## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 68 of Exercise 10E

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68
##### 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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