Question

The area of a AABD in Fig. 13.51 is 24 cm². If DE = 6 cm, and AB || CD, BD || CE,
AE || BC, find

(i) Altitude of the parallelogram BCED.
(ii) Area of the parallelogram BCED.
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Answer 1

First off all I have to assume that the parallelogram is not AABD but triangle ABD

Triangle ABD and BCDE lie on the same base AD and between the same parallels BC and AD.

Hence, ar triangle ABD = 1/2 ar llgm BCED = 24 cm^2

So, ar llgm BCED = 24×2=48cm^2

Now, DE = 6cm.

Therefore, Area = 1/2 × base × height

Calculating, altitude (or height) = 2area/base = 2×48/6 = 16cm

[Please correct me if I am wrong]

Hope this helps!

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