Exercise 2.1 - Chapter 2 Measurements 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $2.1$ : Chapter 2 Measurements Term 1 Class 8th std Maths Guide Samacheer Kalvi Solutions
Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is Answer:
$\pi$
(ii) A line segment which joins any two points on a circle is a
Answer:
Chord
(iii) The longest chord of a circle is
Answer:
Diameter
(iv) The radius of a circle of diameter $24 \mathrm{~cm}$ is
Answer:
$12 \mathrm{~cm}$
(v) A part of circumference of a circle is called as
Answer:
an arc

Question $2 .$
Match the following

Answer:
(i) $-\mathrm{c}$
(ii) $-\mathrm{d}$
(iii) $-\mathrm{e}$
(iv) $-b$
(v) $-\mathrm{a}$

Question $3 .$
Find the central angle of the shaded sectors (each circle is divided into equal sectors).

Question $4 .$
For the sectors with given measures, find the length of the arc, area and perimeter. $(\pi=3.14)$
(i) central angle $45^{\circ} \mathrm{r}=16 \mathrm{~cm}$
Answer:
(i) central angle $45^{\circ} \mathrm{r}=16 \mathrm{~cm}$
Length of the arc $1=\frac{\theta^{\circ}}{360^{\circ}} \times 2 \pi \mathrm{r}$ units
$\begin{aligned}
&\mathrm{I}=\frac{45^{\circ}}{360^{\circ}} \times 2 \times 3.14 \times 16 \mathrm{~cm} \\
&\mathrm{I}=\frac{1}{8} \times 2 \times 3.14 \times 16 \mathrm{~cm} \\
&\mathrm{I}=12.56 \mathrm{~cm}
\end{aligned}$
Area of the sector $=\frac{\theta}{360^{\circ}} \times \pi r^{2}$ sq. units
$\mathrm{A}=\frac{45^{\circ}}{360^{\circ}} \times 3.14 \times 16 \times 16$ $\mathrm{~A}=100.48 \mathrm{~cm}^{2}$ $\mathrm{Perimeter}$ of the sector $\mathrm{P}=1+2 \mathrm{r}$ units $\mathrm{P}=12.56+2(16) \mathrm{cm}$ $\mathrm{P}=44.56 \mathrm{~cm}$
(ii) central angle $120^{\circ}, \mathrm{d}=12.6 \mathrm{~cm}$
Answer:
$\begin{aligned}
&\therefore \mathrm{r}=\frac{12.6}{2} \mathrm{~cm} \\
&\mathrm{r}=6.3 \mathrm{~cm}
\end{aligned}$
Length of the arc $1=\frac{\theta^{*}}{360^{\circ}} \times 2 \pi \mathrm{r}$ units
$\begin{aligned}
&\mathrm{I}=\frac{120^{\circ}}{360^{\circ}} \times 2 \times 3.14 \times 63 \mathrm{~cm} \\
&\mathrm{I}=13.188 \mathrm{~cm} \\
&\mathrm{I}=13.19 \mathrm{~cm}
\end{aligned}$
Area of the sector $\mathrm{A}=\frac{\mathscr{\theta}}{360^{2}} \times \pi \mathrm{r}^{2}$ sq. units
$\begin{aligned}
&\mathrm{A}=\frac{120}{360^{\circ}} \times 3.14 \times 6.3 \times 6.3 \mathrm{~cm}^{2} \\
&\mathrm{~A}=3.14 \times 6.3 \times 2.1 \mathrm{~cm}^{2} \\
&\mathrm{~A}=41.54 \mathrm{~cm}^{2} \\
&\text { Perimeter of the sector } \mathrm{P}=1+2 \mathrm{r} \mathrm{cm} \\
&\mathrm{P}=13.19+2(6.3) \mathrm{cm} \\
&=13.19+1.2 .6 \mathrm{~cm} \\
&\mathrm{P}=25.79 \mathrm{~cm}
\end{aligned}$

Question $5 .$
From the measures given below, find the area of the sectors.
(i) Length of the arc $=48 \mathrm{~m}, \mathrm{r}=10 \mathrm{~m}$
Answer:
Area of the sector $\mathrm{A}=\frac{l r}{2}$ sq. units
$\begin{aligned}
&\mathrm{l}=48 \mathrm{~m} \\
&\mathrm{r}=10 \mathrm{~m} \\
&=\frac{48 \times 10}{2} \mathrm{~m}^{2} \\
&=24 \times 10 \mathrm{~m}^{2} \\
&=240 \mathrm{~m}^{2}
\end{aligned}$
Area of the sector $=240 \mathrm{~m}^{2}$
(ii) length of the $\operatorname{arc}=50 \mathrm{~cm}, \mathrm{r}=13.5 \mathrm{~cm}$
Answer:
Length of the arc $1=12.5 \mathrm{~cm}$
Radius $\mathrm{r}=6 \mathrm{~cm}$
Area of the sector $\mathrm{A}=\frac{l r}{2}$ sq. units
$\begin{aligned}
&A=\frac{12.5 \times 6}{2} \\
&A=12.5 \times 3 \mathrm{~cm}^{2} \\
&A=37.5 \mathrm{~cm}^{2}
\end{aligned}$
Area of the sector $\mathrm{A}=37.5 \mathrm{~cm}^{2}$

