Question

Find a point on the hypotenuse of a right angled triangle from which perpendiculars can be dropped on other sides to form a rectangle of max area

Answer 1

Answer:

Mid point  of hypotenuse

Step-by-step explanation:

Let say ABC is right angled at B

and D is the point on hypotenuse AC

DP ⊥ AB   & DQ ⊥ BC

Δ ADP ≅ ΔABC

=> AD/AC = AP/AB = DP/BC

=> DP = AD * BC/AC

Δ CDQ ≅ ΔABC

=> CD/AC = CQ/BC = DQ/AB

=> DQ = AB * CD/AC

Area of rectangle DP * DQ

=  (AD * BC/AC)(AB * CD/AC)

= (AB * BC  * AD * CD)/(AC²)

= (AB * BC /AC²) ( AD * CD)

= (AB * BC /AC²)( AD * (AC - AD))

differentiating wrt AD as that is only variable

(AB * BC /AC²) ( AD (-1) + (AC - AD)  =0

=> AC = 2AD

=> AD = AC/2

=> D is the mid point of AC

Second differention is - ve so  AD = AC/2 will give max Area

point on the hypotenuse of a right angled triangle from which perpendiculars can be dropped on other sides to form a rectangle of max area is mid point  of hypotenuse