Question

Construct $∆$PQR such that perimeter $∆$PQR is 16cm and $\angle$Q = 90$°$ and QR = 5 cm.

Answer 1

$$\begin{gathered} \mathrm{Perimeter}\left(△\mathrm{PQR}\right)=16\mathrm{cm} \\ \Rightarrow \mathrm{PQ}+\mathrm{PR}+\mathrm{QR}=16 \\ \Rightarrow \mathrm{PQ}+\mathrm{PR}=16-5=11\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw QR = 5 cm.
2. Draw ray QX, such that $\angle \mathrm{RQX}=90°$.
3. With Q as centre and radius 11 cm, draw an arc, intersecting ray QX at point A.
4. Join A and R.
5. Draw the perpendicular bisector of seg AR, intersecting seg AQ at point P.
6. Join P and R.
7. Hence, $△$PQR is the required triangle.