NCERT Solutions for Exercise 12.3 Class 10 Maths Chapter 12 - Areas related to circles

Q2 Find the area of the shaded region in Fig, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and $\angle AOC = 40^\circ$ .

Answer:

The area of a shaded region can be easily found by using the formula of the area of the sector.

Area of the shaded region is given by : Area of sector OAFC - Area of sector OBED

$=\ \frac{40^{\circ}}{360^{\circ}}\times \pi (14)^2\ -\ \frac{40^{\circ}}{360^{\circ}}\times \pi (7)^2$

$=\ \frac{616}{9}\ -\ \frac{154}{9}\ =\ \frac{462}{9}$

$=\ 51.33\ cm^2$

Q3 Find the area of the shaded region in Fig, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Answer:

Area of the shaded region is given by = Area of the square - Area of two semicircles.

Area of square is : $=\ 14^2\ =\ 196\ cm^2$

And the area of the semicircle is:-

$=\ \frac{1}{2}\pi r^2$

$=\ \frac{1}{2}\pi \times 7^2$

$=\ 77\ cm^2$

Hence the area of the shaded region is given by : _{$=\ 196 - 2(77) = 42\ cm^2.$}

Q4 Find the area of the shaded region in Fig, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.

Answer:

Area of the shaded region is given by = Area of triangle + Area of the circle - Area of the sector

Area of the sector is : -

$=\ \frac{60^{\circ}}{360^{\circ}}\times \pi\times 6^2$

or $=\ \frac{132}{7}\ cm^2$

And, the area of the triangle is :

$=\ \frac{\sqrt{3}}{4}a^2\ =\ \frac{\sqrt{3}}{4}\times 12^2\ =\ 36\sqrt{3}\ cm^2$

And, the area of the circle is : $=\ \pi r^2$

or $=\ \pi \times 6^2$

or $=\ \frac{792}{7}\ cm^2$

Hence the area of the shaded region is:-

$=\ 36\sqrt{3}\ +\ \frac{792}{7}\ -\ \frac{132}{7}$

or $=\ \left ( 36\sqrt{3}\ +\ \frac{660}{7} \right )\ cm^2$

Q5 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in Fig. Find the area of the remaining portion of the square.

Answer:

Consider the quadrant in the given figure:- We have an angle of the sector as 90 ^o and radius 1 cm.

Thus the area of the quadrant is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi\times 1^2$

or $=\ \frac{22}{28}\ cm^2$

And the area of the square is : $=\ side^2$

$=\ 4^2$

$=\ 16\ cm^2$

And, the area of the circle is:-

$=\ \pi r^2\ =\ \pi \times 1^2\ =\ \pi\ cm^2$

Hence the area of the shaded region is: = Area of the square - Area of the circle - 4 (Area of quadrant)

or $=\ 16\ -\ \frac{22}{7}\ -\ 4\left ( \frac{22}{28} \right )$

or $=\ \frac{68}{7}\ cm^2$

Q6 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. Find the area of the design.

Answer:

Assume the center of the circle to be point C and AD as the median of the equilateral triangle.

Then we can write:-

$AO\ =\ \frac{2}{3}AD$

or $32\ =\ \frac{2}{3}AD$

Thus $AD\ =\ 48\ cm$

Consider $\Delta$ ABD,

$AB^2\ =\ AD^2\ +\ BD^2$

or $AB^2\ =\ 48^2\ +\ \left ( \frac{AB}{2} \right )^2$

or $AB\ =\ 32\sqrt{3}\ cm$

Thus the area of an equilateral triangle is:-

$=\ \frac{\sqrt{3}}{4}\times \left ( 32\sqrt{3} \right )^2$

or $=\ 768\sqrt{3}\ cm^2$

And the area of the circle is : $=\ \pi r^2\ =\ \pi\times 32^2$

or $=\ \frac{22528}{7} cm^2$

Hence the area of the design is:-

$=\ \left ( \frac{22528}{7}\ -\ 768 \sqrt{3} \right )\ cm^2$

Q7. In Fig, ABCD is a square of side 14 cm. With centers A, B, C, and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Answer:

It is clear from the figure that the area of all sectors is equal (due to symmetry).

Also, the angle of the sector is 90 ^o and the radius is 7 cm.

Thus the area of the sector is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi (7)^2$

or $=\ \frac{77}{2}\ cm^2$

And, the area of the square is : $=\ a^2\ =\ 14^2\ =\ 196\ cm^2$

Hence the area of the shaded region is :

$=\ 196\ -\ 4\times \frac{77}{2}$

$=\ 42\ cm^2$

Q8 Fig. depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(i) the distance around the track along its inner edge

Answer:

The distance around the track is Length of two straight lines + Length of two arcs.

Length of the arc is -

$=\ \frac{1}{2}\times 2\pi r$

$=\ \frac{1}{2}\times 2\pi \times 30\ =\ 30\pi\ m$

Thus the length of the inner track is :

$=\ 106\ +\ 30\pi\ +\ 106\ +\ 30\pi$

$=\ 212\ +\ 60\pi$

$=\ \frac{2804}{7}\ m$

Q8 Fig. depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :

(ii) the area of the track.

Answer:

The area of track = Area of outer structure - Area of inner structure.

Area of outer structure is: = Area of square + Area of 2 semicircles

$=\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ m^2$

And, area of inner structure: = Area of inner square + Area of 2 inner semicircles

$=\ 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2\ m^2$

Thus the area of the track is :

$\\=\ 106\times 80\ +\ \frac{1}{2}\pi (40)^2\ +\ \frac{1}{2}\pi (40)^2\ -\ \left ( 106\times 60\ +\ \frac{1}{2}\pi (30)^2\ +\ \frac{1}{2}\pi (30)^2 \right )\\\\=\ 4320\ m^2$

Hence the area of the track is 4320 m ² .

