Question

24.
In a parallelogram ABCD, if L and M divide AB and DC in the ratio 1:2 then prove that:
ar (Ilgm ALMD) = 1/2 ar (Ilgm LBCM) = ar 1/3(Ilgm ABCD)​

Answer 1

Step-by-step explanation:

Given,

L and M divide AB and DC in the ratio 1:2 Solution,

in llgm ALMD, let us consider AL as the base and a perpendicular AJ as the height

in llgm LBCM, let us consider LB as the base and a perpendicular LR as the height

as we know that both the quadrilaterals lie in the same parallel their heights have to be equal,( LR = AJ )

ar(ALMD)/ar(LBCM)=AL×AJ/LB×LR

as we know that LR = AJ they cancell out

as given in the question AL:LB=1:2

ar(ALMD)/ar(LBCM)=1:2

therefore, ar(ALMD)= 1/2 ar(LBCM)

Hence proved

as the ratio of

L and M divide AB and DC in the ratio 1:2

we are having 3 parts of equal bases and the llgms lie between the same parallels we can say that

(Ilgm ALMD) = ar 1/3(Ilgm ABCD)

since,

ar (Ilgm ALMD) = 1/2 ar (Ilgm LBCM)

ar (Ilgm ALMD) = 1/2 ar (Ilgm LBCM) = ar 1/3(Ilgm ABCD)

Hence proved