To construct a triangle similar to a given ∆ ABC with its sides 7/3 of the corresponding sides of ∆ABC, draw a ray BX making an acute angle with BC and X lies on the opposite side of A with respect of BC. The points B 1 , B 2 ,...........B 1 are located at equal distances on BX, B 3 is joined to C and then a where C’ lines on BC produced. Finally line segment A’C’ is drawn parallel to AC.

Question

Philoid · Accepted Answer

False

Steps of construction

1. Draw a line segment BC.

2. B and C as centers draw two arcs of suitable radius intersecting each other at A.

3. Join BA and CA. ∆ABC is required triangle.

4. From B draw any ray BX downwards making an acute angle CBX.

5. Marked seven points B₁,B₂,B₃,............B₇ on BX (BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅ = B₅B₆ = B₆B₇).

6. Join B₃C and from B₇ draw a line B₇C’B₃C intersecting the extended line segment BC at C’.

7. Draw C’ A’CA intersecting the extended line segment BA at A’.

Then, ∆A’BC’ is the required triangle whose sides are 7/3 of the corresponding sides of ∆ABC.

Given that,

segment B₆C’ B₃C. But since our construction is never possible that segment B₆C’ B₃C because the similar triangle A’BC’ has its sides 7/3 of the corresponding sides of triangle ABC.

So, B₇C’ is parallel to B₃C.