Examples (Revised) - Chapter 11 - Perimeter & Area - Ncert Solutions class 7 - Maths

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Chapter 9 - Perimeter & Area | NCERT Solutions for Class 7 Maths

Example 1

One of the sides and the corresponding height of a parallelogram are $4 \mathrm{~cm}$ and $3 \mathrm{~cm}$ respectively. Find the area of the parallelogram (Fig 9.8).

Solution

Given that length of base $(b)=4 \mathrm{~cm}$, height $(h)=3 \mathrm{~cm}$
Area of the parallelogram $=b \times h$
$
=4 \mathrm{~cm} \times 3 \mathrm{~cm}=12 \mathrm{~cm}^2
$

Example 2

Find the height ' $x$ ' if the area of the parallelogram is $24 \mathrm{~cm}^2$ and the base is $4 \mathrm{~cm}$

Solution

Area of parallelogram $=b \times h$
Therefore, $24=4 \times x$ (Fig 9.9)
or $\frac{24}{4}=x$ or $\quad x=6 \mathrm{~cm}$
So, the height of the parallelogram is $6 \mathrm{~cm}$.

Example 3

The two sides of the parallelogram $\mathrm{ABCD}$ are $6 \mathrm{~cm}$ and $4 \mathrm{~cm}$. The height corresponding to the base $\mathrm{CD}$ is $3 \mathrm{~cm}$ (Fig 9.10). Find the
(i) area of the parallelogram.
(ii) the height corresponding to the base $\mathrm{AD}$.

Solution
(i)
$
\begin{aligned}
\text { Area of parallelogram } & =b \times h \\
& =6 \mathrm{~cm} \times 3 \mathrm{~cm}=18 \mathrm{~cm}^2
\end{aligned}
$
base $(b)=4 \mathrm{~cm}$, height $=x$ (say),
Area $=18 \mathrm{~cm}^2$

$
\begin{aligned}
\text { Area of parallelogram } & =b \times x \\
18 & =4 \times x \\
\frac{18}{4} & =x
\end{aligned}
$

Therefore,
$
x=4.5 \mathrm{~cm}
$

Thus, the height corresponding to base $\mathrm{AD}$ is $4.5 \mathrm{~cm}$.

Example 4

Find the area of the following triangles

Solution
(i)
$
\begin{aligned}
\text { Area of triangle } & =\frac{1}{2} b h=\frac{1}{2} \times \mathrm{QR} \times \mathrm{PS} \\
& =\frac{1}{2} \times 4 \mathrm{~cm} \times 2 \mathrm{~cm}=4 \mathrm{~cm}^2
\end{aligned}
$
(ii)
$
\begin{aligned}
\text { Area of triangle } & =\frac{1}{2} b h=\frac{1}{2} \times \mathrm{MN} \times \mathrm{LO} \\
& =\frac{1}{2} \times 3 \mathrm{~cm} \times 2 \mathrm{~cm}=3 \mathrm{~cm}^2
\end{aligned}
$

Example 5

Find BC, if the area of the triangle $\mathrm{ABC}$ is $36 \mathrm{~cm}^2$ and the height $\mathrm{AD}$ is $3 \mathrm{~cm}$ (Fig 9.12).

Solution

Height $=3 \mathrm{~cm}$, Area $=36 \mathrm{~cm}^2$

Area of the triangle $\mathrm{ABC}=\frac{1}{2} b h$
or
$
36=\frac{1}{2} \times b \times 3 \text { i.e., } \quad b=\frac{36 \times 2}{3}=24 \mathrm{~cm}
$

So,
$
\mathrm{BC}=24 \mathrm{~cm}
$

Example 6

In $\triangle \mathrm{PQR}, \mathrm{PR}=8 \mathrm{~cm}, \mathrm{QR}=4$ $\mathrm{cm}$ and $\mathrm{PL}=5 \mathrm{~cm}$ (Fig 9.13). Find:
(i) the area of the $\triangle \mathrm{PQR}$
(ii) $\mathrm{QM}$

Solution

Solution
(i) $\mathrm{QR}=$ base $=4 \mathrm{~cm}, \mathrm{PL}=$ height $=5 \mathrm{~cm}$
Area of the triangle $\mathrm{PQR}=\frac{1}{2} b h$
$
=\frac{1}{2} \times 4 \mathrm{~cm} \times 5 \mathrm{~cm}=10 \mathrm{~cm}^2
$
(ii) $\mathrm{PR}=$ base $=8 \mathrm{~cm}$
$\mathrm{QM}=$ height $=$ ?
Area $=10 \mathrm{~cm}^2$
$
\begin{aligned}
& \text { Area of triangle }=\frac{1}{2} \times b \times h \quad \text { i.e., } \quad 10=\frac{1}{2} \times 8 \times h \\
& h=\frac{10}{4}=\frac{5}{2}=2.5 \text {. So, } \quad \mathrm{QM}=2.5 \mathrm{~cm} \\
&
\end{aligned}
$

Example 7

What is the circumference of a circle of diameter $10 \mathrm{~cm}$ (Take $\pi=3.14$ )?

