Class 10th, Area Related Circle Exe-12.3 (Solution4)

             Mathematicsraj1.blogspot.com

Maths NCERT Class 10th Chapter 12 – Areas Related to Circles

 Class 10th Maths Chapter 12 Exercise: 12.3 (Q13 to Q16)

..............................................................................................................................................

13. In Fig. 12.31, 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:

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB ,

OB2 = AB2 + OA2

⇒ OB2 = 202 + 202

⇒ OB2 = 400 + 400

⇒ OB2 = 800

⇒ OB = 20√2 cm

Area of the quadrant = (πR2)/4 cm2 = 3.14/4 × (20√2)2 cm2 = 628 cm2

Area of the square = 20 × 20 = 400 cm2

Area of the shaded region = Area of the quadrant – Area of the square

= 628 – 400 cm2 = 228 cm2

14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If ∠AOB = 30°, find the area of the shaded region.

Solution:

Radius of the larger circle, R = 21 cm

Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30°

Area of the larger sector = (30°/360°) × π R2 cm2

= 1/12 × 22/7 × 212 cm2

=231/2cm2

Area of the smaller circle = (30°/360°) × π r2 cm2

= 1/12 × 22/7 × 72 cm2

=77/6cm2

Area of the shaded region = 231/2 – 77/6 cm2

= 616/6 cm2 = 308/3 cm2

15. In Fig. 12.33, 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:

Radius of the the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC2 = AB2 + AC2

⇒ BC2 = 142 + 142

⇒ BC = 14√2 cm

Radius of semicircle = 14√2/2 cm = 7√2 cm

Area of ΔABC = 1/2 × 14 × 14 = 98 cm2

Area of quadrant = 1/4 × 22/7 × 14 × 14 = 154 cm2

Area of the semicircle = 1/2 × 22/7 × 7√2 × 7√2 = 154 cm2

Area of the shaded region = Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 + 98 – 154 cm2 = 98 cm2

16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.

Solution:

AB = BC = CD = AD = 8 cm

Area of ΔABC = Area of ΔADC = 1/2 × 8 × 8 = 32 cm2

Area of quadrant AECB = Area of quadrant AFCD = 1/4 × 22/7 × 82

= 352/7 cm2

Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)

= (352/7 – 32) + (352/7 -32) cm2

= 2 × (352/7 -32) cm2

= 256/7 cm2

See next blog for more solution...... Thank you.