Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

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

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

$a = 24 cm , b = 16 cm, c = 28 cm, d = 20 cm$

Area of trapezium A = $\frac{1}{2}(a + b) \times h_1$

$=\frac{1}{2}(24 + 16) \times 4$

$=\frac{1}{2}\left(49\right) \times 4$

$=80 c{m}^2$

Area of trapezium A = Area of trapezium C $=80 c{m}^2$

Area of trapezium B = Area of trapezium D =$\frac{1}{2}(c + d) \times h_2$

$=\frac{1}{2}(28 + 20) \times 4$

$=\frac{1}{2}\left(48\right) \times 4$

$=96 c{m}^2$