Q9 In Fig, AB, and CD are two diameters of a circle (with center O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Answer:

Firstly, the area of the smaller circle is :

$=\ \pi r^2$

$=\ \pi \times \left ( \frac{7}{2} \right )^2$

$=\ \frac{77}{2}\ cm^2$

Now, the area of $\Delta ABC$ :-

$=\ \frac{1}{2}\times AB\times OC$

or $=\ \frac{1}{2}\times 14\times 7\ =\ 49\ cm^2$

And, the area of the bigger semicircle is :

$=\ \frac{1}{2}\pi r^2$

$=\ \frac{1}{2}\pi \times 7^2$

$=\ 77\ cm^2$

Hence the area of the shaded region is:-

$=\ \frac{77}{2}\ +\ 77\ -\ 49$

$=\ 66.5\ cm^2$

Therefore the area of the shaded region is 66.5 cm ² .

Q10 The area of an equilateral triangle ABC is 17320.5 cm ² . With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use $\pi = 3.14$ and $\sqrt3 = 1.73205$ )

Answer:

Area of an equilateral triangle is:-

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

$\frac{\sqrt{3}}{4}\times a^2\ =\ 17320.5$

$a\ =\ 200\ cm$

Now, consider the sector:- Angle of the sector is 60 ^o and the radius is 100 cm.

Thus the area of the sector:-

$=\ \frac{60^{\circ}}{360^{\circ}}\times \pi \times 100^2$

$=\ \frac{15700}{3}\ cm^2$

Thus the area of the shaded region is :

$=\ 17320.5\ -\ 3\times \frac{15700}{3}$

$=\ 1620.5\ cm^2$

Q11 On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief.

Answer:

Since one side of the square has 3 circles, thus the side of the square is 42 cm.

Area of the square : $=\ a^2\ =\ 42^2\ =\ 1764\ cm^2$

And, area of a circle :

$=\ \pi r^2\ =\ \frac{22}{7}\times 7^2$

$=\ 154\ cm^2$

Hence the area of the remaining portion is : $=\ 1764\ -\ 9\times 154$

$=\ 378\ cm^2$

Q12 In Fig, OACB is a quadrant of a circle with center O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB

Answer:

The quadrant OACB is a sector with angle 90 ^o and radius 3.5 cm.

Thus the area of the quadrant is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 3.5^2$

$=\ \frac{77}{8}\ cm^2$

Hence the area of the quadrant is _{$\frac{77}{8}\ cm^2$} .

Q12 In Fig, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (ii) shaded region.

Answer:

For area of shaded region we need to find area of the triangle.

Area of triangle is:-

$=\ \frac{1}{2}\times 3.5\times 2$

$=\ 3.5\ cm^2$

Hence the area of the shaded region is = Area of the quadrant - Area of triangle

$=\ \frac{77}{8}\ -\ 3.5$

$=\ \frac{49}{8}\ cm^2$

Q13 In Fig, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use $\pi = 3.14$ )

Answer:

In the given figure we need to find the radius of the circle:-

Consider $\Delta$ OAB,

$OB^2\ =\ OA^2\ +\ AB^2$

$=\ 20^2\ +\ 20^2$

$OB\ =\ 20\sqrt{2}\ cm$

Thus area of quadrant:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times 3.14\times (20\sqrt{2})^2$

$=\ 628\ cm^2$

Also, the area of the square is : $=\ 20^2\ =\ 400\ cm^2$

Area of the shaded region is : $=\ 628\ -\ 400\ =\ 228\ cm^2$

Q14 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If $\angle \textup{AOB} = 30^\circ$ , find the area of the shaded region.

Answer:

Area of the shaded region is = Area of larger sector - Area of smaller sector

$=\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times 21^2\ -\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times 7^2$

$=\ \frac{30^{\circ}}{360^{\circ}}\times \pi \times (21^2\ -\ 7^2)$

$=\ \frac{308}{3}\ cm^2$

Hence the area of the shaded region is $\frac{308}{3}\ cm^2.$

Q15 In Fig, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Answer:

Consider $\Delta$ ABC,

$BC^2\ =\ AC^2\ +\ AB^2$

$=\ 14^2\ +\ 14^2$

$BC\ =\ 14\sqrt{2}\ cm$

Area of triangle is :

$=\ \frac{1}{2}\times 14\times 14$

$=\ 98\ cm^2$

Now, area of sector is :

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 14^2$

$=\ 154\ cm^2$

And area of semicircle is : -

$=\ \frac{1}{2} \pi r^2$

$=\ \frac{1}{2} \pi \times (7\sqrt{2})^2$

$=\ 154\ cm^2$

Hence the area of the shaded region is : _{$=\ 154\ -\ (154\ -\ 98)\ =\ 98\ cm^2$}

Q16 Calculate the area of the designed region in fig. common between the two quadrants of circles of radius 8 cm each.

Answer:

It is clear from the figure that the required area (designed area) is the area of the intersection of two sectors.

Area of the sector is:-

$=\ \frac{90^{\circ}}{360^{\circ}}\times \pi \times 8^2$

$=\ \frac{352}{7}\ cm^2$

And, area of the triangle:-

$=\ \frac{1}{2}\times 8\times 8\ =\ 32\ cm^2$

Hence the area of the designed region is :

$=\ 2\left ( \frac{352}{7}\ -\ 32 \right )$

$=\ \frac{256}{7}\ cm^2$