Question
Construct a triangle PQR with sides QR = 7cm, PQ = 6cm and $ \angle\text{PQR} = 60^\circ.$ Then construct another triangle whose sides are $\frac{3}{5}$ of the corresponiding sides of $\triangle\text{PQR}.$
✓
Answer
Steps of Construction:
Step I: Draw a line segment $QR =7 cm$.
Step II: At Q, draw $\angle PQR =60^{\circ}$.
Step III: With Q as centre and radius 6 cm , draw an arc cutting the ray QX at $P$.
Step IV: Join $P R$. Thus, $\triangle PQR$ is the required triangle.
Step V: Below QR , draw an acute angle $\angle YQR$.
Step VI: Along $Q Y$, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $Q R_1=R_1 R_2=R_2 R_3=R_3 R_4=R_4 R_5$.
Step VII: Join RR5.
Step VIII: From $R_3$, draw $R_3 R^{\prime} \| R R_5$ meeting $Q R$ at $R^{\prime}$.
Step IX: From R', draw P'R'\|PR meeting PQ in $P ^{\prime}$.
Here, $\triangle\text{P}'\text{QR}'$ is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle\text{PQR}.$
Step I: Draw a line segment $QR =7 cm$.
Step II: At Q, draw $\angle PQR =60^{\circ}$.
Step III: With Q as centre and radius 6 cm , draw an arc cutting the ray QX at $P$.
Step IV: Join $P R$. Thus, $\triangle PQR$ is the required triangle.
Step V: Below QR , draw an acute angle $\angle YQR$.
Step VI: Along $Q Y$, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $Q R_1=R_1 R_2=R_2 R_3=R_3 R_4=R_4 R_5$.
Step VII: Join RR5.
Step VIII: From $R_3$, draw $R_3 R^{\prime} \| R R_5$ meeting $Q R$ at $R^{\prime}$.
Step IX: From R', draw P'R'\|PR meeting PQ in $P ^{\prime}$.
Here, $\triangle\text{P}'\text{QR}'$ is the required triangle whose sides are $\frac{3}{5}$ of the corresponding sides of $\triangle\text{PQR}.$
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