In a Δ A B C , A B = B C = C A = 2 a and A D ⊥ B C . Prove that A D = a 3 .

Given: ΔABC , AB = BC = CA = 2 a , it means ABC is an equilateral triangle. We know that in an equilateral triangle the altitude also bisects the base into two equal parts. Therefore, BD = CD = a We can observe that ADB and ADC are two right-angled triangles. Now we will use Pythagoras theorem in right-angled triangle ADB . A B 2 = A D 2 + B D 2 Now we will substitute the values of AB and BD we get, ( 2 a ) 2 = A D 2 + a 2 4 a 2 = A D 2 + a 2 3 a 2 = A D 2 ∴ A D = 3 a . Therefore, A D = 3 a .

In a Δ A B C , A B = B C = C A = 2 a and A D ⊥ B C . Prove that A r e a ( Δ A B C ) = 3 a 2 .

We have been given an equilateral triangle and we know that using Pythagoras theorem the area of an equilateral triangle is found to be = 3 4 × s i d e 2 Here, sides is 2 a . ∴ Ar ( Δ A B C ) = 3 4 × ( 2 a ) 2 ∴ Ar ( Δ A B C ) = 3 4 × 4 a 2 Ar ( ΔABC ) = 3 a 2 Therefore, Ar ΔABC = 3 a 2 .