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In the given figure, find the area of the shaded region and also its perimeter if the length of AB and BC (in cm) is 28 and 21 respectively and BCD is a quadrant of circle. AEC is semicircle on AC as diameter (Take $π = \frac{22}{7}$ ).

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85cm

95cm

35cm

25cm

answer is B.

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Detailed Solution

The figure consists of a quadrant, a semicircle and a right angled triangle.
(i) Quadrant:
Quadrant is one-fourth part of the circle.
BC = BD = 21 cm (Given)
⇒Area of quadrant BCD =

\frac{π r^{2}}{4}

⇒Area of quadrant BCD =

\frac{π {(21)}^{2}}{4}

⇒Area of quadrant BCD =

\frac{22}{7} \times \frac{21 \times 21}{4}

⇒Area of quadrant BCD =

\frac{11 \times 3 \times 21}{2}

⇒Area of quadrant BCD =

\frac{693}{2}

⇒Area of quadrant BCD = 346.5cm2.....................… (1)
(ii) Triangle:
In the given figure, the right angle triangle is△ABC with base BC and height AB.
⇒ Area of △ABC =

\frac{1}{2} \times Base \times Height

⇒ Area of △ABC =

\frac{1}{2}

×BC×AB
⇒ Area of △ABC = (

\frac{1}{2}

×21×28) cm2
⇒Area of △ABC = (21×14)cm2
⇒ Area of △ABC = 294cm2......................… (2)
In right triangle △ABC, ∠B = 90∘
⇒AC2=AB2+BC2 (Pythagoras theorem)
AB=28, BC=21
⇒AC2= (28) 2+(21) 2
⇒AC2= 784+441
⇒AC2= 1225
We can write 1225=(35) 2
⇒AC2 = (35) 2
⇒AC2 = (35) 2
⇒AC = 35..................… (3)
(iii) Semicircle:
Semicircle is AEC
Diameter of the semicircle is AC
Since radius of semicircle is half the diameter of the semicircle

\Rightarrow r = \frac{AC}{2}

\Rightarrow r = \frac{35}{2} cm

⇒Area of semicircle AEC

= \frac{π}{2} \times {(\frac{35}{2})}^{2}

⇒Area of semicircle

AEC = \frac{22}{7 \times 2} \times \frac{35 \times 35}{2 \times 2}

⇒Area of semicircle AEC

= (\frac{11 \times 5 \times 35}{2 \times 2}) c m^{2}

⇒Area of semicircle AEC

= (\frac{1925}{4}) c m^{2}

⇒Area of semicircle AEC = 481.25cm2...................… (4)
We know the shaded region is given by subtracting the area of the quadrant from the sum of areas of semicircle and triangle.
Area of shaded area is given by subtracting equation (1) from sum of equations (2) and (4)
⇒Area of shaded region = (481.25+294) − (346.5 ) cm2
⇒Area of shaded region = 428.75cm2
∴ Area of shaded region is 428.75cm2
Now we have to calculate the perimeter of the shaded region.
(i) Quadrant:
Circumference of quadrant =

\frac{2 πr}{4}

r=21 as BC=21cm
⇒Circumference of quadrant

= (2 \times \frac{22}{7} \times \frac{21}{4}) cm

⇒Circumference of quadrant =(11×3)cm
⇒Circumference of quadrant =33cm………………….… (5)
(ii) Semicircle:
Circumference of semicircle

= \frac{2 πr}{2}

⇒Circumference of semicircle =πr

r = \frac{35}{2}

as AC=35cm is the diameter of the semicircle
⇒Circumference of semicircle

= (\frac{22}{7} \times \frac{35}{2}) cm

⇒Circumference of semicircle =(11×5)cm
⇒Circumference of semicircle =55cm……………….… (6)
(iii) AD:
We know AD=AB−BD
AB=28; BD=21
⇒AD= (28−21)cm
⇒AD= 7cm…………...… (7)
So, the perimeter of the shaded area is sum of equations (5), (6) and (7)
⇒Perimeter = (33+55+7)cm
⇒Perimeter = 95cm
∴ Perimeter of the shaded region is 95cm.

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