Bihar Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3 Textbook Questions and Answers.

## BSEB Bihar Board Class 10th Maths Solutions Chapter 12 Areas Related to Circles Ex 12.3

Question 1.

Find the area of the shaded region in the figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

[Annual Paper (Delhi) 2009]

Solution:

Since ROQ is a diameter, therefore ∠RPQ = 90°.

RQ² = RP² + PQ²

or RQ² = 7² + 24² = 49 + 576 = 625

or RQ = \(\sqrt{625}\) cm = 25 cm

∴ Radius r = \(\frac { 1 }{ 2 }\) RQ = \(\frac { 25 }{ 2 }\) cm

Question 2.

Find the area of the shaded region in the figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:

Area of the shaded region = Area of sector AOC – Area of sector OBD

Question 3.

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

Solution:

Area of the square ABCD = (14)² cm² = 196 cm²

Diameter of the semicircles = AD or BC = 14 cm

∴ Radius of each semicircle = \(\frac { 14 }{ 2 }\) cm = 7 cm

∴ Area of the both semicircular regions

= 2 x \(\frac { 1 }{ 2 }\)πr² = πr²

= (\(\frac { 22 }{ 7 }\) x 49) cm² = 154 cm²

∴ Area of the shaded regions = Area of the square ABCD – Area of the both semicircular regions

= (196 – 154) cm² – 42 cm².

Question 4.

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

Solution:

Area of the circular portion

Area of the equilateral ∆ OAB

\(\frac{\sqrt{3}}{4}\)(side)² = (\(\frac{\sqrt{3}}{4}\) x 144) cm²

= 36\(\sqrt{3}\) cm²

Area of the shaded region = Area of circular portion + Area of the equilateral triangle

= \(\left(\frac{660}{7}+36 \sqrt{3}\right)\) cm²

Question 5.

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 the figure. Find the area of the remaining portion of the square.

Solution:

The area of the square ABCD = (side)²

= 4² cm²

= 16 cm²

The sum of the areas of the four quadrants at the four corners of the square

= The area of a circle of radius 1 cm

= \(\frac { 22 }{ 7 }\) x 1² cm²

= \(\frac { 22 }{ 7 }\) cm²

The area of the circle of diameter 2 cm, i.e., radius 1 cm²

= \(\frac { 22 }{ 7 }\) x 1² cm²

= \(\frac { 22 }{ 7 }\) cm²

Area of the remaining portion of the square = The area of the square ABCD – The sum of the areas of 4 quadrants at the four corners of the square – The area of the circle of diameter 2 cm

= (16 – \(\frac { 22 }{ 7 }\) – \(\frac { 22 }{ 7 }\)) cm²

= \(\left(\frac{112-22-22}{7}\right)\) cm²

= \(\frac { 68 }{ 7 }\) cm² – 9.71 cm²

Question 6.

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in the figure. Find the area of the design (shaded region).

Solution:

Let ABC be an equilateral triangle and let 0 be the circumcentre of the circumcircle of radius 32 cm.

Draw OM ⊥ BC.

Now, ∠BOM = \(\frac { 1 }{ 2 }\) x 120° = 60°

So, from ∆ BOM, we have

Hence, area of ∆ BOC

= \(\frac { 1 }{ 2 }\) BC x OM

= \(\frac { 1 }{ 2 }\) x 32\(\sqrt{3}\) x 16 cm²

Area of ∆ ABC = 3 x Area of ∆ BOC

= (3 x \(\frac { 1 }{ 2 }\) x 32\(\sqrt{3}\) x 16) cm²

= 768\(\sqrt{3}\) cm²

∴ Area of the design (i.e., shaded region) = Area of the circle – Area of ∆ ABC

= \(\left(\frac{22528}{7}-768 \sqrt{3}\right)\) cm²

Question 7.

In the figure, ABCD is a square of side 14 cm. With centres 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.

Solution:

The area of the square ABCD = (side)² = 142 cm²

= 196 cm².

The sum of the areas of the four quadrants at the four corners of the square

= The area of a circle of radius \(\frac { 14 }{ 2 }\) cm = 7 cm²

= π(7)² cm² =(\(\frac { 22 }{ 7 }\) x 49) cm² = 154 cm²

Area of the shaded region = The area of the square ABCD – The sum of the areas of four quadrants at the four corners of the square

= (196 – 154) cm² = 42 cm²

Question 8.

Figure 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.

(ii) the area of the track.

Solution:

We have OB = O’C = 30 m

and, AB = CD = 10 m

OA = O’D = (30 + 10) m = 40 m

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

= BC + EH + 2 x circumference of the semicircle of radius OB = 30 m

= ( 106 + 106 + 2 x \(\frac { 1 }{ 2 }\) x 2π(30) ) m

= ( 212 + 2 x \(\frac { 22 }{ 7 }\) x 3o) m

= ( 212 + \(\frac { 1320 }{ 7 }\) )m = \(\left(\frac{1484+1320}{7}\right)\) m

= \(\frac { 2804 }{ 7 }\) m

m = 400.57 m

(ii) Area of the track = Area of the shaded region = Area of rectangle ABCD + Area of rectangle EFGH + 2(Area of the semicircle of radius 40 m – Area of the semicircle with radius 30 m)

Question 9.

In the figure, AB and CD are two diameters of a circle (with centre 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.

