# In Fig. 12.27, 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:**

We use the concepts of areas of circles and areas of triangles to solve the problem.

AB and CD are diameters of the circle with center O

∴ OD = OC = OA = OB = Radius of the circle R = 7 cm

∴ AB = 2R = 14 cm

Radius of shaded circular region, r = OD/2 = 7/2 cm

Area of the shaded smaller circular region = πr²

= π (7/2 cm)^{2}

= 22/7 × 7/2 × 7/2 cm^{2}

= 77/2 cm^{2}

= 38.5 cm^{2}

Area of the shaded segment of larger circular region = Area of semicircle ACB - Area of ΔABC

= 1/2 π(OA)^{2} - 1/2 × AB × OC

= 1/2 πR^{2} - 1/2 × 2R × R

= 1/2 × 22/7 × 7^{2} - 1/2 × 14 × 7

= 77 - 49

= 28 cm^{2}

Area of the shaded region = Area of the shaded smaller circular region + Area of the shaded segment of larger circular region

= 38.5 cm^{2} + 28 cm^{2}

= 66.5 cm^{2}

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 12

**Video Solution:**

## In Fig. 12.27, 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

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 9

**Summary:**

The area of the shaded region if 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 with OA = 7 cm is 66.5 cm^{2}.

**☛ Related Questions:**

- 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 Fig. 12.28). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)
- On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.
- In Fig. 12.30, 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
- 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).