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Diagram of the adjacent picture frame has outer dimensions is equal to $24 cm × 28 cm$

and inner dimensions 16 cm × 20 cm. Find the area of section I, section II, section III and section IV respectively, if the width of each section is same.

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cm 2

,96

cm 2

,80

cm 2

,96

cm 2

cm 2

,80

cm 2

,96

cm 2

,80

cm 2

cm 2

,86

cm 2

,90

cm 2

,86

cm 2

cm 2

,90

cm 2

,86

cm 2

,90

cm 2

answer is A.

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Detailed Solution

It is given,
Question Image

Marking the sections and redrawing the figure,
Question Image

Given,

P Q = R S = 16 cm

P S = Q R = 20 cm

A B = C D = 24 cm

B C = A D = 28 cm

Also, width of each section of the frame is same.
i.e., perpendicular height of trapezium,

⇒ S E = A D − P S 2 ⇒ S E = 28 − 20 2 ⇒ S E = 4 c m

Area of the trapezium;

A r e a = 12 × (sum of parallel sides) × height

Area of section I;

⇒ 12 (P Q + A B) h

⇒ 12 (16 + 24) 4 ⇒ 80 cm 2

Area of section (III) = Area of section (I) =

80 cm 2

.
Area of section (I);

⇒ 12 (Q R + B C) h ⇒ 12 (20 + 28) 4 ⇒ 96 cm 2

Area of section (IV) = Area of section (II) =

96 cm 2

.
Hence option (1) is correct.

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