Please solve the following: Share with your friends Share 0 Gursheen Kaur answered this Dear Student, Solution) 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 Regards! 0 View Full Answer
