Question
Construct a triangle $PQR$ with sides $QR = 7\ cm, PQ = 6\ cm$ 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 $\angle\text{PQR}=60^\circ.$
Step III: With Q as centre and radius $6\ cm$, draw an arc cutting the ray QX at P.
Step IV: Join PR. Thus, $\triangle\text{PQR}$ is the required triangle.
Step V: Below QR, draw an acute angle $\angle\text{YQR}.$
Step VI: Along QY, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $QR_1 = R_1R_2 = R_2R_3 = R_3R_4 = R_4R_5.$
Step VII: Join $RR_5$.
Step VIII: From $R_3$, draw $R_3R' || RR_5$ meeting QR at R'.
Step IX: From $R'$, draw $P'R' || PR$ meeting $PQ$ in $P'$.
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 $\angle\text{PQR}=60^\circ.$
Step III: With Q as centre and radius $6\ cm$, draw an arc cutting the ray QX at P.
Step IV: Join PR. Thus, $\triangle\text{PQR}$ is the required triangle.
Step V: Below QR, draw an acute angle $\angle\text{YQR}.$
Step VI: Along QY, mark five points $R_1, R_2, R_3, R_4$ and $R_5$ such that $QR_1 = R_1R_2 = R_2R_3 = R_3R_4 = R_4R_5.$
Step VII: Join $RR_5$.
Step VIII: From $R_3$, draw $R_3R' || RR_5$ meeting QR at R'.
Step IX: From $R'$, draw $P'R' || PR$ meeting $PQ$ in $P'$.
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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