Answer:
Area of the right section of the frame \[=\frac{(28+20)\times \,\frac{1}{2}\,(24-16)}{2}c{{m}^{2}}\] \[=\frac{48\times 4}{2}c{{m}^{2}}\] \[=96\,c{{m}^{2}}\] Similarly, area of the left section of the frame \[=96\,c{{m}^{2}}\] Area of the upper section of the frame \[=\frac{(24\times 16)\,\frac{1}{2}\,(28-20)}{2}c{{m}^{2}}\] \[=\frac{40\times 4}{2}\,c{{m}^{2}}\] \[=80c{{m}^{2}}\] Similarly, area of the lower section of the frame \[=80c{{m}^{2}}\] .
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