Ex 4.2, 5 - Chapter 4 Class 10 Quadratic Equations

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 4.2, 5
The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Let ABC be right angled triangle,
with altitude = AB, Base = BC & Hypotenuse = AC
Given
Hypotenuse = AC = 13 cm
And,
Altitude is 7 cm less than base
Let Base = BC = x cm
∴ Altitude = AB = Base – 7 = x – 7
Since ABC is a right angled triangle
Using Pythagoras theorem
Hypotenuse2 = Height2 + Base2
(AC) 2 = AB2 + BC2
(13)2 = (x – 7)2 + x2
169 = x2 + 49 – 14x + x2
169 = 2x2 – 14x + 49
0 = 2x2 – 14x + 49 – 169
0 = 2x2 – 14x – 120
2x2 – 14x – 120 = 0
2 (x2 – 7x – 60) = 0
x2 – 7x – 60 = 0
We factorize by
splitting the middle term method
x2 + 5x – 12x – 60 = 0
x (x + 5) – 12 (x + 5) = 0
(x – 12) (x + 5) = 0
So, x = 12 , x = –5 are the roots of the equation
Since x is length, it cannot be negative
Thus, x = 12
Therefore,
Base = x = 12 cm
Altitude = x – 7 = 12 – 7 = 5 cm
x – 12 = 0
x = 12
x + 5 = 0
x = –5

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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