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)
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