Do you like to know the solutions of Chapter 1: Angle and its Measurement – Exercise 1.2 of your Maths Part 1 book? In this article, I will provide detailed solutions for this exercise.
If you have any doubts or queries while going through any solution, feel free to comment below. I will be more than happy to help you out.
Angle and its Measurement Exercise 1.2 Solutions
Q.1 Find the length of an arc of a circle which subtends an angle of $108^\circ$ at the centre, if the radius of the circle is 15 cm.
Solution:
Given radius $r$ = 15 cm,
\begin{align}
\theta &= 108^\circ \\
&= \left( 108 \times \frac{\pi}{180}\right)^c \\
&=\frac{3\pi^c}{5}
\end{align}
\begin{align*}
\text{Length of arc } S &= r\theta \\
&= 15 \times \frac{3\pi^c}{5} \\
S&=9\pi \text{ cm}
\end{align*}
Q.2 The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length, equal to length of radius.
Solution:
Given radius $r$ = 9 cm and AB = $r$.
$\because$ OA = OB = AB = $r$
$\triangle OAB$ is an equilateral triangle.
$\therefore \theta = 60^\circ = \frac{\pi}{3}$
Hence, length of arc $S$ = $r\theta$ = $9 \times \frac{\pi}{3}$ = 3 $\pi$ cm.
Q.3 Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm.
Solution:
Given length of arc $S$ = 15 cm, radius $r$ = 25 cm.|
$\because$ $S$ = $r \theta$
\begin{align*}
\therefore \theta&=\frac{S}{r}=\frac{15}{25}\\
&=\left(\frac{3}{5}\right)^c\\
&=\frac{3}{5}\times\frac{180}{\pi} [\text{Note: conversion to degree as per question}]\\
&=\left(\frac{108}{\pi}\right)^\circ
\end{align*}
Ans: $\theta$ = $\left(\frac{108}{\pi}\right)^\circ$.
Q.4 A pendulum of length 14 cm oscillates through an angle of 18$^\circ$. Find the length of its path.
Solution:
Here, pendulum movement forms arc of a circle whose radius $r$ = 14 cm
\begin{align*}
\theta&=18^\circ=\left(18\times\frac{\pi}{180}\right)^c,\\
&=\left(\frac{\pi}{10}\right)^c.
\end{align*}
\begin{align*}
\therefore \text{Length of pendulum path } S&=r\theta,\\
&=14\times\frac{\pi}{10},\\
&=\frac{7\pi}{5} \text{ cm}
\end{align*}
Q.5 Two arcs of the same lengths subtend angles of $60^\circ$ and $75^\circ$ at the centres of two circles. What is the ratio of radii of two circles?
Solution:
Let $r_1$ and $r_2$ be the radii of the two circles.
$\theta_1$ = 60$^\circ$ = $60 \times \frac{\pi^c}{180}$
$\theta_2$ = 75$^\circ$ = $75 \times \frac{\pi^c}{180}$
As arc lengths are same, we have
\begin{align*}
r_1 \theta_1&=r_2\theta_2\\
\frac{r_1}{r_2}&=\frac{\theta_2}{\theta_1}\\
&=\frac{75 \times \frac{\pi^c}{180}}{60\times \frac{\pi^c}{180}}\\
&=\frac{5}{4}
\end{align*}
Ans: $r_1:r_2$ = 5:4.
Q.6 The area of a circle is 25$\pi$ sq.cm. Find the length of its arc subtending an angle of 144$^\circ$ at the centre. Also find the area of the corresponding sector.
Solution:
Given area of circle $\pi r^2$ = 25$\pi$ sq. cm
\begin{align*}
\therefore r^2&=25\\
r&=5\text{ cm}
\end{align*}
\begin{align*}
\theta&=144^\circ=\left(144\times\frac{\pi}{180}\right)^c\\
&=\frac{4\pi^c}{5}
\end{align*}
\begin{align*}
\text{Length of arc } S&=r\theta\\
&=5\times\frac{4\pi}{5}\\
&=4\pi\text{ cm}
\end{align*}
\begin{align*}
\text{Area of sector }&=\frac{1}{2}r^2\theta\\
&=\frac{1}{2}rS\\
&=\frac{1}{2}\times 5\times 4\pi\\
&=10\pi\text{ sq. cm.}
\end{align*}
Q.7 OAB is a sector of the circle having centre at O and radius 12 cm. If m∠AOB = 45$^\circ$, find the difference between the area of sector OAB and triangle AOB.
