Additional Questions - Chapter 2 - Measurements - Term 2 - 7th Maths Guide Samacheer Kalvi Solutions

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Additional Questions and Answers
Exercise $2.1$
Question $1 .$
A circular disc of radius $28 \mathrm{~cm}$ is divided into two equal parts. What is the perimeter of each semicircular shape disc? Also find the perimeter of the circular disc.
Answer:
Radius of the circular $\operatorname{disc}=28 \mathrm{~cm}$
$\therefore$ circumference of the circle $=2 \pi \mathrm{r}$ units
$\begin{aligned}
&=2 \times \frac{22}{7} \times 28 \mathrm{~cm} \\
&=2 \times 22 \times 4=44 \times 4=176 \mathrm{~cm}
\end{aligned}$
Perimeter of the circular $\operatorname{disc}=176 \mathrm{~cm}$
Now circumference of the semicircular $\operatorname{disc}=\frac{1}{2} \times(2 \pi \mathrm{r})=\pi \mathrm{r}$ units
Perimeter of the semicircular disk $=\pi \mathrm{r}+\mathrm{r}+\mathrm{r}=\frac{22}{7} \times 28+28+28$
$=22 \times 4+28+28=88+28+28=144$
Perimeter of each semicircular disc $=144 \mathrm{~cm}$

Question 2 .
A gardener wants to fence a circular garden of diameter $14 \mathrm{~m}$. Find the length of the rope he needs
to purchase if he makes 3 rounds of fence. Also find the cost of the rope, if it costs $₹ 4$ per metre.
Solution:
Diameter of the circular garden $(\mathrm{d})=14 \mathrm{~m}$
Circumference $=\pi \mathrm{d}=\frac{22}{7} \times 14=44 \mathrm{~m}$
Since the rope makes 3 rounds of fence,
length of the rope needed $=3 \times$ circumference of the garden
$=3 \times 44 \mathrm{~m}=132 \mathrm{~m}$
Cost of rope per meter $=₹ 4 \times 132=₹ 528$

Question $3 .$
Latha wants to put a lace on the edge of a circular table cover of diameter $3 \mathrm{~m}$. Find the length of the lace required and also find the cost if one metre of the lace $\operatorname{costs} ₹ 30$ (Take $\pi=3.15$ )
Solution:
Diameter of the circular table cover $=3 \mathrm{~m}$
Circumference $\mathrm{C}=\pi \mathrm{d}$ units $=3.15 \times 3 \mathrm{~m}=9.45 \mathrm{~m}$
Length of the lace required $=9.45 \mathrm{~m}$
Cost of lace per meter $=₹ 30$
$\therefore$ Cost of $9.45 \mathrm{~m}$ lace $=₹ 30 \times 9.45=₹ 283.50$

Exercise 2.2
Question $1 .$
The circumference of two circles are on the ratio $5: 6$. Find the ratio of their areas.
Solution:
Let the radii of the given circles be $r_{1}$ and $r_{2}$
Let their circumference be $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ respectively
$\mathrm{C}_{1}=2 \pi \mathrm{r}_{1}$ and $\mathrm{C}_{2}=2 \pi \mathrm{r}_{2}$
$\begin{aligned} \text { Since } \mathrm{C}_{1} &: \mathrm{C}_{2}=5: 6 \\ \therefore\left(2 \pi \mathrm{r}_{1}\right):\left(2 \pi \mathrm{r}_{2}\right) &=5: 6 \\ \frac{2 \pi r_{1}}{2 \pi r_{2}} &=\frac{5}{6} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{5}{6} \end{aligned}$
$\begin{aligned}
\therefore\left(2 \pi r_{1}\right):\left(2 \pi r_{2}\right) &=5: 6 \\
\frac{2 \pi r_{1}}{2 \pi r_{2}} &=\frac{5}{6} \Rightarrow \frac{r_{1}}{r_{2}}=\frac{5}{6}
\end{aligned}$
Now let the area of the given circles the $\mathrm{A}_{1}$ and $\mathrm{A}_{2}$

$\begin{aligned}
\mathrm{A}_{1} &=\pi r_{1}^{2} \\
\mathrm{~A}_{2} &=\pi r_{2}^{2} \\
\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}} &=\frac{\pi r_{1}^{2}}{\pi r_{2}^{2}}=\frac{r_{1}^{2}}{r_{2}^{2}} \Rightarrow \frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{5}{6}\right)^{2}=\frac{25}{36} \\
\therefore \mathrm{A}_{1}: \mathrm{A}_{2} &=25: 36
\end{aligned}$
$\therefore$ The areas of two circles are in the ratio $25: 36$.

