin{aligned}
r & =\frac{7}{4} \mathrm{~cm} \\
\text { Volume of the cylinder } & =\mathrm{V}=\pi r^2 h \\
& =\frac{22}{7} \times \frac{7}{4} \times \frac{7}{4} \times 4 \mathrm{~cm}^3=38.5 \mathrm{~cm}^3 .
\end{aligned}
$

Hence the volume of the cylinder is $38.5 \mathrm{~cm}^3$.