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  • Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Draw an isosceles triangle ABC in which AB = AC = 6 cm and BC = 5 cm. Construct a triangle PQR similar to ABC in which PQ = 8 cm. Also justify the Construction.

Answers (1)

Solution

Steps of construction
1. Draw line $B C=5 \mathrm{~cm}$
2. Taking $B$ and $C$ as centres, draw two arcs of equal radius 6 cm intersecting each other at point $A$.
3. Join $A B$ and $A C D A B C$ is required isosceles triangle

4 From B draw ray $B_X$ with an acute angle $C B B^{\prime}$
6. draw $B_1, B_2, B_3, B_4$ at $B X$ with equal distance
7. Join $B_3 C$ and from $B_4$ the draw line $B_4 D \| B_3 C$, intersect extended segment BC at point D .

Then EBD is required triangle. We can name it PQR.

$$
\begin{aligned}
& \text { Justification } \\
& \because B_4 D \| B_3 C \\
& \therefore \frac{B C}{C D}=\frac{3}{1} \Rightarrow \frac{C D}{B C}=\frac{1}{3} \\
& \text { Now } \therefore \frac{B D}{B C}=\frac{B C+C D}{B C}=1+\frac{C D}{B C}=1+\frac{1}{3}=\frac{4}{3} \\
& \text { Also } D E \| C A \\
& \therefore \triangle A B C \sim \triangle D B E \\
& \frac{E B}{A B}=\frac{D E}{C A}=\frac{B D}{B C}=\frac{4}{3}
\end{aligned}
$$

Posted by

infoexpert27

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