Question

The difference between the sides containing a right angle in a right triangle is 14cm. The area of the triangle is 120cm². Calculate the perimeter of a triangle.

Answer 1

$let \: \: sides \: are \: = x \: and \: (x - 14) \\ area \: of \: triangle = \frac{1}{2} x(x - 14) \\ 120 \times 2= x(x - 14) \\ 240 = {x}^{2} - 14x \\ {x}^{2} - 14x - 240 = 0 \\ {x}^{2} - 24x + 10x - 240 = 0 \\ (x - 24)(x + 10) = 0 \\ x = 24 \: \: or \: \: - 10 \: \\ x = 24cm \\ sides \: of \: triangle = 24cm \: and \: (24 - 14)cm \: and \sqrt{ {24}^{2} + {10}^{2} } \\ = 24cm \: \: \: 10cm \: \: \: \: 26cm \\ perimeter \: = 24 + 10 + 26 = 60cm$

Answer 2

we know that, the difference between the sides in right triangle is difference between hypotenuse and base or perpendicular.

Let the perpendicular be x cm
Base be y cm
Hypotenuse be h cm

There will be two Case

(a) When x = 14cm
(b) When y = 14cm

(a) When x =14cm then,

Area of right triangle = xy/2

Area = 120 cm^2

120 = 14×y/2
120 = 7y
y = 120/7
y = 17.14 cm

h^2 = x^2 + y^2
h^2 = (14)^2 + (17.14)^2
h^2 = 196 + 293.77
h^2 = 489.77
h = 22.13 cm

perimeter = x+y+h
= 14+17.14+22.13
= 53.27

similarly in second Case also perimeter will be 53.27 cm

Thanks for watching the answer.