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.
Answers
SOLUTION :
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
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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
AC² = x² + x²
AC² = 2x²
AC = √2x
AC = √2 × 36
AC = 36√2
AC = 36√2
AC = 36 × 1.414
AC = 50.904 cm