## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 66 of Exercise 10E

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66
##### The area of a right triangle is 600 cm2. 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

x2 + 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

x2 + 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

Hypotenuse2 = altitude2 + base2

Hypotenuse2 = (30)2 + (40)2

= 900 + 1600 = 2500

Hypotenuse = 50 cm

Altitude = 30cm

Base = 40cm

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