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