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

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

 Class 10th Maths Chapter 12 Exercise: 12.3 (Q1 to Q4)

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1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Solution:

Here, ∠P is in the semi-circle and so,

∠P = 90°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.

∴ QR = D

Using Pythagorean theorem,

QR2 = PR2 + PQ2

Or, QR2 = 72 + 242

=> QR = 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm

Now, the area of the semicircle = (π R2)/2

= (22/7 × 25/2 × 25/2)/2 cm2

= 13750/56 cm2 = 245.54 cm2

Also, area of the ΔPQR = ½ × PR × PQ

= ½ × 7 × 24 cm2

= 84 cm2

Hence, the area of the shaded region = 245.54 cm2 – 84 cm2

= 161.54 cm2

2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Solution:

Given,

Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (θ/360°) × π r2

So, Area of OAC = (40°/360°) × π r2 cm2

= 68.44 cm2

Area of the sector OBD = (40°/360°) × π r2 cm2

= 1/9 × 22/7 × 72 = 17.11 cm2

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD

= 68.44 cm2 – 17.11 cm2 = 51.33 cm2

3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Solution:

Side of the square ABCD (as given) = 14 cm

So, Area of ABCD = a2

= 14 × 14 cm2 = 196 cm2

We know that the side of the square = diameter of the circle = 14 cm

So, side of the square = diameter of the semicircle = 14 cm

∴ Radius of the semicircle = 7 cm

Now, area of the semicircle = (π R2)/2

= (22/7 × 7 × 7)/2 cm2 =

= 77 cm2

∴ Area of two semicircles = 2 × 77 cm2 = 154 cm2

Hence, area of the shaded region = Area of the Square – Area of two semicircle

= 196 cm2 – 154 cm2

= 42 cm2

4. Find the area of the shaded region in Fig. 12.22, 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:

It is given that OAB is an equilateral triangle having each angle as 60°

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = √3/4 × (OA)2 = √3/4 × 122 = 36√3 cm2

Area of the circle = π R2 = 22/7 × 62 = 792/7 cm2

Area of the sector making angle 60° = (60°/360°) × π r2 cm2

= 1/6 × 22/7 × 62 cm2 = 132/7 cm2

Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector

= 36√3 cm2 + 792/7 cm2 – 132/7 cm2

= (36√3 + 660/7) cm2

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