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Draw a triangle \( \mathrm{ABC} \) with side \( \mathrm{BC}=6 \mathrm{~cm}, \mathrm{AB}=5 \mathrm{~cm} \) and \( \angle \mathrm{ABC}=60^{\circ} \). Then construct a triangle whose sides are \( \frac{3}{4} \) of the corresponding sides of the triangle \( \mathrm{ABC} \).
Ans: Steps of Construction:
(1) Draw a line segment \( B C=6 \mathrm{~cm} \) and at point 3 dnaw an \( \angle A B C=60^{\circ} \)
(2) Cut \( A B=5 \mathrm{~cm} \). Toin \( A C \), We obtain a \( \triangle A B C \).
(2) Drawa ray \( B X \) making an acute angle with BC on the side opposite to the vertexA.
(4) Locate 4 points \( A_{1}, A_{2}, A_{3}, A_{4} \) on the ray \( B x \) So that \( B A_{1}=A_{1} A_{2}=A_{2} A_{3}=A_{3} A_{4} \).
(ङ). Join \( A_{n} \) to \( C \)
(6) At \( A_{3} \) duaw \( A_{3} C^{\prime} \| A_{4} C \), where \( C^{\prime} \) is a point on
Iiq \( A \) : line segment \( B C \).
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