Question

Given that ABCD is a rectangle with AB = 12 cm, AD = 6 cm. M and N are respectively midpoint of segments BC and CD. Find the area of triangle AMN in square centimeters.

Answer 1

You have to draw the geometry yourself.

\(A_{ABCD}=AB.AD=12.6=72\left(cm^2\right)\)

M is the midpoint of segment BC so we have: \(BM=MC=\frac{BC}{2}=\frac{6}{2}=3\left(cm\right)\)

For the midpoint of CD is N, we also have: \(DN=NC=\frac{CD}{2}=\frac{12}{2}=6\left(cm\right)\)

We have:

\(A_{AMN}=A_{ABCD}-\left(A_{ABM}+A_{NCM}+A_{ADN}\right)\\ =72-\left(\frac{1}{2}.AB.BM+\frac{1}{2}.NC.MC+\frac{1}{2}AD.DN\right)\\ =72-\left(\frac{1}{2}.12.3+\frac{1}{2}.6.3+\frac{1}{2}.6.6\right)\\ =72-45\\ =27\left(cm^2\right)\)

Thusly, the area of triangle AMN in square centimeters is 27.