Ex 6.5, 6 - Chapter 6 Class 10 Triangles (Term 1)

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