Question

$△ABC$ is similar to $△PQR$, in $△ABC,AB=3.6cm,BC=4cm$ and $AC=4.2cm$. The corresponding sides of $△ABC$ and $△PQR$ are in the ratio $2:3$. Construct $△PQR$.

Find $m\angle PRQ$ (in deg.)

Answer 1

The sides of $△PQR$ will be $\frac{3}{2}$ times the sides of $△ABC$

$\therefore PQ=\frac{3}{2}\times AB=\frac{3}{2}\times 3.6=5.4cm$

$QR=\frac{3}{2}\times BC=\frac{3}{2}\times 4=6cm$

$PR=\frac{3}{2}\times AC=\frac{3}{2}\times 4.2=6.3cm$

Construction:

Step $1$: Draw base $PR=6.3cm$ with a ruler.

Step $2$: Place the compass on point $P$ and measure an arc with length of the compass as $PQ=5.4cm$.

Step $3$: Place the compass on point $R$ and measure an arc with length of the compass as $QR=6cm$.

Step $4$: Mark the intersection point of the arcs in steps $2$ and $3$ as point $Q$ and draw lines joining $P-Q$ and $R-Q$.

$△PQR$ is the required triangle.

Measure the $\angle PRQ$ using a protractor, which comes out to be ${52}^{\circ }$