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Answer :

We have:

\(h_1=\frac{1}2(28-20)=\frac{1}2\times 8=4cm\)

\(h_2=\frac{1}2(24-16)=\frac{1}2\times 8=4cm\)

We also know, area of trapezium A=\(\frac{1}2\times(a+b)\times h_1\)

\(=\frac{1}2\times(24+16)\times4\)

\(=\frac{1}2\times 48\times 4=80cm^2\)

so, the area of trapezium A=area of trapezium C=80\(cm^2\)

Now, area of trapezium B=Area of trapezium D

\(=\frac{1}2\times(28+20)\times4\)

\(=\frac{1}2\times 48\times 4=96cm^2\)

Hence, we have areas of all the parts as:

Part A=\(80cm^2\)

Part B=\(96cm^2\)

Part C=\(80cm^2\)

Part D=\(96cm^2\)

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