This question is from Geometry. In this question, we need to find the area of the trapezium. CAT Geometry is an important topic with lots of weightage in the **CAT Exam**. In CAT Exam one can generally expect 4-6 questions from Geometry. Make sure you master Geometry problems by solving **CAT Previous Year Paper**.

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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Let's draw the Trapezium.

Now drop a perpendicular from D on to AB touching AB at E.

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 cm^{2}

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" **

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