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Question 24 : In a trepezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trepezium in sq. cm is


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Explanatory Answer

Let's draw the Trapezium.
Tepezium ABCD
Now drop a perpendicular from D on to AB touching AB at E.
Tepezium ABCD after droping perpendicular DE
In ΔAED, ∠AED = 90°, ∠DAE = 45°, therefore ∠EDA = 45°.
Clearly ΔAED is an iscoceles right triangle and Quadrilateral BCDE is a rectangle.
Therefore, BC = DE = 4cm.
Since ΔAED is an iscoceles triangle, AE = ED = 4cm.

Area of the Trepezium = Area of ΔAED + Area of Rectangle BCDE
Area of the Trepezium = \\frac{1}{2}) × AE × ED + DC × BC
Area of the Trepezium = \\frac{1}{2}) × 4 × 4 + 5 × 4
Area of the Trepezium = 8 + 20
Area of the Trepezium = 28 cm2


The question is "In a trepezium ABCD, AB is parallel to DC, BC is perpendicular to DC and ∠BAD = 45°. If DC = 5 cm, BC = 4 cm, the area of the trepezium in sq. cm is"

Hence, the answer is, "28"

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