Question

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD)=24cm^,find ar(triangleABC)

Answer 1

• ar(AMCD)=ar(ACD)+ar(AMC)
• ar(ACD)=ar(ABC)=1/2 ar(ABCD)
• ar(AMC)=1/2ar(ABC)=1/4 ar(ABCD)

• given: ar(AMCD)=24cm²=1/2ar(ABCD)+1/4ar(ABCD)

∴ 24=1/2+1/4 ar(ABCD)

24=3/4 ar(ABCD)

32=ar(ABCD)

∴ar(ABC)=16cm²

Answer 2

Answer:

ar(tri.ACD) = ar(tri.ABC)

ar( tri. AMC) =ar.(tri.MBC)=1/2ar(tri.ABC)

ar(AMCD)=3ar(tri.AMC)=24=3ar(∆AMC)

=ar(∆AMC)=8cm^2

And so,ar(∆ABC)=2ar(∆AMC)=16 cm^2

hope this will help you!!!!!!4