Question $2 .$
Find the area of the circle whose circumference is $88 \mathrm{~cm}$.
Solution:
The circumference of the circle $=88 \mathrm{~cm}$
$\begin{aligned}
&2 \pi \mathrm{r}=88 \mathrm{~cm} \\
&2 \times \frac{22}{7} \times \mathrm{r}=88 \mathrm{~cm} \\
&\mathrm{r}=\frac{88 \times 7}{2 \times 22}=14 \mathrm{~cm} \\
&\mathrm{r}=14 \mathrm{~cm}
\end{aligned}$
Area of the circle $\mathrm{A}=\pi \mathrm{r}^{2}$
$\begin{aligned}
&=\frac{22}{7} \times 14 \times 14 \mathrm{~cm}^{2} \\
&=22 \times 2 \times 14 \mathrm{~cm}^{2} \\
&=616 \mathrm{~cm}^{2}
\end{aligned}$

Question $3 .$
Find the cost of polishing a circular table top of diameter $16 \mathrm{dm}$ if the rate of polishing is ₹ 15 per $\mathrm{dm}^{2}$.
Solution:

Diameter of the circular table top $=16 \mathrm{dm}$
Radius $(\mathrm{r})=\frac{16}{2}=8 \mathrm{~cm}$
Area of the circular table top
$=\pi r^{2}$ sq.units
$=\frac{22}{7} \times 8 \times 8 \mathrm{dm}^{2}=\frac{1408}{7} \mathrm{dm}^{2}$
$=201.14 \mathrm{dm}^{2}$
Rate of the polishing per $\mathrm{dm}^{2}=₹ 15$
$\therefore$ Rate of the polishing $201.14 \mathrm{dm}^{2}$
$\begin{aligned}
&=₹ 15 \times 201.14 \\
&=₹ 3,017
\end{aligned}$

Exercise $2.3$
Question $1 .$
From a circular sheet of radius $5 \mathrm{~cm}$ a circle of radius $3 \mathrm{~cm}$ is removed. Find the area of the remaining sheet.
Solution:
Radius of the outer circle $\mathrm{R}=5 \mathrm{~cm}$
Radius of the inner circle $\mathrm{r}=3 \mathrm{~cm}$
- Area of the inner circle
$=\pi \mathrm{R}^{2}-\pi \mathrm{r}^{2}$ sq. units $=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)$ sq. units
$=\frac{22}{7}\left(5^{2}-3^{2}\right) \mathrm{cm}^{2}=\frac{22}{7}\left(5^{2}-3^{2}\right) \mathrm{cm}^{2}$
$=\frac{22}{7} \times(5+3)(5-3)=\frac{22}{7} \times(8)(2)$
$=\frac{352}{7}=50.28 \mathrm{~cm}^{2}$

Question $2 .$
A picture is painted on a cardborad $8 \mathrm{~cm}$ long and $5 \mathrm{~cm}$ wide such that there is a margin of $1.5 \mathrm{~cm}$ along each of its sides. Find the total area of the margin.
Solution:
Length of the outer rectangle $\mathrm{L}=8 \mathrm{~cm}$
Breadth of the outer rectangle $B=5 \mathrm{~cm}$
Area of the outer rectangle $=\mathrm{L} \times \mathrm{B}$ sq. units $=8 \times 5 \mathrm{~cm}^{2}=40 \mathrm{~cm}^{2}$
Length of the inner rectangle $\mathrm{l}=\mathrm{L}-2 \mathrm{~W}=8-2(1.5) \mathrm{cm}=8-3 \mathrm{~cm}$
$\mathrm{l}=5 \mathrm{~cm}$
Breadth of thqnner rectangle $b=B-2 W=5-2(1.5) \mathrm{cm}$
$=5-3 \mathrm{~cm}=2 \mathrm{~cm}$
$\therefore$ Area of the margin $=$ Area of the outer rectangle
- Area of the inner rectangle
$=(40-10) \mathrm{cm}^{2}=30 \mathrm{~cm}^{2}$

Question $3 .$
A circular piece of radius $2 \mathrm{~cm}$ is cut from a rectangle sheet of length $5 \mathrm{~cm}$ and breadth $3 \mathrm{~cm}$. Find
the area left in the sheet.
Solution:
Radius of the portion removed $\mathrm{r}=2 \mathrm{~cm}$
Area of the circular sheet $=\pi \mathrm{r}^{2}$ sq.units
$=\frac{22}{7} \times 2 \times 2 \mathrm{~cm}^{2}=\frac{88}{7} \mathrm{~cm}^{2}=12.57 \mathrm{~cm}^{2}$
Length of the rectangular sheet $\mathrm{L}=5 \mathrm{~cm}$
Breadth of the rectangular sheet $\mathrm{B}=3 \mathrm{~cm}$
Area of the rectangle $=\mathrm{L} \times \mathrm{B}$ sq. units $=5 \times 3 \mathrm{~cm}^{2}=15 \mathrm{~cm}^{2}$
Area of the sheet left over $=$ Area of the rectangle $-$ Area of the circle
$=15 \mathrm{~cm}^{2}-12.57 \mathrm{~cm}^{2}=2.43 \mathrm{~cm}^{2}$