Exercise 9.2 (Revised) - Chapter 11 - Perimeter & Area - Ncert Solutions class 7 - Maths

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

Ex 9.2 Question 1.

Find the circumference of the circles with the following radius: $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
(a) $14 \mathrm{~cm}$
(b) $28 \mathrm{~mm}$
(c) $21 \mathrm{~cm}$

Answer:

(a) Circumference of the circle $=2 \pi r=2 \times \frac{22}{7} \times 14=88 \mathrm{~cm}$
(b) Circumference of the circle $=2 \pi r=2 \times \frac{22}{7} \times 28=176 \mathrm{~mm}$
(c) Circumference of the circle $=2 \pi r=2 \times \frac{22}{7} \times 21=132 \mathrm{~cm}$

Ex 9.2 Question 2.

Find the area of the following circles, given that: (Take $\pi=\frac{22}{7}$ )
(a) radius $=14 \mathrm{~mm}$
(b) diameter $=49 \mathrm{~m}$
(c) radius $5 \mathrm{~cm}$

Answer:

(a) Area of circle $=\pi r^2=\frac{22}{7} \times 14 \times 14=22 \times 2 \times 14=616 \mathrm{~mm}^2$
(b) Diameter $=49 \mathrm{~m}$

$
\therefore \text { radius }=\frac{49}{2}=24.5 \mathrm{~m}
$
$\therefore$ Area of circle $=\pi r^2=\frac{22}{7} \times 24.5 \times 24.5=22 \times 3.5 \times 24.5=1886.5 \mathrm{~m}^2$
(c) Area of circle $=\pi r^2=\frac{22}{7} \times 5 \times 5=\frac{550}{7} \mathrm{~cm}^2$

Ex 9.2 Question 3.

If the circumference of a circular sheet is $154 \mathrm{~m}$, find its radius. Also find the area of the sheet. (Take $\pi=\frac{22}{7}$ )

Answer:

Circumference of the circular sheet $=154 \mathrm{~m}$
$
\begin{aligned}
& \Rightarrow 2 \pi r=154 \mathrm{~m} \Rightarrow r=\frac{154}{2 \pi} \\
& \Rightarrow r=\frac{154 \times 7}{2 \times 22}=24.5 \mathrm{~m}
\end{aligned}
$

Now, Area of circular sheet $=\pi r^2=\frac{22}{7} \times 24.5 \times 24.5=22 \times 3.5 \times 24.5=1886.5 \mathrm{~m}^2$
Thus, the radius and area of circular sheet are $24.5 \mathrm{~m}$ and $1886.5 \mathrm{~m}^2$ respectively.

Ex 9.2 Question 4.

A gardener wants to fence a circular garden of diameter $21 \mathrm{~m}$. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also, find the costs of the rope, if it cost Rs. 4 per meter. (Take $\pi=\frac{22}{7}$ )

Answer:

Diameter of the circular garden $=21 \mathrm{~m}$
$\therefore$ Radius of the circular garden $=\frac{21}{2} \mathrm{~m}$
Now, Circumference of circular garden $=2 \pi r=2 \times \frac{22}{7} \times \frac{21}{2}=2 \times 11 \times 3=22 \times 3=66 \mathrm{~m}$
The gardener makes 2 rounds of fence so the total length of the rope of fencing

$
=2 \times 2 \pi r=2 \times 66=132 \mathrm{~m}
$

Since the cost of 1 meter rope $=$ Rs. 4
Therefore, cost of 132 meter rope $=4 \times 132=$ Rs. 528

Ex 9.2 Question 5.

From a circular sheet of radius $4 \mathrm{~cm}$, a circle of radius $3 \mathrm{~cm}$ is removed. Find the area of the remaining sheet. (Take $\pi=3.14$ )

Answer:

Radius of circular sheet $(\mathrm{R})=4 \mathrm{~cm}$ and radius of removed circle $(\mathrm{r})=3 \mathrm{~cm}$

Area of remaining sheet $=$ Area of circular sheet - Area of removed circle

$
\begin{aligned}
& =\pi \mathrm{R}^2-\pi r^2=\pi\left(\mathrm{R}^2-r^2\right) \\
& =\pi\left(4^2-3^2\right)=\pi(16-9) \\
& =3.14 \times 7=21.98 \mathrm{~cm}^2
\end{aligned}
$

Thus, the area of remaining sheet is $21.98 \mathrm{~cm}^2$.

Ex 9.2 Question 6.

