#### Question

ABCD is a square. F is the mid-point of AB. BE is one third of BC. If the area of ΔFBE = 108 cm^{2}, find the length of AC.

#### Solution

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 cm^{2}

`rArr1/2xxBExxFB=108`

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

⇒ 𝑥^{2} = 108 × 2 × 3 × 2

⇒ 𝑥^{2} = 1296

`rArrx=sqrt1296=36`cm

In ΔABC, by pythagoras theorem AC^{2} = AB^{2} + BC^{2}

⇒ AC^{2} = 𝑥^{2} + 𝑥^{2}

⇒ AC^{2} = 2𝑥^{2}

⇒ AC^{2} = 2 × (36)^{2}

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

Is there an error in this question or solution?

#### APPEARS IN

Solution 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. Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle.