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# 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=$$96cm^2$$