Question

M and N are midpoints of sides AB and BC of parallelogram ABCD. if the area parallelogram ABCD is 24cm². What is the area of trapezium AMNC?​

Answer 1

Answer:

$\huge\pink {9 \: {cm}^{2} }$

Step-by-step explanation:

Using mid point theorem in ∆ABC

AC || MN

so ∆ABC and ∆MBN are similar triangles with

ratio of sides

$\frac{AB}{MB} = \frac{1}{2}$

Ratio of Area of similar triangles is square of that of sides so

$\frac{Ar(ABC)}{Ar(MBN)} = (\frac{1}{2} ) {}^{2} = \frac{1}{4}$

But area of ∆ABC is half of parallelogram ( diognal ) = 12

So

$Ar(MBN) = \frac{1}{4} Ar(ABC) = \frac{12}{4} = 3$

Now area of trapezium

$Ar(AMNC) \\ \\ = Ar(ABC) - Ar(MBN) \\ \\ = 12 - 3 = 9 \: {cm}^{2}$

Answer 2

Answer:

9cm² is the correct answer