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**Short Answer Type Questions**

**Q.1. Find the area of the shaded region in Fig., if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of circle. [Take π = 3.14] [Delhi 2019]Ans.** O is the centre of the circle.

AC is the diameter.

∴ In ΔADC, ∠D = 90°

∴ AC

AC

⇒ AC

OC = radius of circle

Area of the circle = πr

= 25 x 3.14 = 78.5 cm

Area of the rectangle = 6 x 8 = 48 cm

∴ Area of the shaded region = area of the circle - area of the rectangle

= 78.5 - 48 = 30.5 cm

Ans.

Now, area of the corresponding segment APB of circle

= area of the minor segment

Ans.

Area of sector

Area of sector containing 60°angle

Also, area of sector containing 80° angle

Again, area of sector containing 40° angle

Area of total shaded region

Ans.

OB

OB

OB

Ans.

Area o f each quadrant =

Area of square = (side)

Area of the shaded region = ar(square) - ar(4 quadrants)

= 144 cm

Ans.

OP = r = 5 cm

In ΔOPB

OB

= 5

Area of circumscribed circle with radius (R) = πR

Area of inscribed circle with radius (r)

= πr

Area between inscribed and circumscribed circles of square is shaded area with dots and lines.

= (50π - 25π) cm

Ans.

Short hand covers one complete path in 12 hours Path traced by short hand in 12 hour

Path traced by short hand in 48 hours = 4 x 8π = 32π

Long hand covers one complete path in 1 hour

Path traced by long hand in 1 hour

Path traced by long hand in 48 hours

= 48 x 12π = 576π

Sum of distances travelled by their tips in 48 hours

= 32π + 576π = 608π

Ans.

Areas A, B and C are identical semicircles each having diameter 3 cm.

So, radius of each of these semicircle = 3/2 cm

Area E is a semicircle with diameter

= 3 + 3 +3 = 9 cm

∴ Radius of area E= 9/2 cm

Area D is a complete circle with diameter 4.5 cm

The required shaded area = E + B - A - C - D

Hence, the required shaded area is 12.38 cm

Ans.

radius of larger circle be R cm

also, R - r = 7 cm , ..(i)

difference between the areas of the two circles

πR

Here

R + r = 49 cm ...(ii)

solving (i) and (ii) for R and r

Ans.

= 12.83 cm

CD = DE + EC = 4 + 7 = 11 cm

Area of shaded region = Area of trapezium - Area of sector BGEC

= (31.5)-(12.83) = 18.67 cm

Ans.

∴ Area o f the equilateral triangle =

∴

a

a = 11 x 2 = 22 cm

The wire is bent in the form of circle having radius r cm

∴ Circumference of the circle = perimeter of the equilateral Δ

∴ 2πr = 3 x 22

r = 10.5 cm

Area of circle = πr

Ans.

(ii) Now, we find area of ΔOBD

We have area of ΔOBD =

Hence, area of shaded region = area o f quadrant OACB - area of ΔOBD

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