RS Aggarwal Solutions Class 10 Areas Of Circle Sector And Segment

This chapter of RS Aggarwal deals with teaching students, how to calculate the area of a circle, it’s segment and of its sector. Master all the concepts of this chapter by solving the exercise questions provided in the RS Aggarwal for class 10. To help students understand the concepts and to solve all the questions we have provided the RS Aggarwal class 10 solutions chapter 18 areas of circle, sector, and segment.

Question 1:The difference between the circumference and radius of a circle is 37 cm. Using \(\Pi =\frac{22}{7}\), find the circumference of the circle.

Solution:

Let the radius of the circle be r and circumference be C.

Now,

C – r = 37

\(\Rightarrow\) \(2\Pi r-r=37\)

\(\Rightarrow\) \(r\left ( 2\times \frac{22}{7}-1 \right )=37\)

\(\Rightarrow\) \(r=\frac{37\times 7}{37}=7cm\)

Now, \(C=2\Pi r=2\times \frac{22}{7}\times 7=44cm\)

Hence, the circumference of the circle is 44 cm.

Question 2:The circumference of a circle is 22 cm. Find the area of its quadrant.

Solution:

Let the radius of the circle be r.

Now,

Circumference = 22

\(\Rightarrow\) \(2\Pi r=22\)

\(\Rightarrow\) r = \(\frac{22\times 7}{44}=\frac{7}{2}cm\)

Now, Area of quadrant = \(\frac{1}{4}\Pi r^{2}=\frac{1}{4}\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^{2}=\frac{77}{8}cm^{2}\)

Hence, the area of the quadrant of the circle is \(\frac{77}{8}\) cm2.

Question 3:What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm?

Solution:

Let the diameter of the required circle be d.

Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

\(\Rightarrow\) \(\Pi \left ( \frac{d}{2} \right )^{2}=\Pi \left ( \frac{10}{2} \right )^{2}+\Pi \left ( \frac{24}{2} \right )^{2}\)

\(\Rightarrow\) \(\left ( \frac{d}{2} \right )^{2}=25+144\)

\(\Rightarrow\) \(\left ( \frac{d}{2} \right )^{2}=13^{2}\)

\(\Rightarrow\) \(\frac{d}{2}=13\)

\(\Rightarrow\) d = 26 cm

Hence, the diameter of the circle is 26 cm.

Question 4:If the area of a circle is numerically equal to twice its circumference, then what is the diameter of the circle?

Solution:

Let the diameter of the required circle be d.

Now, Area of circle = 2 x Circumference of the circle

\(\Rightarrow\) \(\Pi \left ( \frac{d}{2} \right )^{2}=2\left ( 2\Pi \times \frac{d}{2} \right )\)

\(\Rightarrow\) \(\left ( \frac{d}{2} \right )^{2}=2d\)

\(\Rightarrow\) d2 = 8d

\(\Rightarrow\) d2 – 8d = 0

\(\Rightarrow\) d(d – 8) = 0

\(\Rightarrow\) d = 8 cm         Since, d \(\neq\) 0]

Hence, the diameter of the circle is 8 cm.

Question 5:What is the perimeter of a square which circumscribes a circle of radius a cm?

Solution:

We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle.

Therefore, Side of square = 2a

Now, Perimeter of the square = 4 x Side of square = 4 x 2a = 8a cm

Hence, the perimeter of the square is 8a cm.

Question 6:Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60o at the centre.

Solution:

We have r = \(\frac{diameter}{2}=\frac{42}{2}\) = 21 cm and \(\Theta\) = 60o

Length of arc = \(\frac{\Theta }{360^{\circ}}\times 2\Pi r=\frac{60^{\circ}}{360^{\circ}}\times 2\times \frac{22}{7}\times 21=22cm\)

Hence, the length of the arc of the circle is 22 cm.

Question 7:Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm.

Solution:

Let the diameter of the required circle be d.

Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm

\(\Rightarrow\) \(\Pi \left ( \frac{d}{2} \right )^{2}=\Pi (4)^{2}+\Pi (3)^{2}\)

\(\Rightarrow\) \(\left ( \frac{d}{2} \right )^{2}=16+9\)

\(\Rightarrow\) \(\left ( \frac{d}{2} \right )^{2}=25=(5)^{2}\)

\(\Rightarrow\) \(\frac{d}{2}=5\)

\(\Rightarrow\) d = 10 cm

Hence, the diameter of the circle is 10 cm.

Question 8:Find the area of a circle whose circumference is \(8\Pi\).

