Question

ABC is an isosceles triangle in which the equal sides are AB and AC if AD is an altitude and /BC/=1/2(AB) prove that /AD/²=15(BD)²

Answer 1

let BD = x
since the altitude in an isosceles right-bisects the base,
BC = 2x
we are given that BC = 1/2 AB
so AB = 4x

now by Pythagoras:
AB^2 = BD^2 + AD^2
16x^2 = x^2 + AD^2
AD^2 = 15x^2 , but we let BD = x
AD^2 = 15 BD^2

Answer 2

LMN is an equilateral triangle and X is the mid-point of LN.prove that MX=3/4MN.

Answer 3

LET BD=X
SINCE THE ALTITUDE IN AN ISOSCELES RIGHT BISECTS THE BASE
BC=2X
WE ARE GIVEN THAT BC=1/2AB
SO AB=4X
BY PYTAGORAS
AB^2=BD^2 + AD^2
15X^2=X^2 + AD^2
AD^2=15X^2 BD=X
/AD/^2=15/BD/^2

Answer 4

True well well