Question

ABCD is a square and P,Q are the midpoints of BC,CD respectively.If AP=a and AQ=b,

Answer 1

a = b

let x be the side of the square

$\vartriangle ADQ:$

AD = x; DQ = $\dfrac{x}{2}$; AQ = b = a

by the Pythagorean theorem:

$x^2 + \dfrac{x^2}{4} = a^2 = b^2$

$4x^2 + x^2 = 4a^2 = 4b^2$

$5x^2 = 4a^2 => x^2 = \dfrac{4a^2}{5} = \dfrac{4b^2}{5}$ - area of a square ABCD