Question

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

• A. 20
• B. 25
• C. 40
• D. 30

Answer 1

The Correct Option is D

Let \(DP =x \)
\(∴ AB =x\)
Now \(DP=CP\)
So, \(CD = 2x\)
Now, if we denote the height of the trapezium as \(h\),
Then, the area of parallelogram \(ABPD\)\(= xh\)
And Area of \(△BPC=\frac{1}{2}xh\)
Now, based on the given condition,
\(xh-\frac{1}{2}xh=10\)
\(\frac{xh}{2}=10\)
\(⇒ xh=20\)
Then, the area of trapezium \(\frac{1}{2}(x+2x)h\)
\(=\frac{3}{2}xh\)

\(=\frac{3}{2} \times 10\)
\(=30\)

So, the correct option is (D): \(30\)