The area of a right – angled triangle is 165 sq meters. Determine its base and altitude if the latter exceeds the former by 7 meters.

Let the base be x m

The altitude will be x + 7 m


Area of triangle = 1/2 base. altitude


= 1/2 x (x + 7) = 165


x2 + 7x = 330


x2 + 7x – 330 = 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 = 7 c = – 330


= 1. – 330 = – 330


And either of their sum or difference = b


= 7


Thus the two terms are 22 and – 15


Difference = 22 – 15 = 7


Product = 22. – 15 = – 330


x2 + 22x – 15x – 330 = 0


x(x + 22) – 15(x + 22) = 0


(x + 22) (x – 15) = 0


x = – 22 or x = 15


Value of x cannot be negative


x = 15


Thus the base be 15m and altitude = 15 + 7 = 22m


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