# in fig. 12.33, 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.

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## In the given 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 shade region.

Click here👆to get an answer to your question ✍️ In the given 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 shade region.

Question

## In the given 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 shade region.

**A**98cm

2

**B**38cm

2

**C**48cm

2

**D**58cm

2 Medium Open in App Solution Verified by Toppr

Correct option is A)

Area of shaded region=area of semicircle of diameter BC-{area of quadrant of radius AB/AC- area of △ABC}

∵ BC is hypotenuse of right angle △ABC

here AB=BC=14 So, BC=14 2 =2×radius⇒radius=7 2

So, Area of semicircle of diameter BC=

2 πr 2 = 2 1 × 7 22 ×(7 2 ) 2 =154cm 2

Area of quadrant of radius AB/AC=

4 πr 2 = 4 1 × 7 22 ×14×14 =154cm 2 Area of △ABC= 2 1 ×h×b = 2 1 ×14×14=98cm 2

Now, area of shaded region=154−{154−98}=98cm

2

Hence, area of shaded region=98cm

2

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## In Fig. 12.33, 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

In Fig. 12.33, 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. We use the area of sector formula to solve the problem. The area of the shaded region is 98 cm^2

In Fig. 12.33, 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**Solution:**

We use the formula for areas of semi-circles and sector of circles to solve the problem.

To find the area of semi-circle, we need to find the radius or diameter (BC) of the semicircle.

ΔABC is a right-angled triangle, right-angled at A (ABC being a quadrant).

AB = AC = 14 cm [Radius of the circle]

Using Pythagoras theorem, we can find the hypotenuse (BC) of ΔABC.

BC2 = AB2 + AC2

= (14 cm)2 + (14 cm)2

BC = √2 × (14 cm)² = 14√2 cm

∴ Radius of semicircle BDC, r = BC/2 = 14√2/2 cm = 7√2 cm

Area of the shaded region = Area of semicircle - (Area of quadrant ABC - Area ΔABC)

= πr2/2 - [90°/360° × π(14)2 - 1/2 × AC × AB]

= π(7√2)2/2 - [π(14)2/4 - 1/2 × 14 × 14]

= [(22 × 7 × 7 × 2)/(7 × 2)] - [(22 × 14 × 14)/(7 × 4) - 7 × 14]

= 154 - (154 - 98) = 98 cm2

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

**Video Solution:**

## In Fig. 12.33, 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.

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.3 Question 15

**Summary:**

The area of the shaded region where ABC is a quadrant of a circle of radius 14 cm and a semicircle drawn with BC as the diameter is 98 cm2.

**☛ Related Questions:**

In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.

In Fig. 12.25, 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.

Fig. 12.26 depicts a racing track whose left and right ends are semicircular. Fig. 12.26 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.

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.

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Guys, does anyone know the answer?