Exercise 2.2 - Chapter 2 Measurements Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.2$
Question $1 .$
Find the area of the dining table whose diameter is $105 \mathrm{~cm}$.
Solution:
Diameter of the dinig table $(\mathrm{d})=105 \mathrm{~cm}$
$\therefore$ Radius $\mathrm{r}=\frac{d}{2}=\frac{105}{2} \mathrm{~cm}$
Area of the circle $=\pi r^{2}=\frac{22}{7} \times \frac{105}{2} \times \frac{105}{2}=8662.5$ sq.cm
Area of the dinning table $=8662.5 \mathrm{~cm}^{2}$
 

Question $2 .$
Calculate the area of the shotput ring whose diameter is $2.135 \mathrm{~m}$.
Solution:
Radius of the shotput ring $\mathrm{r}=\frac{d}{2}=\frac{2.135}{2} \mathrm{~m}$
Area of the circle $=\pi \mathrm{r}^{2}$
$=\frac{22}{7} \times \frac{2.135}{2} \times \frac{2.135}{2}$
$=\frac{25.07}{7}=3.581 \mathrm{~m}^{2}$
$\therefore$ Area of the shotput ring $=3.581 \mathrm{~m}^{2}$
 

Question $3 .$
A sprinkler placed at the centre of a flower garden sprays water covering a circular area. If the area watered is $1386 \mathrm{~cm} 2$, find its radius and diameter.
Solution:
Area of the Circle $=\pi \mathrm{r}^{2}$ sq.units
Area of the circular portion watered $=1386 \mathrm{~cm}^{2}$
$\pi \mathrm{r}^{2}=1386$
$\frac{22}{7} \times \mathrm{r}^{2}=1386$

$\begin{aligned}
&\mathrm{r}^{2}=1386 \times \frac{7}{22}=63 \times 7=9 \times 7 \times 7 \\
&\mathrm{r}^{2}=3^{2} \times 7^{2} \\
&\mathrm{r}=3 \times 7 \\
&\text { Radius }(\mathrm{r})=21 \mathrm{~cm} \\
&\text { Diameter }(\mathrm{d})=2 \mathrm{r}=2 \times 21 \mathrm{~cm} \\
&\text { Diameter }(\mathrm{d})=42 \mathrm{~cm}
\end{aligned}$

Question $4 .$
The circumference of a circular park is $352 \mathrm{~m}$. Find the area of the park.
Solution:
Circumference of a Circle $=2 \pi \mathrm{r}$ units
Given circumference of a circular park $=352 \mathrm{~m}$
$\begin{aligned}
&2 \pi \mathrm{r}=352 \\
&2 \times \frac{22}{7} \times \mathrm{r}=352 \\
&\mathrm{r}=352 \times \frac{7}{22} \times \frac{1}{2}=56 \mathrm{~m}
\end{aligned}$
Area of the park $=\pi r^{2}=\frac{22}{7} \times 56 \times 56$ sq.units
$=22 \times 8 \times 56=9856 \mathrm{~m}^{2}$
$\therefore$ Area of the Circular park $=9856 \mathrm{~m}^{2}$
 

Question $5 .$
In a grass land, a sheep is tethered by a rope of length $4.9 \mathrm{~m}$. Find the maximum area that the sheep can graze.

Solution:
Length of the rope $=4.9 \mathrm{~m}$
Area that the sheep can graze $=$ Area of circle with radius $4.9 \mathrm{~m}$
Area of the circle $=\pi r^{2}$ sq.units
$=\frac{22}{7} \times 4.9 \times 4.9=22 \times 0.7 \times 4.9=75.46$
$\therefore$ Area that the sheep can graze $=75.46 \mathrm{~m}^{2}$
 

Question $6 .$
Find the length of the rope by which a bull must be tethered in order that it may be able to graze an
area of $2464 \mathrm{~m}^{2}$.
Solution:
If the bull is tethered by a rope then the area it can graze is a circular area of radius
$=$ length of the rope
Area of the circle $=2464 \mathrm{~m}^{2}$
$\pi r^{2}=2464 \mathrm{~m}^{2}$
$\frac{22}{7} \times \mathrm{r}^{2}=2464$

$\begin{aligned}
&\mathrm{r}^{2}=2464 \times \frac{7}{22}=122 \times 7=16 \times 7 \times 7 \\
&\mathrm{r}^{2}=42 \times 72 \\
&\mathrm{r}=4 \times 7=28 \mathrm{~m}
\end{aligned}$
 

Question $7 .$
Lalitha wants to buy a round carpet of radius is $63 \mathrm{~cm}$ for her hall. Find the area that will be covered by the carpet.
Solution:
Radius of the round carpet $=63 \mathrm{~cm}$
Area covered by the round carpet $=\pi r^{2}$ sq units
$\mathrm{A}=\frac{22}{7} \times 63 \times 63=22 \times 9 \times 63=12474 \mathrm{~cm}^{2}$
Area covered by the round carpet $=12,474 \mathrm{~cm}^{2}$
 

Question $8 .$
Thenmozhi wants to level her circular flower garden whose diameter is $49 \mathrm{~m}$ at the rate of $₹ 150$ per $\mathrm{m}^{2}$ Find the cost of levelling.
Solution:
Diamter of the circular garden $\mathrm{d}=49 \mathrm{~m}$
Radius $\mathrm{r}=\frac{d}{2}=\frac{49}{2} \mathrm{~m}$
Area of the circular garden $=\pi r^{2}$ sq units
$=\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2} \mathrm{~m}^{2}=1,886.5 \mathrm{~m}^{2}$
Cost of levelling a $\mathrm{m}^{2}$ area $=₹ 150$
$\therefore$ Cost of levelling $1886.5 \mathrm{~m}^{2}=₹ 150 \times 1886.5=₹ 2,82,975$
Cost of levelling the flower garden $=₹ 2,82,975$

Question $9 .$
The floor of the circular swimming pool whose radius is $7 \mathrm{~m}$ has to be cemented at the rate of $₹ 18$ per $\mathrm{m}^{2}$. Find the total cost of cementing the floor.
Solution:
Radius of the circular swimming pool $\mathrm{r}=7 \mathrm{~m}$
Area of the circular swimming pool $\mathrm{A}=\pi \mathrm{r}^{2}$ sq. units
$=\frac{22}{7} \times 7 \times 7 \mathrm{~m}^{2}=154 \mathrm{~m}^{2}$
Cost of cementing a $\mathrm{m}^{2}$ floor $=₹ 18$.
Cost of cementing $154 \mathrm{~m}^{2}$ floor $=₹ 18 \times 154=₹ 2,772$

Question $10 .$
The formula used to find the area of the circle is
(i) $47 \pi r^{2}$
(ii) $\pi \mathrm{r}^{2}$
(iii) $2 \pi r^{2}$
(iv) $\pi r^{2}+2 r$
Answer:
(ii) $\pi \mathrm{r}^{2}$

Question $11 .$
The ratio of the area of a circle to the area of its semicircle is
(i) $2: 1$
(ii) $1: 2$
(iii) $4: 1$
(iv) $1: 4$
Answer:
(i) $2: 1$
 

Question $12 .$
Area of circle of radius ' $\mathrm{n}$ ' units is
(i) $2 \pi \mathrm{r}^{\mathrm{p}}$ sq.units
(ii) $\pi \mathrm{m}^{2}$ sq.units
(iii) $\pi \mathrm{r}^{2}$ sq.units
(iv) $\pi \mathrm{n}^{2}$ sq.units
Answer:
(iv) $\pi \mathrm{n}^{2}$ sq.units