Question

The difference between the sides at the right angles in a right angled triangle is 14 cm.The area of the triangle is 120 cm^2.Calculate the perimeter of the triangle.​

Answer 1

Step-by-step explanation:

Let the sides be b , h & H

1/2 (bh) = 0.5 (bh) = 120

240 = bh

b= 240/h

Difference = b - h = 14

( 240/h ) - h = 14

240 - h^2 = 14h

h^2 + 14h -240 = 0

h= [-14 (+/-)√(196+960)]/2

= -14 (+/-) 34}/2

= 10

b = 240/h = 240/10 = 24

perimeter = a+b+c = 24+10+√676

= 34 + 26 = 60

Answer 2

Answer:-60 cm

Given:

Difference between the sides at right angles=14 cm

Area of ∆=120 cm^2

To find:Perimeter of ∆

Solution:-

Let the sides containing the right angle be x cm and (x-14)cm

Then;

Area if triangle

$\frac{1}{2} \times x \times (x - 14) {cm}^{2}$

Area is given 120 cm^2

$\frac{1}{2}x(x - 14) = 120$

${x}^{2} - 24x + 10x - 240$

$x(x - 24) + 10(x - 24)$

$(x - 24)(x + 10) = 0$

Neglecting x=-10

$x = 24$

--------

One side=24 cm

Other side=24-14

=10 cm

By using Hypotenuse

${ac}^{2} = {ab}^{2} + {bc}^{2}$

${ac}^{2} = {24}^{2} + {10}^{2}$

${ac}^{2} = 576 + 100$

$ac = \sqrt{676}$

$ac = 26 \: cm$

-------------

Perimeter of the triangle=(AB+BC+CA)

=24+10+26

=60 cm