Question

ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ∆FBE = 108 cm², find the length of AC.

Answer 1

Given, 

ABCD is a square. And, F is the mid-point of AB.

BE is one third of BC. 

Area of ∆ FBE = 108 cm² 

Required to find: length of AC

Let’s assume the sides of the square to be x. 

⇒ AB = BC = CD = DA = x cm 

And, AF = FB = \(\frac{x}{2}\) cm 

So, BE = \(\frac{x}{3}\) cm 

Now, the area of ∆ FBE = \(\frac{1}{2}\) x BE x FB 

⇒ 108 = (\(\frac{1}{2}\)) x (\(\frac{x}{3}\)) x (\(\frac{x}{2}\)) 

⇒ x² = 108 x 2 x 3 x 2 = 1296 

⇒ x = √(1296) [taking square roots of both the sides] 

∴ x = 36 cm 

Further in ∆ ABC, by Pythagoras theorem, we have 

AC² = AB² + BC² 

⇒ AC² = x² + x² = 2x² 

⇒ AC² = 2 x (36)² 

⇒ AC = 36√2 

= 36 x 1.414 

= 50.904 cm 

Therefore, the length of AC is 50.904 cm.