Question:hard

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is

Updated On: Jan 15, 2026
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The Correct Option is D

Solution and Explanation

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side

Let the length of DP be \( x \). Since \( DP = AB \), it follows that:

\[ AB = x \]

Given that \( DP = CP \), the total length \( CD \) is:

\[ CD = DP + CP = x + x = 2x \]

Let the height of the trapezium be \( h \).

Step 1: Area of Parallelogram ABPD

The area of a parallelogram is calculated as base × height:

\[ \text{Area}_{ABPD} = AB \times h = xh \]

Step 2: Area of Triangle BPC

Triangle BPC is a right triangle with base \( x \) and height \( h \). Its area is:

\[ \text{Area}_{\triangle BPC} = \frac{1}{2} \times x \times h = \frac{1}{2}xh \]

Step 3: Given Condition

The problem states that the area of the parallelogram minus the area of the triangle equals 10:

\[ xh - \frac{1}{2}xh = 10 \Rightarrow \frac{1}{2}xh = 10 \Rightarrow xh = 20 \]

Step 4: Area of Trapezium ABCD

The trapezium has parallel sides \( AB = x \) and \( CD = 2x \), with a height \( h \). The area is calculated as:

\[ \text{Area}_{ABCD} = \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(x + 2x)h = \frac{1}{2}(3x)h = \frac{3}{2}xh \]

Using the previously determined value \( xh = 20 \):

\[ \text{Area}_{ABCD} = \frac{3}{2} \times 20 = 30 \]

Final Answer:

Option (D): \( \boxed{30} \)

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