Question

Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.

Answer 1

Let the three sides of the triangle be
          base = b
        height = h
hypotenuse = H

b² + h² = H²
⇒(b+h)² - 2bh = H²

Area = 150
⇒(1/2)bh = 150
⇒bh = 300
Perimeter = 60
⇒ b + h + H  = 60
⇒b + h = 60-H
 
Thus (b+h)² - 2bh = H²
⇒(60-H)² -2×300 = H²
⇒ 3600 + H² - 2×60×H - 2×300 = H²
⇒3600 - 120H -600 = 0
⇒ -120H + 3000 = 0
⇒120H = 3000
⇒ H = 3000/120 = 25cm

b+h = 60-25 = 35
⇒300/h + h = 35
⇒h² + 300 = 35h
⇒h² -35h + 300 = 0
⇒ h² -15h -20h +300 = 0
⇒(h-15)(h-20) = 0
⇒h = 15 or 20
⇒b = 20 or 15

Thus Three sides are (15,20,25)

Answer 2

let a right triangle,
in which   , perpendicular = p
                   base = b    , and diagonal  = h
A/q,
            from pythagoras theorem .
p²+b² =  h²  --------------------------(1) 

Area of right tringle  = 1/2×b.p
1/2×b.p = 150
or ,   bp = 300 -------------------(2)
&
           perimeter = p+b+h 

or,            p+b+h = 60  
or,                p+b = 60-h -------------------(3)
Now squaring both side of the equation (3).,
           (p+b)² = (60-h)²
or,        p²+b²+2bp   = 60²+h²-2×60.h

or,        h²+2(300) = 3600+h²-120h

or,        120h = 3600-600 = 3000
 or,             h = 3000/120 = 25 cm

so from eqn (3) ,p+b = 60-25  = 35 ------------(4)
We know that,      (p+b)²-(p-b)² = 4.pb
or,           (35)²+(p-b)² = 4 (300)                       {from eqns (3) & (2)
or,            (p-b)² = 1200 - 1225 = 25
 or,            p-b = (+-) 5    ------(5)
On adding the Eqns (4) & (5)
               2p = 35(+-) 5
so, either          p = (35+5)/2 = 20 cm   ,then b = 35-20 = 15 cm
         or,        p = (35-5)/2 = 15 cm , then b =  35-15 = 20 cm
Therefore sides of the right triangle are 25 cm, 20 cm and 15 cm .