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From a rectangular region ABCD with AB = 20 cm, a right angle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (π = 22/7)

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Very important question of mensuration including
very important concept of it.need solution!
Rd sharma class 10 maths exercise-15.4 mensuration.

we have to find the area of the shaded region from
a rectangular region ABCD with AB = 20 cm, a right
angle AED with AE = 9 cm and DE = 12 cm, is cut off.
On the other end, taking BC as diameter, a semicircle
is added on outside the region.(π = 22/7)

 

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1 Answer

  1. Given,

    Length of the rectangle ABCD = 20 cm

    AE = 9 cm and DE = 12 cm

    The radius of the semi-circle = BC/ 2 or AD/2

    Now, using Pythagoras theorem in triangle AED

    AD = √(AE2 + ED2) = √(92 + 122)

    = √(81 + 144)

    = √(225) = 15 cm

    So, the area of the rectangle = 20 x 15 = 300 cm2

    And, the area of the triangle AED = ½ x 12 x 9 = 54 cm2

    The radius of the semi-circle = 15/2 = 7.5 cm

    Area of semi-circle = ½ π(15/2)2 = ½ x 3.14 x 7.52 = 88.31 cm2

    Thus,

    The area of the shaded region = Area of the rectangle ABCD + Area of semi-circle – Area of triangle AED

    = 300 + 88.31 – 54

    = 334.31 cm2

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