Question 6 .
Find the central angle of each of the sectors whose measures are given below. $\left(\pi=\frac{22}{7}\right)$
(i) area $=462 \mathrm{~cm}^{2}, \mathrm{r}=21 \mathrm{~cm}$
Answer:
area $=462 \mathrm{~cm}^{2}, \mathrm{r}=21 \mathrm{~cm}$
Radius of the Sector $-21 \mathrm{~cm}$
Area of the sector $=462 \mathrm{~cm}^{2}$
$\frac{l r}{2}=462$ $\frac{l \times 21}{2}=462$ $1=\frac{462 \times 2}{21}$ $1=22 \times 2$
Length of the arc $1=44 \mathrm{~cm}$
$\begin{aligned}
\frac{\theta^{\circ}}{360^{\circ}} \times 2 \pi r &=44 \mathrm{~cm} \\
\frac{\theta^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21 &=44 \mathrm{~cm} \\
\theta^{\circ} &=\frac{44 \times 360 \times 7}{2 \times 22 \times 21}
\end{aligned}$
$\theta^{\circ}=120^{\circ}$
$\therefore$ Central angle of the sector $=120^{\circ}$

(ii) length of the arc $=44 \mathrm{~m}, \mathrm{r}=35 \mathrm{~m}$
Answer:
Radius of the sector $=8.4 \mathrm{~cm}$
Area of the sector $=18.48 \mathrm{~cm}^{2}$
$\frac{l r}{2}=18.48$
$\frac{1 \times 8.4}{2}=18.48$
$1=\frac{18.48 \times 2}{84}$

Length of the arc $1=4.4 \mathrm{~cm}$
$\begin{aligned}
\frac{\theta^{\circ}}{360^{\circ}} \times 2 \pi r &=4.4 \mathrm{~cm} \\
\frac{\theta^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 8.4 &=4.4 \mathrm{~cm} \\
\theta^{\circ} &=\frac{4.4 \times 360 \times 7}{2 \times 22 \times 8.4}
\end{aligned}$

$\theta^{\circ}=30^{\circ}$
Central angle $=30^{\circ}$

Question $7 .$
A circle of radius $120 \mathrm{~m}$ is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
Answer:
Radius of the circle $\mathrm{r}=120 \mathrm{~m}$
Number of equal sectors $=8$
$\therefore$ Central angle of each sector $=\frac{360^{\circ}}{n}$
$\begin{aligned}
&\theta^{\circ}=\frac{360^{\circ}}{8} \\
&\theta^{\circ}=45^{\circ}
\end{aligned}$
Length of the arc $1=\frac{\theta}{360^{\circ}} \times 2 \pi$ r units $=\frac{45^{\circ}}{360^{\circ}} \times 2 \pi \times 120 \mathrm{~m}$
Length of the arc $=30 \times \pi \mathrm{m}$
Another method:
$1=\frac{1}{n} \times 2 \pi \mathrm{r}=\frac{1}{8} \times 2 \times \pi \times 120=30 \pi \mathrm{m}$
Length of the arc $=30 \pi \mathrm{m}$

Question $8 .$
A circle of radius $70 \mathrm{~cm}$ is divided into 5 equal sectors. Find the area of each of the sectors.
Answer:
Radius of the sector $r=70 \mathrm{~cm}$
Number of equal sectors $=5$
$\therefore$ Central angle of each sector $=\frac{360^{\circ}}{n}$
$\begin{aligned}
&\theta^{\circ}=360^{\circ} \\
&\theta^{\circ}=72^{\circ}
\end{aligned}$
Area of the sector $=\frac{\theta}{360^{2}} \times \pi \mathrm{r}^{2}$ sq.units $=\frac{72^{\circ}}{360^{\circ}} \times \pi \times 70 \times 70 \mathrm{~cm}^{2}$

$\begin{aligned}
&=14 \times 70 \times \pi \mathrm{cm}^{2} \\
&=980 \pi \mathrm{cm}^{2}
\end{aligned}$
Note: We can solve this problem using $\mathrm{A}=\frac{1}{n} \pi \mathrm{r}^{2}$ sq. units also.

Question $9 .$
Dhamu fixes a square tile of $30 \mathrm{~cm}$ on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector. ( $\pi=3.14)$.

Answer:
Side of the square $=30 \mathrm{~cm}$
$\therefore$ Radius of the sector design $=30 \mathrm{~cm}$
Given the design of a circular quadrant.
Area of the quadrant $=\frac{1}{4} \pi r^{2}$ sq.units
$\begin{aligned}
&=\frac{1}{4} \times 3.14 \times 30 \times 30 \mathrm{~cm}^{2} \\
&=3.14 \times 15 \times 15 \mathrm{~cm}^{2}
\end{aligned}$
$\therefore$ Area of the sector design $=706.5 \mathrm{~cm}^{2}$ (approximately)

Question $10 .$
A circle is formed with 8 equal granite stones as shown in the figure each of radius $56 \mathrm{~cm}$ and whose central angle is $45^{\circ}$. Find the area of each of the granite stones. $\left(\pi=\frac{22}{7}\right)$

Answer:
Number of equal sectors ' $n$ ' $=8$
Radius of the sector ' $r$ ' $=56 \mathrm{~cm}$
Area of each sector $=\frac{1}{n} \pi \mathrm{r}^{2}$ sq. units
$=\frac{1}{8} \times \frac{22}{7} \times 56 \times 56 \mathrm{~cm}^{2}=1232 \mathrm{~cm}^{2}$
Area of each sector $=1232 \mathrm{~cm}^{2}$ (approximately)