question_answer

Diagram of the adjacent 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 same.

Answer:

Area of trapezium AEHD =\[\frac{1}{2}\times h\times \] (sum of parallel sides)

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

\[=2\times 48=96\,\,c{{m}^{2}}\]

Area of trapezium DHGC =\[\frac{1}{2}\times h\times \] (sum of parallel sides)

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

\[=2\times 40=80\,\,c{{m}^{2}}\]

Area of trapezium BFGC =\[\frac{1}{2}\times h\times \] (sum of parallel sides)

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

\[=2\times 48=96\,c{{m}^{2}}\]

Area of trapezium AEFB =\[\frac{1}{2}\times h\times \](sum of parallel sides)

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

\[=\text{ }2\times 40=80\text{ }c{{m}^{2}}\]