Mathematics solution NCERT
Class 9 - Chapter 6: Measuring Space: Perimeter and Area
Question 1.
Find the area of a sector of a circle with radius $7$ cm
if the angle of the sector is $60^\circ$.
Solution:
Given,
$$ r=7\text{ cm} $$ $$ \theta=60^\circ $$We know that the area of a sector is
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2 $$Using
$$ \pi=\frac{22}{7}, $$Substitute the given values.
$$ \begin{aligned} \text{Area} &=\frac{60}{360}\times\frac{22}{7}\times7^2\\[4pt] &=\frac16\times\frac{22}{7}\times49\\[4pt] &=\frac16\times22\times7\\[4pt] &=\frac{154}{6}\\[4pt] &=\frac{77}{3}\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area of the sector}=\frac{77}{3}\text{ cm}^2\approx25.67\text{ cm}^2} $$Question 2.
Find the area of a quadrant of a circle
whose circumference is $44$ cm.
Solution:
Given,
$$ \text{Circumference}=44\text{ cm} $$We know that
$$ C=2\pi r $$Using
$$ \pi=\frac{22}{7}, $$Substitute the given values.
$$ \begin{aligned} 44 &=2\times\frac{22}{7}\times r\\[4pt] 44 &=\frac{44r}{7} \end{aligned} $$Multiply both sides by $7$.
$$ \begin{aligned} 44\times7 &=44r \end{aligned} $$Divide both sides by $44$.
$$ \begin{aligned} r &=7\text{ cm} \end{aligned} $$The area of the circle is
$$ \begin{aligned} \pi r^2 &=\frac{22}{7}\times7^2\\[4pt] &=\frac{22}{7}\times49\\[4pt] &=154\text{ cm}^2 \end{aligned} $$A quadrant is one-fourth of a circle.
Therefore,
$$ \begin{aligned} \text{Area of quadrant} &=\frac14\times154\\[4pt] &=\frac{77}{2}\text{ cm}^2\\[4pt] &=38.5\text{ cm}^2 \end{aligned} $$Answer:
$$ {\text{Area of the quadrant}=38.5\text{ cm}^2} $$Question 3.
The length of the minute hand of a clock is $7$ cm.
Find the area swept by the minute hand in $10$ minutes.
Solution:
Given,
$$ \text{Radius }(r)=7\text{ cm} $$The minute hand completes one full revolution, i.e.,
$$ 360^\circ $$in
$$ 60\text{ minutes}. $$Step 1: Find the angle swept in $10$ minutes.
$$ \begin{aligned} \theta &=\frac{10}{60}\times360^\circ\\[4pt] &=60^\circ \end{aligned} $$Step 2: Find the area of the sector.
We know that
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2 $$Using
$$ \pi=\frac{22}{7}, $$Substitute the values.
$$ \begin{aligned} \text{Area} &=\frac{60}{360}\times\frac{22}{7}\times7^2\\[4pt] &=\frac16\times\frac{22}{7}\times49\\[4pt] &=\frac16\times22\times7\\[4pt] &=\frac{154}{6}\\[4pt] &=\frac{77}{3}\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area swept by the minute hand}=\frac{77}{3}\text{ cm}^2} $$or approximately,
$$ {\text{Area swept}\approx25.67\text{ cm}^2} $$Question 3.
The length of the minute hand of a clock is $7$ cm.
Find the area swept by the minute hand in $10$ minutes.
Solution:
Given,
$$ \text{Radius }(r)=7\text{ cm} $$The minute hand completes one full revolution, i.e.,
$$ 360^\circ $$in
$$ 60\text{ minutes}. $$Step 1: Find the angle swept in $10$ minutes.
$$ \begin{aligned} \theta &=\frac{10}{60}\times360^\circ\\[4pt] &=60^\circ \end{aligned} $$Step 2: Find the area of the sector.
We know that
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2 $$Using
$$ \pi=\frac{22}{7}, $$Substitute the values.
$$ \begin{aligned} \text{Area} &=\frac{60}{360}\times\frac{22}{7}\times7^2\\[4pt] &=\frac16\times\frac{22}{7}\times49\\[4pt] &=\frac16\times22\times7\\[4pt] &=\frac{154}{6}\\[4pt] &=\frac{77}{3}\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area swept by the minute hand}=\frac{77}{3}\text{ cm}^2} $$or approximately,
$$ {\text{Area swept}\approx25.67\text{ cm}^2} $$Question 5.
