NCERT Solutions for Class 10 Maths Chapter 11 Areas Related to Circles

Q2: Find the area of a quadrant of a circle whose circumference is 22 cm.

Answer:

We are given the circumference of the circle.

Thus,

2πr = 22

$r=\frac{11}{\pi }cm$

Also, we know that the area of a sector is given by :

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

It is given that we need to find the area of a quadrant, thus Θ = 90∘

Hence, the area becomes:-

= $\frac{90}{{360}^{\circ }}\times \pi (\frac{11}{\pi }{)}^2$

= $\frac{77}{8}c{m}^2$

Q3: The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

The minute hand rotates 360° in one hour.

We need to find a rotation in 5 minutes. :-

= $\frac{{360}^{\circ }}{60}\times 5={30}^{\circ }$

The area of the sector is given by :

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{30}{{360}^{\circ }}\times \pi {14}^2$

$Area=\frac{154}{3}c{m}^2$

Hence, the area swept by the minute hand in 5 minutes is $\frac{154}{3}c{m}^2$.

Q4: A chord of a circle of radius 10 cm subtends a right angle at the center.
Find the area of the corresponding : (i) minor segment

Answer:

The angle in the minor sector is 90^o.

Thus, the area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi {10}^2$

$Area=\frac{1100}{14}c{m}^2$ = 78.5 cm²

Now the area of a triangle is:-

$Area=\frac{1}{2}bh=\frac{1}{2}\times 10\times 10$ = 50 cm²

Thus, the area of the minor segment
= Area of the sector - Area of a triangle
= 78.5 − 50 = 28.5 cm²

Q4: A chord of a circle of radius 10 cm subtends a right angle at the center.
Find the area of the corresponding : (ii) major sector. (Use π = 3.14)

Answer:

The area of the major sector can be found directly by using the formula :

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

In this case, the angle is 360^o - 90^o = 270°.

Thus, the area is: -

= $\frac{{270}^{\circ }}{{360}^{\circ }}\times \pi \times 10\times 10$

= $\frac{3300}{14}$ = 235.7 cm²

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.
Find: (i) the length of the arc

Answer:

The length of the arc is given by:-

Length of arc = $\frac{\Theta }{360\circ }\times 2\pi r$

= $\frac{{60}^{\circ }}{{360}^{\circ }}\times 2\times \pi \times 21$

$=22cm$

Hence, the length of the arc is 22 cm.

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the center.
Find: (ii) the area of the sector formed by the arc

Answer:

We know that the area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

= $\frac{60}{{360}^{\circ }}\times \pi \times {21}^{\circ }=231c{m}^{\circ }$

Thus, the area of the sector is 231 cm².

Q5: In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.
Find: (iii) area of the segment formed by the corresponding chord

Answer:

For the area of the segment, we need to subtract the area of the triangle attached from the area of the arc.

Thus, consider the triangle:-

It is given that the angle of arc is 60^o, or we can say that all angles are 60^o (since two sides are equal). Hence, it is an equilateral triangle.

Area of the triangle is:-

= $\frac{\sqrt{3}}{4}\times a^2$

= $\frac{\sqrt{3}}{4}\times {21}^2$

= $\frac{441\sqrt{3}}{4}$

Hence, the area of the segment is:-

= $(231-\frac{441\sqrt{3}}{4})c{m}^2$

Q6: A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use π=3.14 and 3=1.73 )

Answer:

The area of the sector is :

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi {15}^2$

= 117.85 cm²

Now consider the triangle; the angle of the sector is 60⁰.

This implies it is an equilateral triangle. (As two sides are equal, so will have the same angle. This is possible only when all angles are equal, i.e., 60^o .)

Thus, the area of the triangle is:-

= $\frac{\sqrt{3}}{4}\times {15}^2$

$=56.25\sqrt{3}=97.31c{m}^2$

Hence area of the minor segment: = $117.85-97.31=20.53c{m}^2$

And the area of the major segment is :

=$\pi r^2-20.53$

$=\pi \times {15}^2-20.53$

$=707.14-20.53$

$=686.6c{m}^2$

Q7: A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle.
(Use π=3.14 and $\sqrt 3=1.73$ )

Answer:

For the area of the segment, we need the area of the sector and the area of the associated triangle.

So, the area of the sector is :

= $\frac{{120}^{\circ }}{{360}^{\circ }}\times \pi \times {12}^2$

= $150.72c{m}^2$

Now, consider the triangle:-

Draw a perpendicular from the centre of the circle to the base of the triangle (let it be h).

