# In $\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}$ and $\mathrm{BM}$ is an altitude. If $\mathrm{AM}=2 x^{2}$ and $\mathrm{CM}=8 x^{2}$, find $\mathrm{BM}$, AB and $\mathrm{BC}$.

Given:

In $\triangle \mathrm{ABC}, \angle \mathrm{B}=90^{\circ}$ and $\mathrm{BM}$ is an altitude.

$\mathrm{AM}=2 x^{2}$ and $\mathrm{CM}=8 x^{2}$

To do:

We have to find $\mathrm{BM}$, AB and $\mathrm{BC}$.

Solution:

In $\triangle ABC$,

By Pythagoras theorem,

$AC^2 =AB^2+BC^2$

$(2x^2+8x^2)^2=AB^2+BC^2$

$(2x^2)^2+(8x^2)^2+2\times2x^2\times8x^2=AB^2+BC^2$

$(2x^2)^2+(8x^2)^2+32x^4=AB^2+BC^2$.........(i)

Similarly,

In $\triangle ABM$,

By Pythagoras theorem,

$AB^2 =AM^2+BM^2$

$AB^2=(2x^2)^2+BM^2$......(ii)

In $\triangle BMC$,

By Pythagoras theorem,

$BC^2 =MC^2+BM^2$

$BC^2=(8x^2)^2+BM^2$......(iii)

From (i), (ii) and (iii),

$(2x^2)^2+(8x^2)^2+32x^4=(2x^2)^2+BM^2+(8x^2)^2+BM^2$

$32x^4=2BM^2$

$16x^4=BM^2$

$BM^2=(4x^2)^2$

$BM=4x^2$

Therefore,

$AB^2=(2x^2)^2+16x^4$

$=4x^4+16x^4$

$=20x^4$

$\Rightarrow AB=\sqrt{20x^4}$

$=2\sqrt5x^2$

$BC^2=(8x^2)^2+16x^4$

$=64x^4+16x^4$

$=80x^4$

$\Rightarrow BC=\sqrt{80x^4}$

$=4\sqrt5x^2$

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Updated on: 10-Oct-2022

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