Question

$△PQR\sim △LTR.$ In $△$
$PQR,PQ=4.2cm,QR=$
$5.4cm,PR=4.8cm$.
Construct $△PQR$ and $△LTR$,
such that $\frac{PQ}{LT}=\frac{3}{4}$

Answer 1

Analysis:
As shown in the figure, Let $R-P-L$ and $R-Q-T$.
$△PQR\sim △LTR\dots$ [Given]
$\therefore \angle PRQ\cong \angle LRT$... [Corresponding angles of similar triangles]
$\frac{PQ}{LT}=\frac{QR}{TR}=\frac{PR}{LR}\dots$ (i) [Corresponding sides of similar triangles]
But, $\frac{PQ}{LT}=\frac{3}{4}$....(ii) [Given]
$\therefore \frac{PQ}{LT}=\frac{QR}{TR}=\frac{PR}{LR}=\frac{3}{4}\dots [$ From (i) and (ii)]
$\therefore$ sides of $LTR$ are longer than corresponding sides of $△$ $PQR$.
If $\mathrm{seg}QR$ is divided into 3 equal parts, then seg TR will be 4 times each part of $\mathrm{seg}QR$.
So, if we construct $△PQR$, point $T$ will be on side RQ, at a distance equal to 4 parts from $R$
Now, point L is the point of intersection of ray RP and a line through $T$, parallel to $PQ$.
$△LTR$ is the required triangle similar to $△PQR$.


Steps of construction:
i. Draw $△PQR$ of given measure. Draw ray $RS$ making an acute angle with side RQ.
ii. Taking convenient distance on the compass, mark 4 points $R1,R2,R3$, and $R4$, such that $RR1=R1R2=R2R3=$ R3R4.
iii. Join R3Q. Draw line parallel to $R3Q$ through $R4$ to intersect ray $RQ$ at $T$.
iv. Draw a line parallel to side $PQ$ through T. Name the point of intersection of this line and ray $RP$ as $L$.
$△LTR$ is the required triangle similar to $△PQR$.