Question

If you are asked to construct $△$APQ $\sim$ $△$ABC with the scale factor $\frac{3}{5}$, which of the following is incorrect?
Given: In $△ABC,AB=4cm$, $BC=3cm$ and $\angle ABC={90}^{\circ }$.
$A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5$.​​​​​​

• A. $AP:PB=5:3$
• B. 5PQ = 3BC
• C. $\angle AQP=\angle ACB$
• D. $\angle A{A}_{3}P=\angle A{A}_{5}P$

Answer 1

The correct option is D $\angle A{A}_{3}P=\angle A{A}_{5}P$

The following steps will give you the information on how to construct the similar triangle to $△$ABC.
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 to $B{A}_5$ passing through $A_3$ by making an angle equal to $\angle A{A}_{5}B$, intersecting AB at the point P.$\frac{AP}{AB}=\frac{3}{5}$.(This is the given scale factor of the smaller triangle which is the ratio of corresponding sides.)
Step 8: Draw a line parallel to BC passing through P, intersecting AC at Q.
Now we have constructed the triangle $△$APQ $\sim$ $△$ABC .
$\Rightarrow$$\angle BAC=\angle PAQ,\angle ABC=\angle APQand\angle ACB=\angle PQA$
Also,
$\frac{AP}{AB}=\frac{PQ}{BC}=\frac{AQ}{AC}$
We know that
$\frac{AP}{AB}=\frac{3}{5}$
Therefore, $\frac{PQ}{BC}=\frac{3}{5}$
$\Rightarrow$ 5PQ = 3BC.
From construction we have, $A_{3}P||A_{5}B$.
$\Rightarrow \angle A{A}_{3}P=\angle A{A}_{5}B$, because they are corresponding angles of parallel lines.
$\Rightarrow$ $\angle A{A}_{3}P\ne \angle A{A}_{5}P$
Consider triangle $A{A}_{5}B$, the line $A_{3}P$ being parallel to $A_{5}B$ cuts the sides AB and $A{A}_5$ in the same proportion.
$\Rightarrow$ $\frac{A{A}_3}{A_3{A}_5}=\frac{AP}{PB}$
$\frac{AP}{PB}=\frac{3}{2}$
$\therefore$ $A_{3}P$ will divide the line AB in the ratio 3:2.
Therefore, $AP:PB\ne 2:3$