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Standard X

Mathematics

Constructing a Similar Triangle with a Scale Factor

To construct ...

Question

To construct a triangle similar to given $Δ A B C$ with its sides $\frac{3}{5}$ of that of $Δ A B C$ , a ray $B X$ is drawn at acute angle with $B C$ . If we mark points $B_{1}, B_{2}, B_{3}, . . . .$ at equal distances from $B$ along $B X$ , then point to be joined in next step is?

B_{3}

to point

C

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B_{4}

to point

C

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B_{5}

to point

C

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B_{6}

to point

C

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Solution

The correct option is C $B_{5}$ to point $C$
Suppose $△ P Q B$ is the required triangle.
$B Q = \frac{3}{5} \times B C \Rightarrow P Q = \frac{3}{5} \times (B Q + Q C) \Rightarrow 5 B Q = 3 B Q + 3 C Q \Rightarrow 2 B Q = 3 C Q$ , i.e, $\frac{B Q}{C Q} = \frac{3}{2}$
Therefore, $Q$ divides $B C$ in ratio $3 : 2$ .
Hence, after drawing the ray $B X$ , the next step will be to join the last point, i.e, $B_{5}$ to point $C$ , and then draw $B_{3} Q$ parallel to $B_{5} C$ to obtain the point $Q$ .

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