A chord of a circle of radius $15$ cm subtends an angle of $60^\circ$
at the centre of the circle. Find the areas of the corresponding
minor and major segments of the circle.
Use $\pi \approx 3.14$ and $\sqrt{3}\approx1.73$.
Solution:
Given,
$$ r=15\text{ cm} $$ $$ \theta=60^\circ $$Step 1: Find the area of the circle.
$$ \begin{aligned} \text{Area of Circle} &=\pi r^2\\[4pt] &=3.14\times15^2\\[4pt] &=3.14\times225\\[4pt] &=706.5\text{ cm}^2 \end{aligned} $$Step 2: Find the area of the minor sector.
We know that
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2 $$Substitute the values.
$$ \begin{aligned} \text{Area of Minor Sector} &=\frac{60}{360}\times706.5\\[4pt] &=\frac16\times706.5\\[4pt] &=117.75\text{ cm}^2 \end{aligned} $$Step 3: Find the area of $\triangle AOB$.
The radii $OA$ and $OB$ are equal.
$$ OA=OB=15\text{ cm} $$Also,
$$ \angle AOB=60^\circ. $$Hence, $\triangle AOB$ is an equilateral triangle.
Area of an equilateral triangle is
$$ \text{Area} = \frac{\sqrt3}{4}a^2 $$Substitute
$$ a=15\text{ cm}. $$ $$ \begin{aligned} \text{Area of }\triangle AOB &=\frac{1.73}{4}\times15^2\\[4pt] &=\frac{1.73}{4}\times225\\[4pt] &=97.3125\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area of }\triangle AOB\approx97.31\text{ cm}^2} $$Step 4: Find the area of the minor segment.
The area of the minor segment is
$$ \text{Area of Minor Segment} = \text{Area of Minor Sector} - \text{Area of }\triangle AOB $$ $$ \begin{aligned} \text{Area of Minor Segment} &=117.75-97.3125\\[4pt] &=20.4375\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area of Minor Segment}\approx20.44\text{ cm}^2} $$Step 5: Find the area of the major segment.
The area of the major segment is
$$ \text{Area of Major Segment} = \text{Area of Circle} - \text{Area of Minor Segment} $$ $$ \begin{aligned} \text{Area of Major Segment} &=706.5-20.4375\\[4pt] &=686.0625\text{ cm}^2 \end{aligned} $$Therefore,
$$ {\text{Area of Major Segment}\approx686.06\text{ cm}^2} $$Answers:
$$ {\text{Area of Minor Segment}\approx20.44\text{ cm}^2} $$ $$ {\text{Area of Major Segment}\approx686.06\text{ cm}^2} $$Question 6.
A sector of a circle has radius $15$ cm and area $150\text{ cm}^2$.
Find the length of the corresponding arc.
Solution:
Given,
$$ r=15\text{ cm} $$ $$ \text{Area of Sector}=150\text{ cm}^2 $$We know that
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2 \qquad\cdots(1) $$Also, the length of the corresponding arc is
$$ \text{Arc Length} = \frac{\theta}{360^\circ}\times2\pi r \qquad\cdots(2) $$Step 1: Find the angle of the sector.
Substitute the given values in equation (1).
$$ \begin{aligned} 150 &=\frac{\theta}{360^\circ}\times\pi\times15^2\\[4pt] &=\frac{\theta}{360^\circ}\times225\pi \end{aligned} $$Therefore,
$$ \begin{aligned} \frac{\theta}{360^\circ} &=\frac{150}{225\pi}\\[4pt] &=\frac{2}{3\pi} \end{aligned} $$Step 2: Find the arc length.
Using equation (2),
$$ \begin{aligned} \text{Arc Length} &=\frac{2}{3\pi}\times2\pi\times15\\[4pt] &=\frac{60\pi}{3\pi}\\[4pt] &=20\text{ cm} \end{aligned} $$Answer:
$$ {\text{Length of the arc}=20\text{ cm}} $$Question 7.
A chord of a circle of radius $r$ subtends an angle of $60^\circ$ at the centre of the circle.
Show that the area of the corresponding minor segment of the circle is equal to
Solution:
Let $O$ be the centre of the circle and $AB$ be the chord.
The chord $AB$ subtends an angle of
at the centre.
The minor segment is obtained by subtracting the area of triangle $AOB$ from the area of the minor sector $AOB$.
