Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles – Exercise 12.3 | Set 1

Question 1. Find the area of the shaded region in Fig., if PQ = 24 cm, PR = 7 cm and O is the center of the circle.

Solution:

In fig. By Pythagoras theorem

RQ²=RP²+PQ²

RQ²=(7)²+(24)²

RQ²=625

RQ=√625

RQ=√5*5*5*5  

RQ=5*5

=25

Radius of circle =25/2cm

Areas of shaded region=Area of semi circle -Area of ∆RPQ

=1/2πR²-1/2*b*h

=(1/2*22/7*25/2)-(1/2*7*24)

=-84

=161.53cm²

Question 2. Find the area of the shaded region in Fig., if radii of the two concentric circles with center O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:

Area of shaded Region=Area of sector AOC – Area of sector BOD

∠AOC = θ

Radius of inner circle = r

Radius of outer circle = R

=θ/360 πR² – θ/360 πr²

= θ/360 π(R²-r²)

=40/(360)*22/7(14*14-7*7)

=40/360*22/7(196-49)

=(22/63) * 147

=154/3

= cm²

Question 3. Find the area of the shaded region in Fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:

Radius =14/2=7

Area of shaded region=Area of square-Area of 2 semi-circles

=side*side-2*1/2πr²

=14*14 – 22/7*7*7

=196-154

=42 cm²

Question 4. 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 center.

Solution:

Area of shaded region=Area of big sector+ Area of equilateral

=θ/360 πr²+√3/4(side)²

=(300/360)*22/7*6*6+(√3/4)*12*12

=5/6*22/7*6*6+36*1.73

=660/7+62.28

=94.28+62.28

=156.56 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 Figure. Find the area of the remaining portion of the square.

Solution:

radius, r = 1cm 

Area of remaining poison=Area of square – Area of 4 quadrants – Area of circle in the middle

=side*side – 4(90/360 πr²) – πr²

=4*4 – 2 πr²

=16-2*22/7*1*1

=16-44/7

=9.72cm²

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 Figure. Find the area of the design.

Solution:

Area of circle= πr²=22/7*32*32=22528/7=3218.28cm²     —–1

Area of ∆ABC=3*Area of ∆BOC

                     =3*1/2*side*side*sin120°

                      =3/2*32*32*sin60°

                      =1536*√3/2

                       =768*1.73

                     =1328.64cm²     ———2

Area of design =Area of circle – Area of ∆ABC

                     =321.28-1328.64

                     =1889.64 cm²

Question 7. 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.

Solution:

Area of shaded region=Area of square – Area of 4 quarters

=side*side – 4(90/360 πr²)

=14*14-22/7*7*7

=196-154

=42cm²

Question 8. Fig. here depicts a racing track whose left and right ends are semi-circular. 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:

i) A distance along inner edge=length of 2 parallel lines+ circumference of 2 circles

                                              =106+106+2πr

                                              =212+2*22/7*30

                                              =212+188.57

                                              =400.57m

ii) Area of track=Area of 2 rectangles+ semi rings

                         =106*10*2+π (R-r)²

                         =2120+22/7((40)²-(30)²)

                         =210+22/7(1600-900)

                         =210+2200

                    =4320 m²

Question 9. 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.

Solution:

Area of smaller circle=πr²

                                 =22/7*7/2*7/2

                                =78/2 =38.5m²

Area of segment=Area of quadrant-Area of ∆BOC

                          =1/4πR²-1/2*BO*OC

                          =1/4*22/7*7*7-1/2*7*7

                          =77/2-49/2

                         =77-49/2

                         =28/2

                         =14 cm²

Area of shaded region=Area of smaller circle+2*Area of segment

                                  =38.5 + 2*14  

                                  =38.5+28

                                 =66.5 cm²

Question 10. The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as center, a circle is drawn with radius equal to half the length of the side of the triangle (see Fig.). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Solution:

Area of equilateral ∆ =1.73205

π=3.14

√3=1.73205

√3/4 (side)²=173205

1.73205/4*(side)²=1.73205

(side)²=173205*4/1.73205

(side)²=173205*100000/10*173205

=√1000*4

=√(100 *100*2*2)  

=100*2

=200cm

∴Radius of each circle=200/2=100  

Area of shaded region=Area of∆ ABC-3*Area of sector  

=1.73205*60/360*3.14*100*100

=1.73205-31400/2

=17320.5-15700

=1620.5cm²

Chapter 12 Areas Related to Circles – Exercise 12.3 | Set 2