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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. - Mathematics

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.

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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 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

Concept: Application of Pythagoras Theorem in Acute Angle and Obtuse Angle
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 7 Triangles
Exercise 7.7 | Q 11 | Page 120
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