Therefore,
$$ { \text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of }\triangle AOB } $$Step 1: Find the area of the sector.
The angle of the sector is
$$ \theta=60^\circ. $$We know that
$$ \text{Area of Sector} = \frac{\theta}{360^\circ}\times\pi r^2. $$Substitute the value of $\theta$.
$$ \begin{aligned} \text{Area of Sector} &=\frac{60}{360}\times\pi r^2\\[4pt] &=\frac16\pi r^2. \end{aligned} $$Step 2: Find the area of $\triangle AOB$.
Since
$$ OA=OB=r, $$and
$$ \angle AOB=60^\circ, $$triangle $AOB$ is an equilateral triangle.
Therefore, each side of the triangle is
$$ r. $$We know that the area of an equilateral triangle is
$$ \frac{\sqrt3}{4}a^2. $$Here,
$$ a=r. $$Hence,
$$ \begin{aligned} \text{Area of }\triangle AOB &=\frac{\sqrt3}{4}r^2. \end{aligned} $$Step 3: Find the area of the minor segment.
Subtract the area of the triangle from the area of the sector.
$$ \begin{aligned} \text{Area of Minor Segment} &=\frac16\pi r^2-\frac{\sqrt3}{4}r^2. \end{aligned} $$Take $r^2$ as the common factor.
$$ \begin{aligned} \text{Area of Minor Segment} &=r^2\left(\frac{\pi}{6}-\frac{\sqrt3}{4}\right). \end{aligned} $$Now take $\pi$ common inside the bracket.
$$ \begin{aligned} \text{Area of Minor Segment} &=\pi r^2 \left( \frac16-\frac{\sqrt3}{4\pi} \right). \end{aligned} $$Hence,
$$ { \text{Area of the corresponding minor segment} = \pi r^2 \left( \frac16-\frac{\sqrt3}{4\pi} \right) } $$Hence Proved.
Question 8.
An equilateral triangle is inscribed in a circle of radius $r$.
Show that the ratio of the area of the triangle to the area of the circle is equal to
Solution:
Let the side of the equilateral triangle be
$$ a. $$The radius of the circumcircle (circumradius) of an equilateral triangle is
$$ R=\frac{a}{\sqrt3}. $$Since the triangle is inscribed in a circle of radius $r$,
$$ R=r. $$Therefore,
$$ \begin{aligned} r &=\frac{a}{\sqrt3} \end{aligned} $$or,
$$ \begin{aligned} a &=\sqrt3\,r. \end{aligned} $$Step 1: Find the area of the equilateral triangle.
We know that
$$ \text{Area of an Equilateral Triangle} = \frac{\sqrt3}{4}a^2. $$Substitute
$$ a=\sqrt3\,r. $$ $$ \begin{aligned} \text{Area of Triangle} &=\frac{\sqrt3}{4}(\sqrt3\,r)^2\\[4pt] &=\frac{\sqrt3}{4}(3r^2)\\[4pt] &=\frac{3\sqrt3}{4}r^2. \end{aligned} $$Step 2: Find the area of the circle.
$$ \begin{aligned} \text{Area of Circle} &=\pi r^2. \end{aligned} $$Step 3: Find the required ratio.
$$ \begin{aligned} \frac{\text{Area of Triangle}} {\text{Area of Circle}} &= \frac{\frac{3\sqrt3}{4}r^2} {\pi r^2}. \end{aligned} $$Cancel the common factor $r^2$.
$$ \begin{aligned} \frac{\text{Area of Triangle}} {\text{Area of Circle}} &= \frac{3\sqrt3}{4\pi}. \end{aligned} $$Using
$$ \pi\approx3.14 \qquad\text{and}\qquad \sqrt3\approx1.732, $$ $$ \begin{aligned} \frac{3\sqrt3}{4\pi} &\approx \frac{3\times1.732}{4\times3.14}\\[4pt] &= \frac{5.196}{12.56}\\[4pt] &\approx0.413. \end{aligned} $$Hence,
$$ { \frac{\text{Area of Triangle}} {\text{Area of Circle}} = \frac{3\sqrt3}{4\pi} \approx0.413 } $$Hence Proved.
Question 9.
A square is inscribed in a circle of radius $r$.
Show that the ratio of the area of the square to the area of the circle is equal to
Solution:
Let the side of the square be
$$ a. $$The diagonal of the square is equal to the diameter of the circle.
