In $11$ cm $A B C D$ , $A B = 10 c m$ . The altitudes corresponding to the sides $A B$ and $A D$ are respectively $7 c m$ and $8 c m$ find $A D$ .

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Solution

Let the altitude on $A B$ is $D E = 7 c m$
$A D$ is $B F = 8 c m$
Area of parallelogram $=$ Base $\times$ height
When we take,
$A B$ as base and $D E$ as height
The area of $A B C D = 10 \times 7$
$= 70 c m^{2}$
When we take $A D$ as base and $B F$ as height
the area of $A B C D = A D \times 8$
As the area remain same, then
$A D \times 8 = A B \times A E = 70 A D = \frac{70}{8} = 8.75 c m$

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