Question

In the given figure, 'O' is a point inside a $\Delta PQR$ such that $\angle POR={90}^{\circ },$Op=6 cm and OR=8 cm. If PQ = 24 cm and $\angle QPR={90}^{\circ },$ then QR=

• A. 28 cm
• B. 25 cm
• C. 26 cm
• D. 32 cm

Answer 1

The correct option is C 26 cm

$\Rightarrow$ $\angle POR={90}^o$ [ Given ]

$\therefore$ $△POR$ is right-angled triangle.

$PO=6cm$ and $OR=8cm$ [ Given ]

$\Rightarrow$ $(PR{)}^2=(PO{)}^2+(OR{)}^2$

$\Rightarrow$ $(PR{)}^2=(6{)}^2+(8{)}^2$

$\Rightarrow$ $(PR{)}^2=36+64$

$\Rightarrow$ $(PR{)}^2=100$

$\therefore$ $PR=10cm$

Now,

$\Rightarrow$ $\angle QPR={90}^o$

$\therefore$ $△PQR$ is right-angled triangle.

$\Rightarrow$ $PQ=24cm$ [ Given ]

$\Rightarrow$ $(QR{)}^2=(PQ{)}^2+(PR{)}^2$

$\Rightarrow$ $(QR{)}^2=(24{)}^2+(10{)}^2$

$\Rightarrow$ $(QR{)}^2=576+100$

$\Rightarrow$ $(QR{)}^2=676$

$\Rightarrow$ $QR=26cm$