Saima wants to put a lace on the edge of a circular table cover of diameter $1.5 \mathrm{~m}$. Find the length of the lace required and also find its cost if one meter of the lace costs Rs. 15 . (Take $\pi=3.14$ )

Answer:

Diameter of the circular table cover $=1.5 \mathrm{~m}$
$\therefore$ Radius of the circular table cover $=\frac{1.5}{2} \mathrm{~m}$
Circumference of circular table cover $=2 \pi r=2 \times 3.14 \times \frac{1.5}{2}=4.71 \mathrm{~m}$
Therefore the length of required lace is $4.71 \mathrm{~m}$.
Now the cost of $1 \mathrm{~m}$ lace $=$ Rs. 15

Then the cost of $4.71 \mathrm{~m}$ lace $=15 \times 4.71=$ Rs. 70.65
Hence, the cost of $4.71 \mathrm{~m}$ lace is Rs. 70.65.

Ex 9.2 Question 7.

Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Answer:

Diameter $=10 \mathrm{~cm}$
$
\therefore \text { Radius }=\frac{10}{2}=5 \mathrm{~cm}
$

According to question,

Perimeter of figure $=$ Circumference of semi-circle + diameter
$
\begin{aligned}
& =\pi r+\mathrm{D}=\frac{22}{7} \times 5+10=\frac{110}{7}+10 \\
& =\frac{110+70}{7}=\frac{180}{7}=25.71 \mathrm{~cm}
\end{aligned}
$

Thus, the perimeter of the given figure is $25.71 \mathrm{~cm}$.

Ex 9.2 Question 8.

Find the cost of polishing a circular table-top of diameter $1.6 \mathrm{~m}$, if the rate of polishing is Rs. $15 / \mathrm{m}^2$. (Take $\pi=3.14$ )

Answer:

Diameter of the circular table top $=1.6 \mathrm{~m}$
$\therefore$ Radius of the circular table top $=\frac{1.6}{2}=0.8 \mathrm{~m}$
Area of circular table top $=\pi r^2=3.14 \times 0.8 \times 0.8=2.0096 \mathrm{~m}^2$
Now cost of $1 \mathrm{~m}^2$ polishing $=$ Rs. 15
Then cost of $2.0096 \mathrm{~m}^2$ polishing $=15 \times 2.0096=$ Rs. 30.14 (approx.)
Thus, the cost of polishing a circular table top is Rs. 30.14 (approx.)

Ex 9.2 Question 9.

Shazli took a wire of length $44 \mathrm{~cm}$ and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides? Which figure encloses more area, the circle or the square? (Take $\left.\pi=\frac{22}{7}\right)$

Answer:

Total length of the wire $=44 \mathrm{~cm}$
$\therefore$ the circumference of the circle $=2 \pi r=44 \mathrm{~cm}$
$
\Rightarrow 2 \times \frac{22}{7} \times r=44 \Rightarrow r=\frac{44 \times 7}{2 \times 22}=7 \mathrm{~cm}
$

Now Area of the circle $=\pi r^2=\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2$
Now the wire is converted into square.

Then perimeter of square $=44 \mathrm{~cm}$
$
\Rightarrow 4 \times \text { side }=44 \Rightarrow \text { side }=\frac{44}{4}=11 \mathrm{~cm}
$

Now area of square $=$ side $\mathrm{x}$ side $=11 \times 11=121 \mathrm{~cm}^2$
Therefore, on comparing the area of circle is greater than that of square, so the circle enclosed more area.

Ex 9.2 Question 10.

From a circular card sheet of radius $14 \mathrm{~cm}$, two circles of radius $3.5 \mathrm{~cm}$ and a rectangle of length $3 \mathrm{~cm}$ and breadth $1 \mathrm{~cm}$ are removed (as shown in the adjoining figure). Find the area of the remaining sheet. (Take $\pi=\frac{22}{7}$ )

Answer:

Radius of circular sheet $(\mathrm{R})=14 \mathrm{~cm}$ and Radius of smaller circle $(r)=3.5 \mathrm{~cm}$
Length of rectangle $(l)=3 \mathrm{~cm}$ and breadth of rectangle $(b)=1 \mathrm{~cm}$
According to question,
Area of remaining sheet $=$ Area of circular sheet - (Area of two smaller circles + Area of rectangle)
$
\begin{aligned}
& =\pi \mathrm{R}^2-\left[2\left(\pi r^2\right)+(l \times b)\right] \\
& =\frac{22}{7} \times 14 \times 14-\left[\left(2 \times \frac{22}{7} \times 3.5 \times 3.5\right)-(3 \times 1)\right] \\
& =22 \times 14 \times 2-[44 \times 0.5 \times 3.5+3] \\
& =616-80 \\
& =536 \mathrm{~cm}^2
\end{aligned}
$

Therefore the area of remaining sheet is $536 \mathrm{~cm}^2$.

Ex 9.2 Question 11.

A circle of radius $2 \mathrm{~cm}$ is cut out from a square piece of an aluminium sheet of side $6 \mathrm{~cm}$. What is the area of the left over aluminium sheet? (Take $\pi=3.14$ )

Answer:

Radius of circle $=2 \mathrm{~cm}$ and side of aluminium square sheet $=6 \mathrm{~cm}$

According to question,

Area of aluminium sheet left $=$ Total area of aluminium sheet - Area of circle
$
=\text { side } \times \text { side }-\pi r^2=6 \times 6-\frac{22}{7} \times 2 \times 2=36-12.56=23.44 \mathrm{~cm}^2
$

Therefore, the area of aluminium sheet left is $23.44 \mathrm{~cm}^2$.

