The area of a right triangle is 600 cm^{2}. If the base of the triangle exceeds the altitude by 10 cm, find the dimensions of the triangle.

Let the altitude of the given triangle x cm

Thus the base of the triangle will be (x + 10)cm

Area of triangle =

x (x + 10) = 1200

x^{2} + 10x – 1200 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 10 c = – 1200

= 1. – 1200 = – 1200

And either of their sum or difference = b

= 10

Thus the two terms are 40 and – 30

Difference = 40 – 30 = 10

Product = 40. – 30 = – 1200

x^{2} + 40x – 30x – 1200 = 0

x(x + 40) – 30(x + 40) = 0

(x + 40)(x – 30) = 0

x = – 40, 30

x = 30 (altitude cannot be negative)

Thus the altitude of the given triangle is 30cm and base of the triangle = 30 + 10 = 40cm

Hypotenuse^{2} = altitude^{2} + base^{2}

Hypotenuse^{2} = (30)^{2} + (40)^{2}

= 900 + 1600 = 2500

Hypotenuse = 50 cm

Altitude = 30cm

Base = 40cm

66