Question 3
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

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Solution

Steps of Construction:
Step I:A triangle ABC with sides 5 cm, 6 cm and 7 cm is drawn.
Step II:A ray BX making an acute angle with BC is drawn opposite to vertex A.
Step III:7 points, $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6} a n d B_{7}$ are marked on BX.
Step IV;Point $B_{5}$ is joined with C to draw $B_{5} C$ .
Step V: $B_{7} C$ ' is drawn parallel to B5C and C'A' is parallel to CA.
Thus A'BC' is the required triangle.

Justification
$Δ A B^{'} C^{'} Δ A B C$ by AAA similarity condition
$∴ \frac{A B}{A^{'} B} = \frac{A C}{A^{'} C^{'}} = \frac{B C}{B C^{'}}$
$a n d \frac{B C}{B C^{'}} = \frac{B B 5}{B B 7} = \frac{5}{7}$
$∴ \frac{A^{'} B}{A B} = \frac{A^{'} C^{'}}{A C} = \frac{B C^{'}}{B C} = \frac{7}{5}$

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Question 3 Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 75 of the corresponding sides of the first triangle.

Question 3
Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.