Question

construct an equilateral triangle of the side 4.8 cm and the another triangle whose sides are 7/4 of the corresponding sides of the first triangle

Answer 1

It's very simple first construct an equilateral triangle and then the common procedure

Answer 2

A'BC' is the constructed triangle whose sides are 7/4 to the given equilateral triangle.

Step-by-step explanation:

• Construct an equilateral triangle ABC with AB = BC = CA = 4.8 cm
• Now draw an angle at B such that angle ∠CBD = 60 as ∠ABC = 60 (angles of an equilateral triangle is 60)
• As it is given that it should be divided in 7/4 parts so draw an arc from B with any measurement keeping pointed end of the compass at B. Let that arc be $\text{BB}_{1}$
• Similarly draw 7 arcs with same measurement as $\text{BB}_{1}$ on BD and name them as $\text{B}_{2}, \text{B}_{3}, \text{B}_{4}, \text{B}_{5}, \text{B}_{6}, \text{B}_{7}$
• Join $\text{B}_{7}\text{C}$ and extend the line BC.
• Extend the line BC and join $\text{B}_{7}\text{C'}$ such that   is parallel $\text{B}_{4}\text{C}$
• (Parallel lines can be drawn by measuring the angle using a protractor   ∠$\text{B}\text{B}_{4}\text{C}$ and marking the same angle in the required point $\text{B}\text{B}_{7}\text{C}$ and drawing a straight line $\text{B}_{7}\text{C}$ from it.)
• Now draw C’A’ parallel to CA
• Thus A’BC’ is the required triangle.

To  Learn More....

1. Construct a triangle ABC in which BC = 3.4 cm AB -AC = 1.5 cm and ∠B = 45°

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2. Construct a triangle with sides 4.4 cm, 5.5 cm and 6.6 CM draw a perpendicular bisector of a long side.

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