Some Important Questions..!!!

Q1. The area of an equilateral triangle is 49 3 cm^2.. Taking each vertex of a centre, a circle is drawn with radius equal to half the length of the side of the triangle. Find the area of the shaded portion.                                                                                                                        (CBSE 2009)

                                                                                                                         

Area of ABC = 49√3 cm²

√3 / 4 a2 = 49√3

a² = 49 * 4 = 14cm

radius of each circle = 7cm

Required area = Area of ABC – 3 * ( area of sector )

{ 49√3 – 3( 60/ 360 * 22/ 7 * 7²)}cm²

(49√3 – 77)cm²

(49 * 1.73 – 77)cm²

7.77cm²

Q2:AB and CD are two diameters of a circle intersecting each other at centre O and OD is the diameter of the smaller circle. If OA= 7cm, find the area of the shaded region.       (CBSE 2010)

Area of shaded region = (Area of circle with OD as diameter) + 

Area of semicircle with AB as diameter – Area of ABC

∏ * (7/ 2)2 + 1/ 2 * ∏ * (7)² – 1/ 2 * AB * OC

{ ∏/ 4 * 49 + ∏/ 2 * 49 – 1/ 2 * 14 * 7}cm²

( 3∏ / 4 * 49 – 49)cm²

(3/ 4 * 22/ 7 * 49 – 49)cm²

231- 98 / 2 cm²

66.5cm²

Q3:Find the area of the shaded region if PQ= 24cm, PR= 7cm and O is the centre of the circle.

                                                                                                                                           (CBSE 2009)

As, RPQ is angle in a semicircle

therefore, it is a right angle.

Using Pythagoras Theorm:-

RQ² = RP² + PQ²

RQ² = 7² + 24²

RQ² = 625

RQ= 25cm

Radius of circle = 1/ 2 RQ = 25/ 2cm

Area of shaded region = Area of semicircle – Area of RPQ

= 1/ 2 ∏r² – 1/ 2 * PR * PQ

= {1/ 2 * 22/ 7 * (25/ 2)² – 1/ 2 * 7 * 24}cm²

= {6875/ 28 – 84}cm²

= 4523/ 28cm² 

Q4: In the adjoining figure, ABC is a right angled triangle at A. Find the area of the shaded region if AB= 6cm, BC= 10cm and O is the centre of circle.                                                 (CBSE 2009)

Using Pythagoras Theorm in ABC:-

BC² = AB² + AC²

AC² = BC² – AB²

AC² = 100 – 36 = 64

AC= 8cm

Therefore, Area of triangle ABC = 1/ 2 * AB * AC

= 1/ 2 * 6 * 8 = 24cm²

Area of ABC = area of OBC + area of OCA + area of OAB

24 = 1/ 2 (BC* r) + 1/ 2 (CA * r) + 1/ 2 (AB * r)

24 = 1/ 2 r ( BC + CA + AB)

24 = 1/ 2 r (10 + 8 +6)

24 = 12r

r = 2cm

Area of shaded region = Area of ABC – Area of circle

= 24 - ∏r²

= (24 – 22/ 7 * 4)cm²

= 80/ 7cm²

                                          More important questions will be posted soon..!!!