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 Section BEFC of the frame, if the width of each section is same.

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Solution

Given that, the width of each Section is same.
Therefore, IB = BJ= CK =CL = DM = DN = AO =AP
IL= IB +BC + CL
28= IB + 20 +CL
IB+CL =28cm - 20 cm = 8cm
IB =CL = 4cm
Hence, IB =BJ=CK=CL=DM=DN=AO=AP=4cm
Area of Section BEFC =Area of section DGHA
= $\frac{1}{2} (20 + 28) (4)] c m^{2} = 96 c m^{2}$

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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.

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Diagram of the adjacent picture frame has outer dimensions = 24 cm

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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 Section BEFC of the frame, if the width of each section is same.

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

Given that, the width of each Section is same. Therefore, IB = BJ= CK =CL = DM = DN = AO =AP IL= IB +BC + CL 28= IB + 20 +CL IB+CL =28cm - 20 cm = 8cm IB =CL = 4cm Hence, IB =BJ=CK=CL=DM=DN=AO=AP=4cm Area of Section BEFC =Area of section DGHA =12(20+28)(4)]cm2=96cm2

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 Section BEFC of the frame, if the width of each section is same.

Given that, the width of each Section is same.
Therefore, IB = BJ= CK =CL = DM = DN = AO =AP
IL= IB +BC + CL
28= IB + 20 +CL
IB+CL =28cm - 20 cm = 8cm
IB =CL = 4cm
Hence, IB =BJ=CK=CL=DM=DN=AO=AP=4cm
Area of Section BEFC =Area of section DGHA
= $\frac{1}{2} (20 + 28) (4)] c m^{2} = 96 c m^{2}$