Question

In the figure: PQ = 10 cm, QS = 6 cm and RQ = 9 cm.
The length of side PR is.

• A. 15 cm
• B. 17 cm
• C. 19 cm

Answer 1

The correct option is B 17 cm

Given that,
$\angle PSQ={90}^{\circ },$
PQ = 10 cm, QS = 6 cm and RQ = 9 cm.

In the right-angled triangle $\Delta PSQ$,

$P{Q}^2=P{S}^2+Q{S}^2$
[By Pythagoras theorem]

${10}^2=P{S}^2+6^2$

$\Rightarrow P{S}^2=100-36$

$\Rightarrow P{S}^2=64$

$\Rightarrow PS=8cm$

Now, In the right-angled triangle $\Delta PSR$,

$P{R}^2=P{S}^2+R{S}^2$
[By Pythagoras theorem]

$P{R}^2=8^2+{15}^2$
[Since $RS=RQ+QS=9+6=15$]

$\Rightarrow P{R}^2=64+225$

$\Rightarrow P{R}^2=289$

$\Rightarrow PR=17cm$