Question

∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \[\frac{AC}{LN} = \frac{4}{7}\]. Construct ∆ABC and ∆LBN.

Answer 1

As shown in the figure,

Let B – C – N and B – A – L.

∆ABC ~ ∆LBN  ...[Given]

∴ ∠ABC ≅ ∠LBN  ...[Corresponding angles of similar triangles]

`"AB"/"LB"="BC"/"BN"="AC"/"LN"`  ...(i) [Corresponding sides of similar triangles]

But, `"AC"/"LN"=4/7`  ...(ii) [Given]

∴ `"AB"/"LB"="BC"/"BN"="AC"/"LN"`  ...[From(i) and (ii)]

∴ Sides of ∆LBN are longer than corresponding sides of ∆ABC.

∴ If seg BC is divided into 4 equal parts, then seg BN will be 7 times each part of seg BC.

So, if we construct ∆ABC, point N will be on side BC, at a distance equal to 7 parts from B.

Now, point L is the point of intersection of ray BA and a line through N, parallel to AC.

∆LBN is the required triangle similar to ∆ABC.

Steps of construction:

1. Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.
2. Taking convenient distance on compass, mark 7 points B₁, B₂, B₃, B₄, B₅, B₆ and B₇ such that BB₁ = B₁B₂ = B₂B₃ B₃= B4₄ = B₄B₅ = B₅B₆ = B₆B₇.
3. Join B₄C. Draw line parallel to B₄C through B₇ to intersects ray BC at N.
4. Draw a line parallel to side AC through N. Name the point of intersection of this line and ray BA as L.

∆LBN is the required triangle similar to ∆ABC.