Solution:
Given radius $r$ =12 cm
Let m$\angle$AOB=$\theta$ = 45$^\circ$=$\frac{\pi^c}{4}$.
Draw BM $\perp$ OA
Let $h$ = height of $\triangle$OAB
Then, $h$ = OB$\times\sin\theta$ = $r\sin\theta$.
\begin{align*}
\text{Area of }\triangle\text{AOB}&=\frac{1}{2}\times\text{OA}\times h\\
&=\frac{1}{2}rh\\
&=\frac{1}{2}r^2\sin\theta\\
\text{Area of sector OAB}&=\frac{1}{2}r^2\theta
\end{align*}
\begin{align*}
\text{Difference of area }&=\text{Area of sector OAB}-\text{Area of }\triangle{AOB}\\
&=\frac{1}{2}r^2\theta-\frac{1}{2}r^2\sin\theta\\
&=\frac{1}{2}r^2(\theta-\sin\theta)\\
&=\frac{1}{2}\times 144 \left(\frac{\pi}{4}-\sin\frac{\pi}{4}\right)\\
&=72\left(\frac{\pi}{4}-\frac{1}{\sqrt2}\right)\\
&=18(\pi-2\sqrt2)\text{ sq. cm}
\end{align*}
Q.8 OPQ is the sector of a circle having centre at O and radius 15 cm. If m∠POQ = 30$^\circ$, find the area enclosed by arc PQ and chord PQ.
Solution:
Given radius $r$ =15 cm
Let m$\angle$POQ=$\theta$ = 30$^\circ$=$\frac{\pi^c}{6}$.
Draw QM $\perp$ OP
Let $h$ = height of $\triangle$POQ
Then, $h$ = OQ$\times\sin\theta$ = $r\sin\theta$.
\begin{align*}
\text{Area of }\triangle\text{POQ}&=\frac{1}{2}\times\text{OP}\times h\\
&=\frac{1}{2}rh\\
&=\frac{1}{2}r^2\sin\theta\\
\text{Area of sector OPQ}&=\frac{1}{2}r^2\theta
\end{align*}
\begin{align*}
\text{Area enclosed by arc PQ and chord PQ}&=\text{Area of sector OPQ}-\text{Area of }\triangle{POQ}\\
&=\frac{1}{2}r^2\theta-\frac{1}{2}r^2\sin\theta\\
&=\frac{1}{2}r^2(\theta-\sin\theta)\\
&=\frac{1}{2}\times 255\left(\frac{\pi}{6}-\sin\frac{\pi}{6}\right)\\
&=\frac{225}{2}\left(\frac{\pi}{6}-\frac{1}{2}\right)\\
&=\frac{75}{4}(\pi-3)\text{ sq. cm}
\end{align*}
Q.9 The perimeter of a sector of the circle of area 25π sq.cm is 20 cm. Find the area of sector.
Solution:
\begin{align*}
\text{Given, Area of circle }\pi r^2&=25\pi\text{ sq. cm}\\
r^2&=25\\
\therefore r&=5\text{ cm}
\end{align*}
\begin{align*}
\text{Given, perimeter of sector }&=20\text{ cm}\\
\text{OA}+\text{length(arc AB)}+\text{BO}&=20\\
r+S+r&=20\\
S&=10\text{ cm}
\end{align*}
\begin{align*}
\text{Area of sector}&=\frac{1}{2}rS\\
&=\frac{1}{2}\times 5 \times10\\
&=25\text{ sq. cm}
\end{align*}
Q.10 The perimeter of the sector of the circle of area 64π sq.cm is 56 cm. Find the area of the sector.
Solution:
\begin{align*}
\text{Given, Area of circle }\pi r^2&=64\pi\text{ sq. cm}\\
r^2&=64\\
\therefore r&=8\text{ cm}
\end{align*}
\begin{align*}
\text{Given, perimeter of sector }&=56\text{ cm}\\
r+S+r&=56\\
S&=56-2r\\
S&=40\text{ cm}
\end{align*}
\begin{align*}
\text{Area of sector}&=\frac{1}{2}rS\\
&=\frac{1}{2}\times 8 \times40\\
&=160\text{ sq. cm}
\end{align*}
Summary
We have seen solutions of all questions of exercise 1.2. If you have any doubts related to these solutions, then you can drop a comment below!
Happy learning!
