Triangle ABC is similar to triangle PQR- If AD and PM are altitudes of the two triangles- then AB - PQ - AD - PMA- AB - PQ - AD - PMB- AB - PM - AL - PC- A C-P D-m-P ND- AB - PR - AD - PM

The correct option is A AB/PQ=AD/PM Given, Δ ABC ∼Δ PQR AD and PM are altitude of these two triangles Since Δ ABC ∼Δ PQR, we have, ∠ B =∠ Q AB/PQ=BC/QR Now in ...

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Standard X

Mathematics

Similar Triangles

Triangle ABC ...

Question

Triangle ABC is similar to triangle PQR. If AD and PM are altitudes of the two triangles, then $\frac{A B}{P Q} = \frac{A D}{P M}$

$\frac{A B}{P M} = \frac{A D}{P Q}$

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$\frac{A B}{P R} = \frac{A D}{P M}$

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$\frac{A B}{P Q} = \frac{A D}{P M}$

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$\frac{A C}{P Q} = \frac{A D}{P M}$

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Solution

The correct option is C
$\frac{A B}{P Q} = \frac{A D}{P M}$

Given, $Δ A B C \sim Δ P Q R$

AD and PM are altitude of these two triangles

Since $Δ A B C \sim Δ P Q R$ , we have,

$∠ B = ∠ Q$

$\frac{A B}{P Q} = \frac{B C}{Q R}$

Now in $Δ A B D$ and $Δ P Q M$

$∠ B = ∠ Q$

$∠ D = ∠ M$

$∴ Δ A B D \sim Δ P Q M$

$∴ \frac{A B}{P Q} = \frac{A D}{P M}$

Suggest Corrections

Similar questions

Q. State true or false:

Triangle

A B C

is similar to triangle

P Q R

. If

A D

and

P M

are altitudes of the two triangles, then

\frac{A B}{P Q} = \frac{A D}{P M} .

Triangle ABC is similar to triangle PQR. If AD and PM are corresponding medians of the two triangles, prove that : $\frac{A B}{P Q} = \frac{A D}{P M}$ .