In a trapezium ABCD, AB is parallel to DC, BC is perpendicular to DC and \(∠BAD=45°\). If DC=5cm , BC=4 cm, the area of the trapezium in sq cm is ? [This Question was asked as TITA]
\(25 \ cm^2\)
\(26 \ cm^2\)
\(28\ cm^2\)
\(30\ cm^2\)
The Correct Option is C
Solution and Explanation
Given:
\( \angle BAD = 45^\circ \), \( DC = 5\,\text{cm} \), \( BC = 4\,\text{cm} \), and \( BC \perp DC \).
AB is parallel to DC, and we are to find the area of trapezium ABCD.
Step 1: Since \( BC \perp DC \), BC is the height of the trapezium.
So, height \( h = BC = 4\,\text{cm} \).
Step 2: Triangle \( \triangle ABD \) is right-angled at A, and \( \angle BAD = 45^\circ \).
Let’s find the length of \( AD \) using the trigonometric identity: \[ \tan \angle BAD = \frac{AD}{BC} \] Substituting the values: \[ \tan 45^\circ = \frac{AD}{4} \Rightarrow 1 = \frac{AD}{4} \Rightarrow AD = 4\,\text{cm} \]
Step 3: Since \( AB = AD + DC \) (as they lie along a straight line), \[ AB = AD + DC = 4 + 5 = 9\,\text{cm} \]
Step 4: Now use the formula for the area of a trapezium: \[ \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{Height} \] Here, the parallel sides are \( AB = 9\,\text{cm} \) and \( DC = 5\,\text{cm} \), and height is \( 4\,\text{cm} \).
So, \[ \text{Area} = \frac{1}{2} \times (9 + 5) \times 4 = \frac{1}{2} \times 14 \times 4 = 28\,\text{cm}^2 \]
Final Answer: The area of trapezium ABCD is: \[ \boxed{28\,\text{cm}^2} \]
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