Question

In ∆XYZ , P is any point on XY and PQ perpendicular to XZ. If XP = 4cm, XY = 16cm and XZ = 24 cm, find XQ.
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Answer with explanation .. ​

Answer 1

Given :

• In ∆XYZ , P is any point on XY and PQ perpendicular to XZ.
• XP = 4 cm
• XY = 16 cm
• XZ = 24 cm

To find :

• Value of XQ.

Solution :

∆ XYZ is a right triangle.

• $\angle\:XYZ=90\degree$

P is any point of XY.

PQ _|_ XZ

Then,

• $\angle\:XQP=90\degree$

In ∆ XYZ and ∆ XQP ,

$\sf{\angle\:XYZ=\angle\:XQP=90\degree}$

$\sf{\angle\:YXZ=\angle\:PXQ\:\:(Common\:angles)}$

Therefore,

∆ XYZ ≈ ∆ XQP

By properties of similar triangle →

$\implies\sf{\dfrac{XP}{XZ}=\dfrac{XQ}{XY}}$

$\implies\sf{\dfrac{4}{24}=\dfrac{XQ}{16}}$

$\implies\sf{24\:XQ=4\times\:16}$

$\implies\sf{XQ=\dfrac{4\times\:16}{24}}$

$\implies\sf{XQ=\dfrac{8}{3}}$

Therefore, XQ = 8/3 cm.