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In the given figure , AC is the diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, what is the area of the shaded region to the nearest whole number?

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28 c m^{2}

29 c m^{2}

30 c m^{2}

45 c m^{2}

answer is C.

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Detailed Solution

Given that, AC is the diameter of a circle with radius

5 cm

.
And

AB = BC, CD = 8 cm

Diameter =

2 \times Radius

\Rightarrow

AC = 2(5)

\Rightarrow AC = 10 cm

Area of the circle =

π r^{2}

\Rightarrow

Area of the circle

= π (5)^{2}

= 25 π c m^{2}

.
We are aware that any angle that the diameter of a circle subtends is a right angle.
Hence,

△ ABC

is a right angled triangle with

∠ B = 90^{°}

.
Pythagoras theorem states that the square of the hypotenuse side in a right-angled triangle equals the sum of the squares of the other two sides.

∴ A C^{2} = A B^{2} + B C^{2}

[Pythagoras theorem] .

\Rightarrow 100 = 2 A B^{2}

\Rightarrow A B^{2} = 50

\Rightarrow AB = \sqrt{50}

\Rightarrow AB = 5 \sqrt{2}

∴ AB = BC = 5 \sqrt{2}

Area of the triangle =

\frac{1}{2} (Base) (Height)

\Rightarrow

Area of

△ ABC

= \frac{1}{2} \times AB \times BC

\Rightarrow

Area of

△ ABC

\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2}

\Rightarrow

Area of

△ ABC

= 25 c m^{2}

Similarly,

△ ADC

is a right angled triangle with

∠ D = 90^{°}

∴ A C^{2} = A D^{2} + D C^{2}

\Rightarrow 10^{2} = A D^{2} + 8^{2}

\Rightarrow 100 = A D^{2} + 64

\Rightarrow A D^{2} = 36

\Rightarrow AD = 6

Area of

△ ADC

= \frac{1}{2} \times AD \times DC

\Rightarrow

Area of

△ ADC

\frac{1}{2} \times 6 \times 8

= 24 c m^{2}

Area of shaded region = Area of the circle - Area of

△ ABC

- Area of

△ ADC

\Rightarrow

Area of shaded region =

25 π - 49 = 29.5 c m^{2}

\Rightarrow

Area of shaded region

\sim

30 c m^{2}

So, option 3 is correct.

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In the given figure , AC is the diameter of a circle with radius 5cm. If AB = BC and CD = 8 cm, what is the area of the shaded region to the nearest whole number?

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