Question

answer anyone of the above​

Answer 1

Answer:

Given , a ∆ ABC , in which BC = 5 cm , AC= 4 cm & AC = 6 cm

Steps of construction:

1. Draw a line segment BC =5 cm.

2.Taking B & C as draw two  arcs of radii 4 cm and 6 cm to intersect each other at A

3. Join BA and CA. Thus ∆ABC is the given Triangle.

4. Now from B draw any ray BX making an acute ∠CBX with base BC on the side opposite to the vertex A.

• 5. Along BX Mark 3 points B1,B2,B3,B4 on BX such that BB1 = B1B2 = B2=B3 =B4.

6.Join B3C and from B2

draw a line B2N || B3C intersecting BC at N.

7. From point N, draw MN || CA intersecting AB at M. Thus , ∆MBN is the required triangle whose sides are 4/3 of the corresponding sides of ∆ABC.

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