ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2 .
Given, ΔABC is an isosceles triangle right angled at C.
In ΔACB, ∠C = 90°
AC = BC (By isosceles triangle property)
AB2 = AC2 + BC2 [By Pythagoras theorem]
= AC2 + AC2 [Since, AC = BC]
AB2 = 2AC2
In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm2. Find ar(∆COD)
Find the length of the altitude of an equilateral triangle of side 2a cm.
A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.
In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.
If ∆ABC ∼ ∆DEF such that 2 AB = DE and BC = 6 cm, find EF.