$A M = 8 x^{2}, M C = 2 x^{2}$ , then find BM and AB.

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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 $A M = 8 x^{2}$ and $M C = 2 x^{2}$ , consider,

$B M^{2} = A M \times M C = 8 x^{2} \times 2 x^{2} = 16 x^{4} \Rightarrow B M = \sqrt{16 x^{4}} = 4 x^{2}$

Similarly, with $A M = 8 x^{2}$ and $M C = 2 x^{2}$ , consider,

$A B^{2} = A C \times A M \Rightarrow A B^{2} = (A M + M C) \times A M \Rightarrow A B^{2} = (8 x^{2} + 2 x^{2}) \times 8 x^{2} \Rightarrow A B^{2} = 10 x^{2} \times 8 x^{2} \Rightarrow A B^{2} = 80 x^{4} \Rightarrow A B = \sqrt{80 x^{4}} \Rightarrow A B = 4 x^{2} \sqrt{5}$

Hence, $B M = 4 x^{2}$ and $A B = 4 x^{2} \sqrt{5}$ .

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Similar questions

△ A B C, ∠ A B C = 90^{o}

AB = 4 BC

△ 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 the given figure, ∠ABC = $90^{\circ}$ and BM is a median, AB = 4 cm and BC = 3 cm. Then, choose the correct option(s).

Δ

∠

= 90

¯ ¯¯¯¯¯¯¯¯ ¯ B M ⊥ ¯ ¯¯¯¯¯¯ ¯ A C

M \in ¯ ¯¯¯¯¯¯ ¯ A C

-

= 7

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

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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