# Class 10 NCERT Solutions- Chapter 12 Areas Related to Circles – Exercise 12.3 | Set 2

### Chapter 12 Areas Related to Circles – Exercise 12.3 | Set 1

**Question 11. 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.**

**Solution:**

Side of square=6*Radius

=6*7

=42 cm

Area of shaded region=Area of square – Area of 9 circles

=side*side – 9πr

^{2}=42*42 – 9*22/7*7*7

=1764-1386

=378 cm

^{2}

The area of the remaining portion of the handkerchief =378 cm^{2}

**Question 12. 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**

**(i) quadrant OACB, (ii) shaded region.**

**Solution:**

(i)Area of shaded region=Area of quadrant=1/4πr

^{2}=1/4*22/7*3.5*3.5

=38.5/4

=9.625 cm

^{2}

(ii)Area of shaded region=Area of quadrant -Area of ∆BOD=9.625-1/2*BO*OD

=9.625-1/2*3.5*2

=9.625-3.5

=6.125 cm

^{2}

**Question 13. In Fig., 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:**

By Pythagoras theorem,

OB

^{2}=DA^{2}+AB^{2}BO

^{2}=(20)^{2}+(20)^{2}OB

^{2}=400+400OB

^{2}=800OB=√800

OB=√(2*2*2*2*5*5)

OB=2*2*5√2

OB=20√2

Area of shaded region=Area of quadrant -Area of square

=1/4πr

^{2}– side*side=1/4*3.14*20√2*20√2-20*20

=1/4*3.14*800-400

=1*3.14

=22cm

^{2}=1/4*3.14

The area of shaded region is =1/4*3.14

**Question 14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If ∠AOB = 30°, find the area of the shaded region.**

**Solution:**

Area of shaded region=Area of sector AOB-Area of sector COD

=θ/(360°) πR

^{2}-θ/(360°) πr^{2}=θ/(360°) π[R

^{2}-r^{2}]=30°/360*22/7[(21)

^{2}-(7)^{2}]=1/12*22/7*28*14

=308/3cm

^{2}=102.66cm

^{2 }

The area of shaded region 102.66cm^{2}

**Question 15. 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.**

**Solution:**

Area of segment=Area of quadrant -Area of ∆BAC

=1/4πr

^{2}-1/2*AC*AB=1/4*22/7*14*14-1/2*14*14

=11*14-98

=154-98

=56 cm

^{2}Semicircle R=?

In rt. ∆BAC, By Pythagoras theorem,

BC

^{2}=AB^{2}+BC^{2}BC

^{2}= (14)^{2}+(14)^{2}BC=√((14)

^{2}+(14)^{2})BC=√((14)

^{2}[1+1] )BC=14√2

∴Diameter of semicircle=14√2cm

then radius R of semicircle=14√2/2=7√2cm

Area of semicircle =1/2πR

^{2}=1/2*22/7*7√2*7√2

=22*7

=154 cm

^{2}Area of shaded region=Area of semicircle-Area of segment

=154-56 cm

^{2}=98 cm

^{2}

The area of shaded region is =98cm^{2}

**Question 16. Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each.**

**Solution:**

Area of design=Area of 2 quadrant -Area of square

=2*1/4πr

^{2}-side*side=1/2*22/7*8*8-8*8

=704/7-64

=100.57-64

=36.57cm

^{2}

Area designed region in figure is 36.57cm^{2}

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