Q. 6 D5.0( 1 Vote )

# m∠<s

AB =

AM = 2.5

Applying Pythagoras theorem in ΔABM

AB2 = AM2 + BM2

BM2 = 10 – 6.25 = 3.75

Let CM = x

AC = x + 2.5

Applying Pythagoras theorem in ΔCBM

BC2 = BM2 + CM2

BC2 = x2 + 3.75 …Equation (i)

Applying Pythagoras theorem in ΔABC

AC2 = AB2 + BC2

(x + 2.5)2 = BC2 + 10

x2 + 6.25 + 5x = BC2 + 10

BC2 = x2 – 3.75 + 5x …Equation (ii)

Equating Equation (i) and Equation (ii)

x2 + 3.75 = x2 – 3.75 + 5x

5x = 7.5

x = 1.5

So CM = 1.5

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