In figure below, $ \mathrm{ABCD} $ is a parallelogram, $ \mathrm{AE} \perp \mathrm{DC} $ and $ \mathrm{CF} \perp \mathrm{AD} $. If $ \mathrm{AB}=16 \mathrm{~cm}, \mathrm{AE}=8 \mathrm{~cm} $ and $ \mathrm{CF}=10 \mathrm{~cm} $, find $ \mathrm{AD} $.
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Given:
$ABCD$ is a parallelogram, $AE \perp DC$ and $CF \perp AD$.
$AB = 16\ cm, AE = 8\ cm$ and $CF = 10\ cm$.
To do:
We have to find $AD$.
Solution:
We know that,
Area of a parallelogram $=$ Base $\times$ Altitude
Therefore,
Area of parallelogram $ABCD = AB \times AE$
$= 16 \times 8$
$= 128\ cm^2$
Now,
Altitude $CF = 10\ cm$
This implies,
Area $=$ Base $(AD)\times$ Altitude$(CF)$
$128 = 10 \times AD$
$AD = \frac{128}{10}$
$AD = 12.8\ cm$
Hence, $AD = 12.8\ cm$.
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