Solution
$
\begin{aligned}
\text { Diameter of the circle }(d) & =10 \mathrm{~cm} \\
\text { Circumference of circle } & =\pi d \\
& =3.14 \times 10 \mathrm{~cm}=31.4 \mathrm{~cm}
\end{aligned}
$

So, the circumference of the circle of diameter $10 \mathrm{~cm}$ is $31.4 \mathrm{~cm}$.
Example 8

What is the circumference of a circular disc of radius $14 \mathrm{~cm}$ ?
$
\left(\text { Use } \pi=\frac{22}{7}\right)
$

Solution
$
\begin{aligned}
\text { Radius of circular disc }(r) & =14 \mathrm{~cm} \\
\text { Circumference of disc } & =2 \pi r \\
& =2 \times \frac{22}{7} \times 14 \mathrm{~cm}=88 \mathrm{~cm}
\end{aligned}
$

So, the circumference of the circular disc is $88 \mathrm{~cm}$.
Example 9

The radius of a circular pipe is $10 \mathrm{~cm}$. What length of a tape is required to wrap once around the pipe $(\pi=3.14)$ ?

Solution

Radius of the pipe $(r)=10 \mathrm{~cm}$
Length of tape required is equal to the circumference of the pipe.
$
\begin{aligned}
\text { Circumference of the pipe } & =2 \pi r \\
& =2 \times 3.14 \times 10 \mathrm{~cm} \\
& =62.8 \mathrm{~cm}
\end{aligned}
$

Therefore, length of the tape needed to wrap once around the pipe is $62.8 \mathrm{~cm}$.

Example 10

Find the perimeter of the given shape (Fig 9.23) (Take $\pi=\frac{22}{7}$ ).
Solution

In this shape we need to find the circumference of semicircles on each side of the square. Do you need to find the perimeter of the square also? No. The outer boundary, of this figure is made up of semicircles. Diameter of each semicircle is $14 \mathrm{~cm}$.

We know that:
$
\begin{aligned}
\text { Circumference of the circle } & =\pi d \\
\text { Circumference of the semicircle } & =\frac{1}{2} \pi d \\
& =\frac{1}{2} \times \frac{22}{7} \times 14 \mathrm{~cm}=22 \mathrm{~cm}
\end{aligned}
$

Circumference of each of the semicircles is $22 \mathrm{~cm}$
Therefore, perimeter of the given figure $=4 \times 22 \mathrm{~cm}=88 \mathrm{~cm}$

Example 11

Sudhanshu divides a circular disc of radius $7 \mathrm{~cm}$ in two equal parts.
What is the perimeter of each semicircular shape disc? (Use $\pi=\frac{22}{7}$ )

Solution

To find the perimeter of the semicircular disc (Fig 9.24), we need to find
(i) Circumference of semicircular shape
(ii) Diameter

Given that radius $(r)=7 \mathrm{~cm}$. We know that the circumference of circle $=2 \pi r$

So, the circumference of the semicircle $=\frac{1}{2} \times 2 \pi r=\pi r$
$=\frac{22}{7} \times 7 \mathrm{~cm}=22 \mathrm{~cm}$
So, the diameter of the circle
$=2 r=2 \times 7 \mathrm{~cm}=14 \mathrm{~cm}$
Thus, perimeter of each semicircular disc $=22 \mathrm{~cm}+14 \mathrm{~cm}=36 \mathrm{~cm}$

Example 12

Find the area of a circle of radius $30 \mathrm{~cm}$ (use $\pi=3.14$ ).
Solution

Radius, $r=30 \mathrm{~cm}$
Area of the circle $=\pi r^2=3.14 \times 30^2=2,826 \mathrm{~cm}^2$
Example 13

Diameter of a circular garden is $9.8 \mathrm{~m}$. Find its area.
Solution

Diameter, $d=9.8 \mathrm{~m}$. Therefore, radius $r=9.8 \div 2=4.9 \mathrm{~m}$
Area of the circle $=\pi r^2=\frac{22}{7} \times(4.9)^2 \mathrm{~m}^2=\frac{22}{7} \times 4.9 \times 4.9 \mathrm{~m}^2=75.46 \mathrm{~m}^2$
Example 14

The adjoining figure shows two circles with the same centre. The radius of the larger circle is $10 \mathrm{~cm}$ and the radius of the smaller circle is $4 \mathrm{~cm}$.

Find: (a) the area of the larger circle
(b) the area of the smaller circle
(c) the shaded area between the two circles. $(\pi=3.14)$

Solution
(a) Radius of the larger circle $=10 \mathrm{~cm}$
$
\begin{aligned}
\text { So, area of the larger circle } & =\pi r^2 \\
& =3.14 \times 10 \times 10=314 \mathrm{~cm}^2
\end{aligned}
$
(b) Radius of the smaller circle $=4 \mathrm{~cm}$
Area of the smaller circle $=\pi r^2$
$
=3.14 \times 4 \times 4=50.24 \mathrm{~cm}^2
$
(c) Area of the shaded region $=(314-50.24) \mathrm{cm}^2=263.76 \mathrm{~cm}^2$