Answer :

__Steps of construction__:

1. Draw a line segment AB=5 cm.

2. From point B, draw ∠ABY=60°.

To measure angle at B:

a. With B as centre and with any radius, draw another arc cutting the line AB at D.

b. With D as centre and with the same radius, draw an arc cutting the first arc (drawn in step a) at point E.

c. Draw a ray BY passing through E which forms an angle of 60° with the line AB.

3. With B as centre and radius of 6 cm draw an arc intersecting the line BY at C.

4. Join AC, ∆ABC is the required triangle.

5. From A, draw any ray AX downwards making an acute angle.

6. Mark 7 points A_{1},A_{2},A_{3},A_{4},A_{5},A_{6} and A_{7} on AX such that AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}=A_{5}A_{6}=A_{6}A_{7}

7. Join A_{7}B and from A_{5} draw A_{5}M||A_{7}B intersecting AB at M.

8. From point M draw MN││BC intersecting AC at N. Then, ∆AMN is the required triangle whose sides are equal to of the corresponding sides of ∆ABC.

__Justification__:

Let AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5} = A_{5}A_{6} = A_{6}A_{7} = x

As per the construction A_{5}M││A_{7}B

Now,

Also MN || BC,

Therefore rAMC is congruent to rAMN

Thus,

Hence, the new triangle rAMN is similar to the given triangle rABC and its sides are times of the corresponding sides of rABC.

Rate this question :