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Sum

Construct a triangle with sides 5 cm, 6 cm, and 7 cm and then another triangle whose sides are `3/5` of the corresponding sides of the first triangle.

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#### Solution

1. Draw a line AB = 5 cm and draw a ray from A and taking A as centre cut an arc at C of 6 cm and taking B as centre cut an arc of 7 cm at C

2. Draw AX such that ∠BAX is an acute angle.

3. Cut 5 equal arcs AA_{1}, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, and A_{4}A_{5}.

4. Join A_{5} to B and draw a line through A_{3} parallel to A_{5}B which meets AB at B'.

Here, `AB' = 3/5AB`

5. Now draw a line through B' parallel to BC which joins AC at C'.

Here, `B'C' = 3/5BC "and" AC' = 3/5AC`

Thus, AB'C' is the required triangle.

Concept: Similarity of Triangles

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