Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 66 of Exercise 10E
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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
minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
