Question

XYZ is a right angled triangle with angle=90,XY=9cm,and YZ=12cm.OX is perpendicular to YZ.Find OX​

Answer 1

Answer :-

• OX = 9√7/4 cm

Given :-

• Δ XYZ is right angle triangle at X
• XY = 9 cm
• YZ = 12 cm
• OX ⊥ YZ

To Find :-

• OX = ?

Explanation :-

➠ In this figure,

$\rm \implies \triangle YOX \sim \triangle YXZ$

$\boxed { \rm \therefore \dfrac{YO}{YX} = \dfrac{OX}{XZ} = \dfrac{YX}{YZ} \qquad ...........(1) }$

➠ In Δ XYZ,

Δ XYZ is right angle at X ; By applying Pythagoras theorem, we get

$\rm \implies (12)^2 - (9)^2 = (XZ)^2$

$\rm \implies 144 - 81 = (XZ)^2$

$\rm \implies (XZ)^2 = 63$

$\boxed { \bf \implies \sqrt{63} \; cm }$

➠ From Equation (1),

$\rm \implies OX = \dfrac{XZ \times YX}{YZ}$

$\rm \implies OX = \dfrac{\sqrt{63} \times 9}{12}$

$\red{\underline { \boxed { \bf \implies OX = \dfrac{ 9 \sqrt{7}}{4} \; cm }}}$