All these RS Aggarwal class 10 solutions chapter 18 - Areas of Circle, Sector and Segment are solved by Byju's top ranked professors as per CBSE guidelines.

**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 \(\frac{77}{8}\) cm^{2}.

**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

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

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

\(\Rightarrow\) d^{2} = 8d

\(\Rightarrow\) d^{2} ??? 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 60**^{o}** at the centre.**

**Solution:**

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

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 4^{2} = 16\(\Pi\) cm^{2}

Hence, the area of the circle is 16\(\Pi\) cm^{2}.

**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

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

\(\Rightarrow\) r^{2} = 64 + 36

\(\Rightarrow\) r^{2} = 10^{2}

\(\Rightarrow\) 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 30**^{o}**. [Take \(\Pi\) = 3.14]**

**Solution:**

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

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 cm^{2}.

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

**Solution:**

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

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 72**^{o}**.**

**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 \(\Theta\) = 72^{o}

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 cm^{2}.

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm^{2}. But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm^{2}.

**Question 19:** **A pendulum swings through an angle of 30**^{o}** 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,

\(\Theta\) = 30^{o}

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 = 360^{o}

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

We have:

\(\Theta\) = 120^{o} and r = 15 cm

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

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

= \(3.14\times 15\times 15\times \frac{120}{360}\) = 235.5 cm^{2}

**Question 21:** **A sector of 56**^{o}**, cut out from a circle, contains 17.6 cm**^{2}**. Find the radius of the circle.**

**Solution:**

Area of the sector = 17.6 cm^{2}

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

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

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

\(\Rightarrow\) r^{2} = 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 cm**^{2}**. Find the central angle of the sector.**

**Solution:**

Given:

Area of the sector = 63 cm^{2}

Radius = 10.5 cm

Now,

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

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

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

\(\Rightarrow\) \(\Theta\) = 72^{o}

Therefore, Central angle of the sector = 72^{o}

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