Solution:

Let the radius of the circle be r.

Now,

Circumference = \(8\Pi\)

\(\Rightarrow\) \(2\Pi r=8\Pi\)

\(\Rightarrow\) r = 4 cm

Now, Area of circle = \(\Pi r^{2}\) = \(\Pi\) x 42 = 16\(\Pi\) cm2

Hence, the area of the circle is 16\(\Pi\) cm2.

Question 9:Find the perimeter of a semicircular protractor whose diameter is 14 cm.

Solution:

Perimeter of a semi-circular protractor = Circumference of semi-circular protractor + Diameter of semi-circular protractor

= \(\frac{1}{2}(2\Pi r)+d\) = \(\Pi r+d\) = \(\Pi \frac{d}{2}+d\)

= \(\frac{22}{7}\times 7+14\) = 22 + 14 = 36 cm

Hence, the perimeter of a semi-circular protractor is 36 cm.

Question 10:Find the radius of a circle whose perimeter and area are numerically equal.

Solution:

Let the radius of the required circle be r.

Now, Area of circle = Perimeter of the circle

\(\Rightarrow\) \(\Pi r^{2}=2\Pi \times r\)

\(\Rightarrow\) \(r^{2}=2r\)

\(\Rightarrow\) \(r^{2}-2r=0\)

\(\Rightarrow\) r(r – 2) = 0

\(\Rightarrow\) r – 2 = 0          Since, r \(\neq\) 0]

\(\Rightarrow\) r = 2 units

Hence, the radius of the circle is 2 units.

Question 11:The radii of two circles are 19 cm and 9 cm. Find the radius of the circle which has the circumference equal to the sum of the circumferences of the two circles.

Solution:

Let the radius of the required circle be r.

Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm

\(\Rightarrow\) \(2\Pi r=2\Pi \times 19+2\Pi \times 9\)

\(\Rightarrow\) r = 19 + 9

\(\Rightarrow\) r = 28 cm

Hence, the radius of the required circle is 28 cm.

Question 12:The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having the area equal to the sum of the areas of the two circles.

Solution:

Let the radius of the required circle be r.

Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm

\(\Rightarrow\) \(\Pi r^{2}=\Pi (8)^{2}+\Pi (6)^{2}\)

\(\Rightarrow\) r2 = 64 + 36

\(\Rightarrow\) r2 = 102

\(\Rightarrow\) r = 10 cm

Hence, the radius of the circle is 10 cm.

Question 13:Find the area of the sector of a circle having radius 6 cm and of angle 30o. [Take \(\Pi\) = 3.14]

Solution:

We have r = 6 cm and \(\Theta\) = 30o

Now, Area of sector = \(\frac{\Theta }{360^{\circ}}\times \Pi r^{2}=\frac{30^{\circ}}{360^{\circ}}\times 3.14\times 36=9.42cm^{2}\)

Hence, the area of the sector of the circle is 9.42 cm2.

Question 14:In a circle of radius 21 cm, an arc subtends an angle of 60o at the center. Find the length of the arc.

Solution:

We have r = 21 cm and \(\Theta\) = 60o

Length of arc = \(\frac{\Theta }{360^{\circ}}\times 2\Pi r=\frac{60^{\circ}}{360^{\circ}}\times 2\times \frac{22}{7}\times 21=22cm\)

Hence, the length of the arc of the circle is 22 cm.

Question 15:The circumference of two circles are in the ratio 2 : 3. What is the ratio between their areas?

Solution:

Let the radii of the two circles be r and R respectively, the circumference of the circles be c and C respectively and the areas of the two circles be a and A respectively.

Now,

\(\frac{c}{C}=\frac{2}{3}\)

\(\Rightarrow\) \(\frac{2\Pi r}{2\Pi R}=\frac{2}{3}\)

\(\Rightarrow\) \(\frac{r}{R}=\frac{2}{3}\)

Now, the ratio between their areas is given by

\(\frac{a}{A}=\frac{\Pi r^{2}}{\Pi R^{2}}\)

= \(\left ( \frac{r}{R} \right )^{2}\) = \(\left ( \frac{2}{3} \right )^{2}\) = \(\frac{4}{9}\)

Hence, the ratio between their areas is 4 : 9.

Question 16:The areas of two circles are in the ratio 4 : 9. What is the ratio between their circumferences?

Solution:

Let the radii of the two circles be r and R respectively, the circumference of the circles be c and C respectively and the areas of the two circles be a and A respectively.

