

Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at May 29, 2018 by Teachoo
Transcript
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
Ex 6.5
Ex 6.5, 2 Important Not in Syllabus - CBSE Exams 2021
Ex 6.5, 3 Important Not in Syllabus - CBSE Exams 2021
Ex 6.5, 4
Ex 6.5, 5
Ex 6.5, 6 You are here
Ex 6.5, 7
Ex 6.5, 8 Important
Ex 6.5, 9
Ex 6.5, 10
Ex 6.5, 11 Important
Ex 6.5, 12 Important
Ex 6.5, 13
Ex 6.5, 14
Ex 6.5, 15 Important
Ex 6.5, 16
Ex 6.5, 17
About the Author