Therefore,
$$ \begin{aligned} a\sqrt2 &=2r. \end{aligned} $$Divide both sides by $\sqrt2$.
$$ \begin{aligned} a &=\frac{2r}{\sqrt2}\\[4pt] &=\sqrt2\,r. \end{aligned} $$Step 1: Find the area of the square.
We know that
$$ \text{Area of Square}=a^2. $$Substitute
$$ a=\sqrt2\,r. $$ $$ \begin{aligned} \text{Area of Square} &=(\sqrt2\,r)^2\\[4pt] &=2r^2. \end{aligned} $$Step 2: Find the area of the circle.
$$ \begin{aligned} \text{Area of Circle} &=\pi r^2. \end{aligned} $$Step 3: Find the required ratio.
$$ \begin{aligned} \frac{\text{Area of Square}} {\text{Area of Circle}} &= \frac{2r^2}{\pi r^2}. \end{aligned} $$Cancel the common factor $r^2$.
$$ \begin{aligned} \frac{\text{Area of Square}} {\text{Area of Circle}} &= \frac{2}{\pi}. \end{aligned} $$Using
$$ \pi\approx3.14, $$ $$ \begin{aligned} \frac{2}{\pi} &\approx\frac{2}{3.14}\\[4pt] &\approx0.637. \end{aligned} $$Hence,
$$ { \frac{\text{Area of Square}} {\text{Area of Circle}} = \frac{2}{\pi} \approx0.637 } $$Hence Proved.
Question 10.
A regular hexagon is inscribed in a circle of radius $r$.
Show that the ratio of the area of the hexagon to the area of the circle is equal to
Can you see why the answer is exactly twice the answer to Question 8?
Solution:
Let the circle have radius
$$ r. $$A regular hexagon inscribed in a circle is divided into
$$ 6 $$equal equilateral triangles.
Each side of the regular hexagon is equal to the radius of the circle.
Therefore,
$$ {\text{Side of the hexagon}=r.} $$Step 1: Find the area of one equilateral triangle.
We know that
$$ \text{Area of an Equilateral Triangle} = \frac{\sqrt3}{4}a^2. $$Here,
$$ a=r. $$Therefore,
$$ \begin{aligned} \text{Area of one triangle} &=\frac{\sqrt3}{4}r^2. \end{aligned} $$Step 2: Find the area of the hexagon.
The hexagon consists of six congruent equilateral triangles.
Hence,
$$ \begin{aligned} \text{Area of Hexagon} &=6\times\frac{\sqrt3}{4}r^2\\[4pt] &=\frac{3\sqrt3}{2}r^2. \end{aligned} $$Step 3: Find the area of the circle.
$$ \begin{aligned} \text{Area of Circle} &=\pi r^2. \end{aligned} $$Step 4: Find the required ratio.
$$ \begin{aligned} \frac{\text{Area of Hexagon}} {\text{Area of Circle}} &= \frac{\frac{3\sqrt3}{2}r^2} {\pi r^2}. \end{aligned} $$Cancel the common factor $r^2$.
$$ \begin{aligned} \frac{\text{Area of Hexagon}} {\text{Area of Circle}} &= \frac{3\sqrt3}{2\pi}. \end{aligned} $$Using
$$ \pi\approx3.14 \qquad\text{and}\qquad \sqrt3\approx1.732, $$ $$ \begin{aligned} \frac{3\sqrt3}{2\pi} &\approx \frac{3\times1.732}{2\times3.14}\\[4pt] &= \frac{5.196}{6.28}\\[4pt] &\approx0.827. \end{aligned} $$Hence,
$$ { \frac{\text{Area of Hexagon}} {\text{Area of Circle}} = \frac{3\sqrt3}{2\pi} \approx0.827 } $$Why is this answer exactly twice the answer to Question 8?
In Question 8, the inscribed equilateral triangle has area
$$ \frac{3\sqrt3}{4}r^2. $$The regular hexagon is made up of
$$ 6 $$equilateral triangles, whereas the inscribed equilateral triangle is equivalent to only
$$ 3 $$of those equal triangular parts.
Hence,
$$ \text{Area of Hexagon} = 2\times \text{Area of the Inscribed Equilateral Triangle}. $$Therefore,
$$ \frac{3\sqrt3}{2\pi} = 2\times \frac{3\sqrt3}{4\pi}. $$Thus, the required ratio for the hexagon is exactly twice the ratio obtained in Question 8.
Hence Proved.