Book: RS Aggarwal - Mathematics
Chapter: 10. Quadratic Equations
Subject: Maths - Class 10th
Q. No. 68 of Exercise 10E
Listen NCERT Audio Books to boost your productivity and retention power by 2X.
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
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.
