Question

The difference between the sides at right angles in a right angled triangle is 14 cm. The area of the triangle is 120sq.cm. find the perimeter.

Answer 1

Let one side at right angle be x
the other must be x-14

area of a triangle =(1/2) × (height) (base)
120 = (1/2)(x)(x-14)
120 = (x^2 -14x)/2
240 = x^2 -14x

x^2 -14x -240 = 0
x^2 -24x +10x -240 =0
x (x-24) + 10 (x-24) =0
(x-24)(x+10)=0

now, find x
x-24 =0
x = 24

again
x+10 =0
x = -10

as the side couldn't be -v so the value of x as a side must be 24cm

so the measure of one side we got is x =24cm and the other side = x-14 = 24 -14 =10cm

apply pythagoras theorem for third side as it is a right angled triangle

[let hypotaneous be a]

(24)^2 + (10^2) = a^2
576 +100 = a^2
676 = a^2
a = 26 cm

so the hypataneous = 26cm

perimeter of a triangle = sum of all sides
= (24 + 10 + 26 ) cm
= 60cm