Ex 9.2 Question 12.

The circumference of a circle is $31.4 \mathrm{~cm}$. Find the radius and the area of the circle. (Take $\pi=3.14$ )

Answer:

The circumference of the circle $=31.4 \mathrm{~cm}$
$
\begin{aligned}
& \Rightarrow 2 \pi r=31.4 \Rightarrow 2 \times 3.14 \times r=31.4 \\
& \Rightarrow r=\frac{31.4}{2 \times 3.14}=5 \mathrm{~cm}
\end{aligned}
$

Then area of the circle $=\pi r^2=3.14 \times 5 \times 5$
$
=78.5 \mathrm{~cm}^2
$

Therefore, the radius and the area of the circle are $5 \mathrm{~cm}$ and $78.5 \mathrm{~cm}^2$ respectively.

Ex 9.2 Question 13.

A circular flower bed is surrounded by a path $4 \mathrm{~m}$ wide. The diameter of the flower bed is $66 \mathrm{~m}$. What is the area of this path? (Take $\pi=3.14$ )

Answer:

Diameter of the circular flower bed $=66 \mathrm{~m}$
$\therefore$ Radius of circular flower bed $(r)=\frac{66}{2}=33 \mathrm{~m}$
$\therefore$ Radius of circular flower bed with $4 \mathrm{~m}$ wide path (R) $=33+4=37 \mathrm{~m}$
According to the question,
Area of path $=$ Area of bigger circle - Area of smaller circle
$=\pi \mathrm{R}^2-\pi r^2=\pi\left(\mathrm{R}^2-r^2\right)$
$=\pi\left[(37)^2-(33)^2\right]$
$=3.14[(37+33)(37-33)]\left[\because a^2-b^2=(a+b)(a-b)\right]$
$=3.14 \times 70 \times 4$
$=879.20 \mathrm{~m}^2$
Therefore, the area of the path is $879.20 \mathrm{~m}^2$.

Ex 9.2 Question 14.

A circular flower garden has an area of $314 \mathrm{~m}^2$. A sprinkler at the centre of the garden can cover an area that has a radius of $12 \mathrm{~m}$. Will the sprinkler water the entire garden? (Take $\pi=3.14)$

Answer:

Circular area by the sprinkler $=\pi r^2=3.14 \times 12 \times 12$

$
=3.14 \times 144=452.16 \mathrm{~m}^2
$

Area of the circular flower garden $=314 \mathrm{~m}^2$

Since Area of circular flower garden is smaller than area by sprinkler.
Therefore the sprinkler will water the entire garden.

Ex 9.2 Question 15.

Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take $\pi=3.14$ )

Answer:

Radius of outer circle $(r)=19 \mathrm{~m}$
$\therefore$ Circumference of outer circle $=2 \pi r=2 \times 3.14 \times 19=119.32 \mathrm{~m}$
Now radius of inner circle $\left(r^{\prime}\right)=19-10=9 \mathrm{~m}$
$\therefore$ Circumference of inner circle $=2 \pi r^{\prime}=2 \times 3.14 \times 9=56.52 \mathrm{~m}$
Therefore the circumferences of inner and outer circles are $56.52 \mathrm{~m}$ and $119.32 \mathrm{~m}$ respectively.

Ex 9.2 Question 16.

How many times a wheel of radius $28 \mathrm{~cm}$ must rotate to go $352 \mathrm{~m}$ ? $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$

Answer:

Let wheel must be rotate $n$ times of its circumference.
Radius of wheel $=28 \mathrm{~cm}$ and Total distance $=352 \mathrm{~m}=35200 \mathrm{~cm}$
$\therefore$ Distance covered by wheel $=n \mathrm{x}$ circumference of wheel
$
\begin{aligned}
& \Rightarrow 35200=n \times 2 \pi r \\
& \Rightarrow 35200=n \times 2 \times \frac{22}{7} \times 28 \\
& \Rightarrow n=\frac{35200 \times 7}{2 \times 22 \times 28}
\end{aligned}
$

$\Rightarrow n=200$ revolutions
Thus wheel must rotate 200 times to go $352 \mathrm{~m}$.

Ex 9.2 Question 17.

The minute hand of a circular clock is $15 \mathrm{~cm}$ long. How far does the tip of the minute hand move in 1 hour? (Take $\pi=3.14$ )

Answer:

In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle $(r)=15 \mathrm{~cm}$
Circumference of circular clock $=2 \pi r=2 \times 3.14 \times 15=94.2 \mathrm{~cm}$
Therefore, the tip of the minute hand moves $94.2 \mathrm{~cm}$ in 1 hour.