Question

the length of the two sides a right triangle containing the right angles differ by 2 centimetre if the area of a triangle is 24 cm square find the perimeter of a triangle

Answer 1

Let the base be a

Perpendicular = a+2 

Area = base * Perpendicular / 2

=> 24 = a(a+2) / 2

=> 48 = a^2 + 2a

=> a^2 +2a -48 = 0

=> a^2 +8a - 6a -48 = 0

=> a(a+8) - 6(a+8) = 0

=> (a+8)(a-6) = 0

a = 6  and -8

But Length cannot be negative,

so,  a = 6

Base = 6 cm

Perpendicular = 8 cm

By Pythagores theoram

H = sqrt (8^2 +6^2)

H = sqrt 64 + 36

H = sqrt 100

Hypotenuse = 10 cm

Perimeter = 10 + 8  + 6

 =  24 cm

Answer 2

Let the length of one side be x and the other side will be x + 2.

Given that Area of triangle = 24cm^2.

1/2 * b * h = 24

1/2 * x * x + 2 = 24

x^2 + 2x = 48

x^2 + 2x - 48 = 0

x^2 - 6x + 8x - 48 = 0

x(x - 6) + 8(x - 6) = 0

(x - 6)(x + 8) = 0

x = 6 (or) x = -8.

Since the value of x cannot be negative. So, x = 6.

Therefore,

One side is x = 6cm.

Another side is x = x + 2

                            = 8cm.



Now,

Hypotenuse^2 = 8^2 + 6^2

                         = 64 + 36

                         = 100

$Hypotenuse = \sqrt{100}$

                             = 10cm.



                             
Therefore the perimeter of the triangle = 8 + 6 + 10

                                                                   = 24cm.


Hope this helps!