Ex 6.5,6
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Given:
Equilateral triangle ABC with each side 2a
Altitude AD is drawn such that AD BC
To find: AD
Solution:
In ADB and ADC
AB = AC
AD = AD
ADB= ADC
Hence ADB ADC
Hence , BD = DC
BD = DC
BD = DC = 1/2BC
BD = DC = 2 /2
BD = DC = a
Hence BD = a
Hence in right
Using Pythagoras theorem
(Hypotenuse)2 = (Height )2 + (Base)2
(AB)2 = (AD)2 + (BD)2
(2a)2 = (AD)2 + a2
4a2 = (AD)2 + a2
4a2 a2 = AD2
3a2 = AD2
AD2 = 3a2
AD = 3 a
AD = a 3
Thus, length of altitude AD = a 3
Similarly ,
length of altitude BE = a 3
length of altitude CF = a 3

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.