# In the diagram,NM is parallel to BC,and NL is parallel to AC.Given that AN/AB =2/3,calculate the numerical value of the ratio: area of trapezium BNMC/area of triangle ABC. Thanks

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We are given `Delta ABC` with N on AB, M on AC and NM||AC. Also AN:AB=2:3.

Now `Delta ABC` ~ `Delta ANM` by AA~. (The parallel lines mean that `/_ANM cong /_ABC` by corresponding angles, and `/_A cong /_A` )

Since the triangles are similar, their sides are proportional. The scale factor is given as 2:3. Then the ratio of their areas is 4:9. (The ratio of areas of similar figures is the square of the scale factor.)

Without loss of generality, let the area of triangle ABC be 9. Then the area of triangle ANM is 4 so the area of the trapezium BNMC is 9-4=5.

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The ratio of the areas of BNMC to ABC is 5:9

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By the way, the notations on your diagram are incorrect. The ratio of AN to AB is given as 2 to 3. Then let AN=2x and BN=x. The ratio of AN:NB=2:1 not 2:3 as in your diagram.

Further, BL:LC is also 2:1 so you would mark them as 2y:y. (You do not know that AN=LC, only that the ratios remain the same.) You could also mark AM:MC as 2z:z

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