Question

Given: △KMN, ABCD is a square KN=a, MP ⊥ KN , MP=h .Find: AB

Answer 1

Solution :-
Given


Let the side of square ABCD be x units.

∠CDN=90° -> ∠CDP=90°

CP||MP||AB

In ΔMNP and ΔCND

∠NCD = ∠NMP     ( Since corresponding angles )

∠NDC = ∠NPM     ( Since corresponding angles )

ΔMNP similar to ΔCND

ΔKAP similar to ΔKPM by similarity rule as above

Therefore,

ND / NP = CD / MP

KA / KP = AB / PM

ND / NP = x / h

KA / KP = x / h

Now, NP / ND = h / x - 1

        KP / KA = h / x - 1

PD / ND  = h/x -1 , Therefore PD = (h/x - 1) x ND

AP / KA = h/x -1 , Therefore AP = (h/x - 1) x KA

Adding above 2,

(KA + ND) (h/x - 1) =  AP + PD

(KN - ND) = x / (h/x - 1 )

a - x = x / (h/x - 1)

x ( h + a) = ah

x = ah / (a + h)

=============
@GauravSaxena01