In $△ A B C, ∠ A B C = 90^{o}, B M ⊥ A C$ , $B M = x + 2, A M = x + 7, C M = x$ , find $x$

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Solution

It is given that $∠ A B C = 90^{0}$ and $B M ⊥ A C$ as shown in the above figure:

Now with $B M = (x + 2)$ , $C M = x$ and $A M = (x + 7)$ , consider,

$B M^{2} = C M \times A M \Rightarrow (x + 2)^{2} = x \times (x + 7) \Rightarrow x^{2} + 4 x + 4 = x^{2} + 7 x \Rightarrow x^{2} + 4 x + 4 - x^{2} - 7 x = 0 \Rightarrow - 3 x + 4 = 0 \Rightarrow - 3 x = - 4 \Rightarrow x = \frac{4}{3}$

Hence, $x = \frac{4}{3}$ .

Suggest Corrections

△ A B C, m ∠ B = 90^{o}, ¯ ¯¯¯¯¯¯¯¯ ¯ B M ⊥ ¯ ¯¯¯¯¯¯ ¯ A C, M \in A C

A M - M C = 7

{A B}^{2} - {B C}^{2} = 175

A C

In $Δ A B C$ ; $B M ⊥ A C$ and $C N ⊥ A B$ ;

Then, $\frac{A B}{A C} = \frac{B M}{C N} = \frac{A M}{A N}$

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