Question

ABCD is a square. F is the midpoint of AB. BE is one third of BC. If the area of
triangle FBE is 108 cm2 . Find the length of AC.

Answer 1

 In square ABCD, let AB = BC = CD = DA = xSince F is midpoint of AB, BF = x/2 Given that BE = (1/3) BC =(x/3) Area of ΔFBE = 108 sq cm Hence AB = BC = CD = DA = 36 cm Recall that diagonal of a square is √2 times its side ⇒ AC = √2AB ∴ AC = 36√2 cm

Answer 2

Given : ABCD is a square. F is the midpoint of AB. BE = ⅓ BC. ar∆FBE = 108 cm²

Let the side of a square be x cm . Therefore AB = BC = CD = DA = x cm

Since, F is the mid-point of AB

Then, AF = FB = AB/2 = x/2  

Since, BE =  ⅓ BC  = x/3

BE = x/3

ar ∆FBE = 108 cm²

½ × BE × FB = 108 cm²

½ × x/2 × x/3 = 108

x² = 108 × 2 × 3 × 2

x²  = 1296

x = √1296  

x = 36 cm

AB = BC = CD = DA = 36 cm

In ∆ABC,

AC² = AB² + BC²

[By using pythagoras theorem]

AC² = x² + x²

AC² = 2x²

AC = √2x

AC = √2 × 36

AC = 36√2

AC = 36√2  

AC = 36 × 1.414  

[Value of √2 = 1.414]

AC = 50.904 cm

Hence, the length of AC is 50.904 cm

Step-by-step explanation:

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