Question

draw a triangle ABC with side BC = 6cm, AB =5cm and ABC =60 then construct a triangle whose side are 3/4 of the corresponding side of the triangle ABC ​

Answer 1

Step-by-step explanation:

Steps of constructions:

1. Draw a line BC = 6 cm.
2. At B, make ∠C = 60° and cut an arc at A on the same line so that BA = 5 cm. Join AC, ΔABC is obtained.
3. Draw the ray BX such that ∠CBX is acute.
4. Mark 4 (since 4 > in 3/4) points B₁, B₂, B₃, B₄ on BX such that BB₁ = B₁B₂ = B₂B₃ = B₃B₄
5. Join B₄ to C and draw B₃C' parallel to B₄C to intersect BC at C'.
6. Draw C'A' parallel to CA to intersect BA at A’.

Now, ΔA'BC' is the required triangle similar to ΔABC

where BA'/BA = BC'/BC = C'A'/CA = 3/4

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