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In triangle ABC, M is the midpoint of BC such that CM = MB. In triangle AMB, L is the midpoint of AB such that AL = 2LB. The area of triangle ABC is 7.2 sq.cm
Find the area of triangle ALM.

Hi Srinidhi,
Please find below the solution to the asked query:



We have, ABC as the given ∆ in which M is the mid point of BC or AM is median to BC.We know that, median of a ∆ divides it into 2 ∆'s of equal areas.In ∆ABC, AM is median to BC, then       ar∆AMC = ar∆AMB  ......1Now, ar∆ABC = 7.2 cm2   given⇒ar∆AMC + ar∆AMB = 7.2⇒ar∆AMB + ar∆AMB = 7.2     Using 1⇒2ar∆AMB = 7.2⇒ar∆AMB = 3.6 cm2   .......2Now, we  have L as a point on AB such that AL = 2 LB.Now, from point M, draw MQ⊥AB.    ar∆ALM = 12×base×height⇒ar∆ALM = 12×AL×MQ      ........3and ar∆MLB = 12×LB×MQ       .......4dividing 3 by 4, we get   ar∆ALMar∆MLB = 12×AL×MQ12×LB×MQ⇒ar∆ALMar∆MLB = ALLB⇒ar∆ALMar∆MLB = 2LBLB   as, AL = 2LB⇒ar∆ALMar∆MLB = 2⇒ar∆MLB = 12ar∆ALM    ..........5Now, from 2, we have    ar∆AMB = 3.6 cm2  ⇒ar∆ALM + ar∆MLB = 3.6⇒ar∆ALM +  12ar∆ALM  = 3.6   using 5⇒32ar∆ALM = 3.6⇒ar∆ALM = 3.6 × 23⇒ar∆ALM = 2.4 cm2

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