Now,

\(\frac{a}{A}=\frac{4}{9}\)

\(\Rightarrow\) \(\frac{\Pi r^{2}}{\Pi R^{2}}=\left ( \frac{2}{3} \right )^{2}\)

\(\Rightarrow\) \(\frac{r}{R}=\frac{2}{3}\)

Now, the ratio between their circumferences is given by

\(\frac{c}{C}=\frac{2\Pi r}{2\Pi R}\)

= \(\frac{r}{R}\)

= \(\frac{2}{3}\)

Hence, the ratio between their circumference is 2 : 3.

Question 17:A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.

Solution:

Let the side of the square be a and radius of the circle be r.

We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.

Therefore,  \(\sqrt{2a}=2r\)

\(\Rightarrow\) a = \(\sqrt{2r}\)

Now,

\(\frac{Area\, of\, a\, circle}{Area\, of\, a\, square}=\frac{\Pi r^{2}}{a^{2}}\)

= \(\frac{\Pi r^{2}}{(\sqrt{2r})^{2}}\) = \(\frac{\Pi r^{2}}{(2r)^{2}}\) = \(\frac{\Pi }{2}\)

Hence, the ratio of the areas of the circle and the square is \(\Pi\) : 2.

Question 18:The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72o.

Solution:

Let the radius of the circle be r.

Now,

Circumference = 8

\(\Rightarrow\) \(2\Pi r\) = 8

\(\Rightarrow\) r = \(\frac{14}{11}\) cm

We have r = \(\frac{14}{11}\) cm and \(\Theta\) = 72o

Area of sector = \(\frac{\Theta }{360^{\circ}}\times \Pi r^{2}=\frac{72^{\circ}}{360^{\circ}}\times \frac{22}{7}\times \left ( \frac{14}{11} \right )^{2}=1.02cm^{2}\)

Hence, the area of the sector of the circle is 1.02 cm2.

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm2. But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm2.

Question 19:A pendulum swings through an angle of 30o and describes an arc 8.8 cm in length. Find the length of the pendulum.

Solution:

Given:

Length of the arc = 8.8 cm

And,

\(\Theta\) = 30o

Now,

Length of the arc = \(\frac{2\Pi r\Theta }{360^{\circ}}\)

\(\Rightarrow\) 8.8 = \(\frac{2\times \frac{22}{7}\times r\times 30}{360}\)

\(\Rightarrow\) r = \(\frac{8.8\times 360\times 7}{44\times 30}\)

\(\Rightarrow\) r = 16.8 cm

Therefore,  Length of the pendulum = 16.8 cm

Question 20:The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes. [Take \(\Pi\) = 3.14]

Solution:

Angle inscribed by the minute hand in 60 minutes = 360o

Angle inscribed by the minute hand in 20 minutes = \(\frac{360}{60}\times 20=120^{\circ}\)

We have:

\(\Theta\) = 120o and r = 15 cm

Therefore,  Required area swept by the minute hand in 20 minutes = Area of the sector with r = 15 cm and \(\Theta\) = 120o

= \(\frac{\Pi r^{2}\Theta }{360}\)

= \(3.14\times 15\times 15\times \frac{120}{360}\) = 235.5 cm2

Question 21:A sector of 56o, cut out from a circle, contains 17.6 cm2. Find the radius of the circle.

Solution:

Area of the sector = 17.6 cm2

Area of the sector = \(\frac{\Pi r^{2}\Theta }{360}\)

\(\Rightarrow\) 17.6 = \(\frac{22}{7}\times r^{2}\times \frac{56}{360}\)

\(\Rightarrow\) r2 = \(\frac{17.6\times 7\times 360}{22\times 56}\)

\(\Rightarrow\) r2 = 36

\(\Rightarrow\) r = 6 cm

Therefore,  Radius of the circle = 6 cm

Question 22:The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector.

Solution:

Given:

Area of the sector = 63 cm2

Radius = 10.5 cm

Now,

Area of the sector = \(\frac{\Pi r^{2}\Theta }{360}\)

\(\Rightarrow\) 69.3 = \(\frac{22}{7}\times 10.5\times 10.5\times \frac{\Theta }{360}\)

\(\Rightarrow\) \(\Theta =\frac{69.3\times 7\times 360}{22\times 10.5\times 10.5}\)

\(\Rightarrow\) \(\Theta\) = 72o

Therefore, Central angle of the sector = 72o


Practise This Question

Statement 1: A tangent is perpendicular to the radius at the point of contact.
Statement 2: A line from the centre to any other point on the tangent has a length greater than the radius of the circle.