11.Diagram of the picture frame has outer dimensions = \(24 cm \times 28 cm\) and inner dimensions \(16 cm \times 20 cm\). Find the area of each section of the frame, if the width of each section is the same.

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=\(96cm^2\)