Question

30 POINTS, Im pretty hopeless at this point and i know someone is just gonna rob the points thanks anyways

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

Answer 1

`(ah)/(a+h)`

Explanation:

Consider similar triangles BMC and KMN.

In these triangles,

`(BC)/(KN)=(MO)/(MP),`

where O is the point of intersection of BC and MP.

Note that segment OP has the length the same as the side of the square ABCD. Hence,

`(BC)/(a)=(h-BC)/(h),\\ \\BC\cdot h=a(h-BC),\\ \\BCh=ah-aBC,\\ \\BC(h+a)=ah,\\ \\BC=(ah)/(a+h).`