Solution:

Area of the sector OCB = \(\left(\frac{90^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 7 \times 7\right)\) cm²

= \(\frac { 77 }{ 2 }\) cm²

Area of ∆ OCB = \(\frac { 1 }{ 2 }\) x OC x OB

= \(\frac { 1 }{ 2 }\) x 7 x 7 cm²

= \(\frac { 49 }{ 2 }\) cm²

The area of the segment BPC = Area of the sector OCB – Area of the A OCB

= (\(\frac { 77 }{ 2 }\) – \(\frac { 49 }{ 2 }\)) cm²

= \(\frac { 28 }{ 2 }\) cm²

= 14 cm²

Similarly, the area of the segment AQC

= 14 cm²

Also,the area of the circle with DO as diameter

= (\(\frac { 22 }{ 7 }\) x \(\frac { 7 }{ 2 }\) x \(\frac { 7 }{ 2 }\)) cm² = \(\frac { 77 }{ 2 }\) cm²

Hence,the total area of the shaded region

= (14 + 14 + \(\frac { 77 }{ 2 }\)) cm²

= \(\left(\frac{28+28+77}{2}\right)\) cm²

= \(\frac { 133 }{ 2 }\) cm²

= 66.5 cm²

Question 10.

The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region.

(Use π = 3.14 and \(\sqrt{3}\) = 1.73205)

Solution:

Let each side of the triangle be a cm. Then,

Thus, radius of each circle is \(\frac { 200 }{ 2 }\) cm = 100 cm.

Now, area of shaded region

= Area of ∆ ABC – 3 x (Area of a sector of angle 60° in a circle of radius 100 cm)

= [17320.5 – 3(\(\frac{60^{\circ}}{360^{\circ}}\) x 3.14 x 100 x 100)] cm²

= (17320.5 – 15700) cm² = 1620.5 cm²

Question 11.

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

Solution:

Side of the square ABCD

= AB – 3 x diameter of circular design

= 3 x (2 x 7) cm – 42 cm

∴ Area of the square ABCD

= (42 x 42) cm² = 1764 cm²

Area of one circular Resign

= πr² = (\(\frac { 22 }{ 7 }\) x 7 x 7) cm²

= 154 cm²

∴ Area of 9 such designs

= (9 x 154) cm² = 1386 cm²

∴ Area of the remaining portion of the handkerchief

= Area of the square ABCD – Area of 9 circular designs

= (1764 – 1386) cm² – 378 cm²

Question 12.

In the figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB,

(ii) shaded region.

Solution:

Hence, area of the shaded region = Area of quadrant – Area of ∆ BOD

= (\(\frac { 77 }{ 8 }\) – \(\frac { 7 }{ 2 }\)) cm² = (\(\frac { 77-28 }{ 8 }\)) cm²

= \(\frac { 49 }{ 8 }\) cm² = 6.125 cm²

Question 13.

In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution:

Radius of the quadrant = OB = \(\sqrt{\mathrm{OA}^{2}+\mathrm{AB}^{2}}\)

= \(\sqrt{20^{2}+20^{2}}\) cm

= 20\(\sqrt{1+1}\) cm

= 20\(\sqrt{2}\) cm

∴ Area of quadrant OPBQ

= \(\frac { 1 }{ 4 }\)πr²

= \(\frac { 1 }{ 4 }\) x 314 x (20\(\sqrt{2}\))² cm²

= ( \(\frac { 1 }{ 4 }\) x 3.14 x 80o) cm² = 628 cm²

Area of the square OABC = (20) cm² = 400 cm²

Hence, area of the shaded region = Area of quadrant – Area of square OABC

= (628 – 400) cm² = 228 cm².

Question 14.

AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠AOB = 30°, find the area of the shaded region.

Solution:

Let A_{1} and A_{2} be the areas of sector OAB and OCD respectively. Then,

A_{1} = Area of a sector of angle 30° in a circle of radius 21 cm

A_{2} = Area of a sector of angle 30° in a circle of radius 7 cm

Question 15.

In the figure, 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.

[Annual Paper (Foreign) 2008]

Solution:

Here the radius of the quadrant with A as the centre is 14 cm.

Then, the area of the quadrant ABMC

Area of the segment BMC of the circle = Area of the quadrant ABMC – Area of ∆ BAC

= (154 – 98) cm²

= 56 cm²

Now, since AC = AB = 14 cm and ∠BAC = 90°, therefore by Pythagoras Theorem,

BC = \(\sqrt{\mathrm{AC}^{2}+\mathrm{AB}^{2}}\)

= \(\sqrt{14^{2}+14^{2}}\)

= 14\(\sqrt{2}\) cm

∴ Radius of the semicircle BNC = \(\frac { 1 }{ 2 }\) x 14\(\sqrt{2}\) cm = 7\(\sqrt{2}\) cm

∴ Area of the semicircle BNC

= (7\(\sqrt{2}\) )² cm²

= \(\left(\frac{1}{2} \times \frac{22}{7} \times 98\right)\) cm² = 154 cm²

Hence, the area of the shaded region

= the area of the region between two arcs BMC and BNC = The area of semicircle BNC – The area of the segment of the circle BMC

= (154 – 56) cm² = 98 cm²

Question 16.

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

Solution:

Here, 8 cm is the radius of each of the quadrants ABMD and BNDC.

Sum of their areas

Area of the square ABCD = (8 x 8) cm² = 64 cm²

Area of the designed region = Area of the shaded region

= Sum of the areas of quadrants – Area of the square ABCD

= \(\left(\frac{704}{7}-64\right)\) cm² = \(\left(\frac{704-448}{7}\right)\) cm²

= \(\frac{256}{7}\) cm²

= 36.57 cm²