Using geometry, we can write,

$\frac{h}{r}=cos{60}^{\circ }$

or h = 6 cm

Similarly, $\frac{b2}{r}=sin{60}^{\circ }$

or $b=12\sqrt{3}$ cm

Thus, the area of the triangle is :

= $\frac{1}{2}\times 12\sqrt{3}\times 6$

= 62.28 cm²

Hence, the area of segment is: = 150.72 − 62.28 = 88.44 cm².

Q8: A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig.). Find (i) the area of that part of the field in which the horse can graze.

Answer:

The part grazed by the horse is given by = Area of the sector

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

= $\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times 5^2$

= 19.62 m²

Q8: A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see Fig.). Find (ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use π=3.14 )

Answer:

When the length of the rope is 10 m, the area grazed will be:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$\frac{{90}^{\circ }}{{360}^{\circ }}\times \pi \times {10}^2$

= 25π m²

Hence, the change in the grazing area is given by :

= $25\pi -\frac{25\pi }{4}=58.85m^2$

Q9: A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. Find: (i) the total length of the silver wire required.

Answer:

The total wire required will be for 5 diameters and the circumference of the brooch.

The circumference of the brooch:-

= $2\pi r=2\times \pi \times \frac{35}{2}$

= $110mm$

Hence, the total wire required will be = $110mm+5\times 35mm=285mm.$

Q9: A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in Fig. 12.12. Find : (ii) the area of each sector of the brooch.

Answer:

The total number of lines present in the brooch is 10 (lines starting from the centre).

Thus, the angle of each sector is 36^o.

The area of the sector is given by:

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{36}^{\circ }}{{360}^{\circ }}\times \pi \times (\frac{35}{2}{)}^2$

$Area=\frac{385}{4}m{m}^2$

Q10: An umbrella has 8 ribs that are equally spaced (see Fig. ). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Answer:

It is given that the umbrella has 8 ribs, so the angle of each sector is 45^o.

Thus, the area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{45}^{\circ }}{{360}^{\circ }}\times \pi \times {45}^2$

$Area=\frac{22275}{28}$

Hence, the area between two consecutive ribs is $\frac{22275}{28}c{m}^2$

Q11: A car has two wipers that do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Answer:

The area cleaned by one wiper is:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{115}^{\circ }}{{360}^{\circ }}\times \pi \times {25}^2$

$Area=\frac{158125}{252}c{m}^2$

Hence, the required area (area cleaned by both blades) is given by:-

= $2\times \frac{158125}{252}c{m}^2$

= $\frac{158125}{126}c{m}^2$

Q12: To warn ships of underwater rocks, a lighthouse spreads a red-colored light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Answer:

The area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

In this case, the angle is 80^o.

Thus, the area is:-

$Area=\frac{{80}^{\circ }}{{360}^{\circ }}\times \pi \times {16.5}^2$

$=189.97K{m}^2$

Q13: A round table cover has six equal designs as shown in Fig. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm ². (Use 3=1.7 )

Answer:

The angle of each of the six sectors is 60° at the centre. (∵$\frac{{360}^{\circ }}{6}={60}^{\circ }$)

Area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

$Area=\frac{{60}^{\circ }}{{360}^{\circ }}\times \pi \times {78}^2$

$Area=410.66c{m}^2$

And the area of the equilateral triangle associated with the segment:-

= $\frac{\sqrt{3}}{4}\times a^2$

$=34\times 282=333.2c{m}^2$

Hence, the area of segment is: = $410.66-333.2=77.46c{m}^2$

Thus, the total area of design is: = $6\times 77.46=464.76c{m}^2$

So, the total cost for the design is: = $0.35\times 464.76=Rs.162.66$

Q14: Tick the correct answer in the following :

Area of a sector of angle p (in degrees) of a circle with radius R is

(A) $\frac{p}{180}\times 2\pi R$

(B) $\frac{p}{180}\times \pi R^2$

(C) $\frac{p}{360}\times 2\pi R$

(D) $\frac{p}{720}\times 2\pi R^2$

Answer:

We know that the area of the sector is given by:-

$Area=\frac{\Theta }{{360}^{\circ }}\times \pi r^2$

= $\frac{p}{{360}^{\circ }}\times \pi r^2$

Hence, option (d) is correct.