Question

If you are asked to construct APQ ~ ABC in the scale factor $\frac{3}{5}$. In $△ABC$, AB = 4 cm, BC = 3 cm and $\angle ABC$ = ${90}^{\circ }$.

Answer 1

The following steps will give you the information on how to construct the similar triangle.

Step 1: Draw a line AB = 4 cms.

Step2: Draw a line BC = 3 cm, perpendicular to AB passing through B.

Step 3: Join AC.

Step 4: Draw a ray AX, making an acute angle with line AB.

Step 5: Mark 5 points $A_1$, $A_2$, $A_3$,$A_4$ and $A_5$ such that $A_1$ $A_2$ = $A_2$ $A_3$ = $A_3$ $A_4$ = $A_4$ $A_5$.

Step 6: Join B$A_5$.

Step 7: Draw a line parallel B$A_5$ passing through $A_3$ by making an angle equal to ∠A$A_5$B, intersecting AB at the point P.$\frac{AP}{AB}$ = $\frac{3}{5}$.

Step 8: Draw a line parallel to BC passing through P, intersecting AC at Q.

Now we have constructed the triangle APQ ~ ABC.

By basic proportionality theorem,

The corresponding angles are equal. Therefore in $△ABC$ and $△APQ$, $\angle BAC$ = $\angle PAQ$, $\angle ABC$ = $\angle AQP$ and $\angle ACB$ = $\angle QPA$.

And the corresponding sides are proportional. Therefore in $△ABC$ and $△APQ$,

$\frac{AP}{AB}$ = $\frac{PQ}{BC}$ = $\frac{AQ}{AC}$

In the question it is given that the scale factor is = $\frac{3}{5}$

Therefore,

$\frac{PQ}{BC}$ = $\frac{3}{5}$.

5PQ = 3BC

From construction we have, $A_3$P || $A_5$B. ∠A$A_3$P = ∠A$A_5$P, because they are corresponding angles of parallel lines. Consider $△A{A}_{5}B$, the line $A_{3}P$ being parallel to $A_{5}B$ cuts the sides AB and $A{A}_5$ in proportion.

i.e.

$\frac{A{A}_3}{A_3{A}_5}$ = $\frac{AP}{PQ}$ ​
$\frac{AP}{PQ}$ = $\frac{3}{2}$

Therefore $A_3$P will divide the line AP in the ratio 3:2 .