**Unless stated otherwise,
**

* Sol.* Since ROQ is a diameter, therefore,

In rt

Therefore, Radius

Area of the semi circle

and area of

Area of the shaded region

= Area of the semi circle – Area ( RPQ)

**Q.2 Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and **

* Sol. *Area of the shaded region

= Area of sector AOC – Area of sector OBD

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

* Sol.* Area of the square ABCD =

Diameter of the semicircles = AD or BC = 14 cm

Therefore, Radius of each semicircle = 7 cm

Area of the semicircular regions

Therefore, Area of the shaded portion

= Area of the square ABCD – Area of the semicircular regions

=

**Q.4 Find the area of the shaded region in figure, 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.**

* Sol.* Area of the circular portion

= Area of the circle – Area of the sector

where r = 6

Area of the equilateral

Therefore, Area of the shaded region

**Q.5 From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.**

* Sol.* The area of the whole square ABCD

The sum of the area of the four quadrants at the four corners of the square

= The area of a circle of radius 1 cm

The area of the circle of diameter 2 cm,

i.e., radius

Therefore, Area of the remaining portion

= The area of the square ABCD

– The sum of the area of 4 quadrants at the four corners of the square

– The area of the circle of diameter 2 cm

**Q.6 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region).**

* Sol. *Let ABC be an equilateral triangle and let O be the circumcentre of the circumcircle of radius 32 cm.

Area of the circle

Area of ABC = 3 × Area of BOC

Therefore, Area of the design (i.e., shaded region)

= Area of the circle – Area of ABC

**Q.7 In figure , ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.**

* Sol. *The area of the whole square

The sum of the area of the four quadrants at the four corners of the square

= The area of a circle of radius

Area of the shaded portion = The area of the square ABCD

– The sum of the area of four quadrants at the four corners of the square

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**Q.8 Figure depicts a racking track whose left and right ends are semicircular.**

** The distance between the two inner parallel line segment is 60 m and they are each 106 m long. If the track is 10 m wide, find **

** (i) The distance around the track along its inner edge. **

** (ii) The area of the track. **

* Sol. *We have, OB = O'C = 30 m

and AB = CD = 10 m

OA = O'D = (30 + 10) m = 40 m

= BC + EH + 2 × circumference of the semicrircle of radius OB = 30 m

** (ii)** Area of the track

= Area of the shaded region

= Area of rectangle ABCD + Area of rectangle EFGH

+ 2 (Area of the semicircle of radius 40 m

– Area of the semicircle with radius 30 m)

**Q.9 In figure 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.**

* Sol. * Area of the sector

Area of OCB

Therefore, The area of the segment BPC

Similarly the area of the segment AQC =

Also, the area of the circle with DO as diameter

Hence, the total area of the shaded region

**Q.10 The area of an equilateral triangle ABC is 17320.5 . 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 figure). Find the area of the shaded region.
(Use )**

** **

* Sol*. Let each side of the triangle be a cm. Then,

Area =

a = 200

Thus, radius of each circle is 100 cm.

Now, required area

= Area of ABC – 3 ×(Area of a sector of angle 60º in a circle of 100 cm)

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**Q.11 On a square hand kerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief. **

* Sol.* Side of the square ABCD = AB

= 3 × diameter of circular design

= 3 × (2 × 7) cm = 42 cm

Therefore, Area of the square ABCD

Area of one circular design

Therefore, Area of 9 such designs

Therefore, Area of the remaining portion of the handkerchief

= Area of the square ABCD – Area of 9 circular designs

**Q.12 In figure OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm , find the area of the
**

**(ii)** Area of

Hence, area of the shaded region

= Area of quadrant – Area of AOD

* Sol.* Radius of the quadrant = OB =

Therefore, Area of quadrant OPBQ

Area of the square OABC

Hence, area of the shaded region

= Area of quadrant – Area of square OABC

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**Q.14 AB and CD are respectively across of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If , find the area of the shaded region.**

* Sol. * Let be the areas of sectors OAB and OCD respectively. Then,

= Area of a sector of angle 30º in a circle of radius 21 cm.

= Area of a sector of angle 30º in a circle of radius 7 cm

Area of the shaded region

**Q.15 In figure , 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.**

* Sol. *Let 14 cm be the radius of the quadrant with A as the centre.

Then the area of the quadrant ABMC

Area of BAC

Therefore, Area of the segment of the circle, BMC

= Area of the quadrant ABMC – Area of BAC

= (154 – 98)

Now, since AC = AB = 14 cm and

Therefore, By Pyrthagoras Theorem,

Therefore, Radius of the semicircle BNC =

Area of the semicircle BNC =

Hence, the area of the region between two arcs BMC and BNC

= The area of the shaded region

= The area of semicircle BNC – The area of the segment of the circle BMC

**Q.16 Calculate the area of the the designed region in figure common between the two quadrants of circle of radius 8 cm each. **

* Sol. *Here, 8 cm is the radius of the quadrants ABMD and BNDC.

Sum of their areas

Area of the square ABCD

Area of the designed region

= Area of the shaded region

= Sum of the area of quadrants

– Area of the square ABCD

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