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प्रश्न
∆ABC ~ ∆LBN. In ∆ABC, AB = 5.1 cm, ∠B = 40°, BC = 4.8 cm, \[\frac{AC}{LN} = \frac{4}{7}\]. Construct ∆ABC and ∆LBN.
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उत्तर
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:
- Draw ∆ABC of given measure. Draw ray BD making an acute angle with side BC.
- 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.
- Join B4C. Draw line parallel to B4C through B7 to intersects ray BC at N.
- 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.
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