Question

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

Answer 1

Since, ABCD is a square

Then, AB = BC = CD = DA = x cm

Since, F is the mid-point of AB

Then, AF = FB = `x/2`cm

Since, BE is one third of BC

Then, BE = `x/3`cm

We have, area of ΔFBE = 108 cm²

`rArr1/2xxBExxFB=108`

`rArr1/2xx x/2xx x/3=108`

⇒ x² = 108 × 2 × 3 × 2

⇒ x² = 1296

`rArrx=sqrt1296=36` cm

In ΔABC, by pythagoras theorem AC² = AB² + BC²

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC² = 2 × (36)²

⇒ AC = 36`sqrt2` = 36 × 1.414 = 50.904 cm