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
- 20
- 25
- 40
- 30
The Correct Option is D
Solution and Explanation
Let the length of DP be \( x \). Then, since \( DP = AB \), we have:
\[ AB = x \]
Also given that \( DP = CP \), so the full length \( CD \) becomes:
\[ CD = DP + CP = x + x = 2x \]
Let the height of the trapezium be \( h \).
Step 1: Area of Parallelogram ABPD
Area of parallelogram is given by 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 \( h \), so its area is:
\[ \text{Area}_{\triangle BPC} = \frac{1}{2} \times x \times h = \frac{1}{2}xh \]
Step 3: Given Condition
It is given that the area of parallelogram minus the area of triangle is 10:
\[ xh - \frac{1}{2}xh = 10 \Rightarrow \frac{1}{2}xh = 10 \Rightarrow xh = 20 \]
Step 4: Area of Trapezium ABCD
Trapezium has parallel sides \( AB = x \) and \( CD = 2x \), and height \( h \), so:
\[ \text{Area}_{ABCD} = \frac{1}{2}(AB + CD) \times h = \frac{1}{2}(x + 2x)h = \frac{1}{2}(3x)h = \frac{3}{2}xh \]
We already found that \( xh = 20 \), so:
\[ \text{Area}_{ABCD} = \frac{3}{2} \times 20 = 30 \]
Final Answer:
Option (D): \( \boxed{30} \)
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