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 distance between AE and BD.

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

Therefore, the distance between AE and BD is 4 m.