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**NCERT TEXTBOOK QUESTIONS SOLVED****Page No. 234****EXERCISE 12.3**

**Unless stated otherwise, use Ï€ = 22/7****Q 1. Find the area of the shaded region in Fig. PQ = 24 cm, PR = 7 cm and O is the centre of the circle.****Sol. **Since O is the centre of the circle,

âˆ´ QOR is a diameter.

â‡’ âˆ RPQ = 90Â° [Angle in a semi-circle]

Now, in right âˆ† RPQ,

RQ^{2} = PQ^{2} + PR^{2}

â‡’ RQ^{2} = 24^{2} + 7^{2} = 576 + 49 = 625

â‡’ RQ

âˆ´ Area of Î” RPQ

Now, area of semi-circle

âˆ´ Area of the shaded portion = 245.54 cm^{2} âˆ’ 84 cm^{2} = 161.54 cm^{2}.**Page No. 235****Q 2. Find the area of the shaded region in figures, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and âˆ AOC = 40Â°.****Sol.** Radius of the outer circle = 14 cm

Here, Î¸ = 40Â°

âˆ´ Area of the sector

Radius of the inner circle = 7 cm

Here, also Î¸ = 40Â°

âˆ´ Area of the sector BOD

Now, area of the shaded region = [Area of sector AOC] âˆ’ [Area of sector BOD]**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 semi-circles.****Sol.** Side of the square = 14 cm

âˆ´ Area of the square ABCD=14 Ã— 14 cm^{2} = 196 cm^{2}

Now, diameter of the circle = (Side of the square) = 14 cm

â‡’ Radius of each of the circles = 14/2 = 7cm

âˆ´ Area of the semi-circle

Area of the semi-circle BPC

âˆ´ Area the shaded region = [Area of the square] âˆ’ [Area of semicircle APD + Area of semicircle BPC]

= 196 âˆ’ [77 + 77] cm^{2} = 196 âˆ’ 154 cm^{2} = 42 cm^{2}.**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 circle with radius = 6 cm.

Area of equilateral triangle, having side a = 12 cm, is given by

âˆµ Each angle of an equilateral triangle = 60Â°

âˆ AOB = 60Â°

âˆ´ Area of sector COD

Now, area of the shaded region, = [Area of the circle] + [Area of the equilateral triangle] âˆ’ [Area of the sector COD]

**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.** Side of the square = 4 cm

âˆ´ Area of the square ABCD = 4 Ã— 4 cm^{2} = 16 cm^{2}

âˆµ Each corner has a quadrant circle of radius 1 cm.

âˆ´ Area of all the 4 quadrant squares

Diameter of the middle circle = 2 cm

â‡’ Radius of the middle circle = 1 cm

âˆ´ Area of the middle circle = Ï€r^{2}

Now, area of the shaded region = [Area of the square ABCD] âˆ’ [(Area of the 4 quadrant circles) + (Area of the middle circle)]**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.****Sol. **Area of the circle having radius r = 32 cm.

= Ï€r^{2}

â€˜Oâ€™ is the centre of the circle,

âˆ´ AO = OB = OC = 32 cm

â‡’ âˆ AOB = âˆ BOC = âˆ AOC = 120Â°

Now, in Î” AOB, âˆ 1 = 30Â°

âˆµ âˆ 1 + âˆ 2 = 60Â°

Also OA = OB â‡’âˆ 1 = âˆ 2

If OM âŠ¥ AB, then

â‡’

Also,

â‡’

â‡’

Now, area of Î” AOB, =

Since area Î” ABC = 3 Ã— [area of âˆ† AOB] =

Now, area of the design = [Area of the circle] âˆ’ [Area of the equilateral triangle]**Page No. 236****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.** Side of the square ABCD = 14 cm

âˆ´ Area of the square ABCD = 14 Ã— 14 cm^{2} = 196 cm^{2}.

