Answer 1

Given ∠AED = 90°

ED = x, DC = `"a"/2`    ...(D is the mid point of BC)

∴ EC = `x + "a"/2`, BE = `"a"/2 - x`

∴ In the right ∆AED

AD² = AE² + ED²

p² = h² + x²

In the right ∆AEC,

AC² = AE² + EC² 

b² = `"h"^2 + (x + "a"/2)^2`

= `"h"^2 + x^2 + "a"^2/4 + 2 xx x xx "a"/2`

b² = `"p"^2 + "a"^2/4 + "a"x`

b² = `"p"^2 + "a"x + 1/4 "a"^2`