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Q 78
.

D is the mid point of BC and AE ⊥ BC.

In right ∆ ABE,

AB² = AE² + BE² … (1) (Pythagoras Theorem)

In right ∆ ADE,

AD² = AE² + ED² … (2) (Pythagoras Theorem)

From (1) and (2), we get

Since, AB= c, AD= p, BC = a, DE = x ……………….(Given)

Substituting the above values, we get

\Rightarrow c^{2} + a x - \frac{a^{2}}{4} = p^{2}

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