âˆµ Circles touch each other

âˆ´ Radius of a circle = 14/2 = 7 cm

Now, area of a sector of radius 7 cm and sector angle Î¸ as 90Â°

â‡’ Area of 4 sectors =

âˆ´ Area of the shaded region = [Area of the square ABCD] âˆ’ [Area of the 4 sectors]

= 196 cm^{2} âˆ’ 154 cm^{2} = 42 cm^{2}.**Q 8. The figure depicts a racing track whose left and right ends are semicircular.****The distance between the two inner parallel line segments 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.**

(ii) Now, area of the track

= Area of the shaded region = (Area of rectangle ABCD) + (Area of rectangle EFGH) [ The track is 10 m wide]

â‡’ Area of the track

= (106 Ã— 10 m

= 2120 m

OA =7 cm

â‡’ AB = 2 OA = 2 Ã— 7 = 14 cm

OC = OA = 7 cm

âˆµ AB and CD are perpendicular to each other

â‡’ OC âŠ¥ AB

âˆ´ Area Î” ABC

Again OD = OA = 7 cm

âˆ´ Radius of the small circle

âˆ´ Area of the small circle

Radius of the big circle 14/2 cm = 7 cm

âˆ´ Area of semi-circle OABC

= 11 Ã— 7 cm

Now, Area of the shaded region = [Area of the small circle] + [Area of the big semi-circle OABC] âˆ’ [Area of Î” ABC]

âˆµ Î” ABC is an equilateral triangle and area of an

â‡’ Radius of each circle = 200/2 = 100 cm

Since each angle of an equilateral triangle is 60Â°,

âˆ´âˆ A = âˆ B = âˆ C = 60Â°

âˆ´ Area of a sector having angle of sector as 60Â° and radius 100 cm.

âˆ´ Area of 3 equal sectors

Now, area of the shaded region = [Area of the equilateral triangle ABC] âˆ’ [Area of 3 equal sectors]

= 17320.5 cm

âˆ´ The side of the square ABCD

= 3 Ã— diameter of a circle

= 3 Ã— (2 Ã— radius of a circle) = 3 Ã— (2 Ã— 7 cm)

= 42 cm

â‡’ Area of the square ABCD = 42 Ã— 42 cm

Now, area of one circle =

âˆµ There are 9 squares

âˆ´ Total area of 9 circles = 154 Ã— 9 = 1386 cm

âˆ´ Area of the remaining portion of the handkerchief = 1764 âˆ’ 1386 cm

(i) quadrant OACB, (ii) shaded region.

âˆ´ Area of the quadrant

[âˆµ OB = 3.5 cm = radius and OD = 2 cm (given)]

âˆ´ Area of the shaded region

= (Area of the quadrant OACB) âˆ’ (Area of âˆ† BOD)

âˆ´ Area of the square = 20 x 20 = 400 cm

Diagonal of the square =âˆš2 x (side of the square)

Radius of the quadrant of circle = Diagonal of square = 20âˆš2 cm

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

=628 - 400 = 228 cm

R = 21 cm

and sector angle Î¸ = 30Â°

âˆ´ Area of the sector OAB

Again, radius of the smaller circle r =7 cm

Here also, the sector angle is 30Â°

âˆ´ Area of the sector OCD

âˆ´ Area of the shaded region

Therefore, area of the quadrant ABPC

Area of right

â‡’ Area of segment BPC = 154 cm

Now, in right Î” ABC,

AC

â‡’ 14

â‡’ 196 + 196 = BC

â‡’ BC

âˆ´ Radius of the semi-circle BQC

âˆ´ Area of the semi-circle BQC

Now, area of the shaded region = [Area of segment BQC] âˆ’ [Area of segment BPC]

= 154 cm

âˆ´ Area of the square (ABCD) = 8 Ã— 8 cm

= 64 cm

Now, radius of the quadrant ADQB = 8 cm

âˆ´ Area of the quadrant ADQB

Similarly, area of the quadrant BPDC

âˆ´ Sum of the two quadrants

Now, area of design = [Sum of the areas of the two quadrants] âˆ’ [Area of the square ABCD]

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