Question

In Fig. 7.241, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, then perimeter of ∆QSR is_______.

Answer 1

In right ∆PQS,

$$\begin{gathered} {\mathrm{PQ}}^2={\mathrm{PS}}^2+{\mathrm{QS}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow {\mathrm{QS}}^2={\left(6\right)}^2-{\left(4\right)}^2 \\ \Rightarrow {\mathrm{QS}}^2=36-16=20 \\ \Rightarrow \mathrm{QS}=\sqrt{20}=2\sqrt{5}\mathrm{cm} \end{gathered}$$

Now,

∠P + ∠PQS = 90º .....(1)

∠PQS + ∠RQS = 90º .....(2)

From (1) and (2), we have

∠P + ∠PQS = ∠PQS + ∠RQS

⇒ ∠P = ∠RQS

In ∆PQS and ∆QRS,

∠P = ∠RQS (Proved)

∠PSQ = ∠QSR (90º each)

∴ ∆PQS $~$ ∆QRS (AA similarity)

$\Rightarrow \frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\mathrm{QS}}{\mathrm{RS}}=\frac{\mathrm{PS}}{\mathrm{QS}}$ (If two triangles are similar, then the ratio of their corresponding sides is proportional)

$$\begin{gathered} \Rightarrow \frac{6\mathrm{cm}}{\mathrm{QR}}=\frac{2\sqrt{5}\mathrm{cm}}{\mathrm{RS}}=\frac{4\mathrm{cm}}{2\sqrt{5}\mathrm{cm}} \\ \Rightarrow \frac{6\mathrm{cm}}{\mathrm{QR}}=\frac{2}{\sqrt{5}}\mathrm{and}\frac{2\sqrt{5}\mathrm{cm}}{\mathrm{RS}}=\frac{2}{\sqrt{5}} \\ \Rightarrow \mathrm{QR}=\frac{6\times \sqrt{5}}{2}=3\sqrt{5}\mathrm{cm}\mathrm{and}\mathrm{RS}=\frac{2\sqrt{5}\times \sqrt{5}}{2}=5\mathrm{cm} \end{gathered}$$

∴ Perimeter of ∆QSR = QS + QR + SR $=2\sqrt{5}+3\sqrt{5}+5=5\sqrt{5}+5=5\left(\sqrt{5}+1\right)\mathrm{cm}$

In Fig. 7.241, PQR is a right triangle right angled at Q and QS ⊥ PR. If PQ = 6 cm and PS = 4 cm, then perimeter of ∆QSR is$\underline{5\left(\sqrt{5}+1\right)\mathrm{cm}}$.