Question

The figure alongside represents the horizontal cross-section of a pond. ABDE is an isosceles trapezium in which AE = 1 m, ED = 5 m, BD = 7 m. BCD is a semicircle on diameter BD. The sides of the pond are vertical and the depth of the water in the pond is 50 cm. Find the area of the sides in contact with water.

Answer 1

We have,

The figure alongside represents the horizontal cross-section of a pond. 

ABDE is an isosceles trapezium in which AE = 1 m, ED = 5 m, BD = 7 m.

BCD is a semicircle on diameter BD.

The sides of the pond are vertical and the depth of the water in the pond is 50 cm.

Now draw perpendiculars AM and EN on BD,

Then, BM = 3 m, DN = 3 m and MN = 1 m

Now, in ΔABM

By Pythagoras theorem

AB² = AM² + BM²

5² = AM² + 3³

AM² = 25 – 9 = 16

AM = 4 m

Similarly, EN = 4 m

Area of trapezium ABDE is `1/2 (AE + BD) xx AM`

= `1/2 1 + 7 xx 4`

= 16 m²

And the area of semicircle BCD is `1/2 π(r)^2`

= `1/2 xx 22/7 xx 7/2 xx 7/2`

= 19.25 m²

Then the area of the given cross-section

= Area of trapezium ABDE + Area of semicircle BCD

= 16 + 19.25

= 35.25 m²