Question

Ritu wants to make a trapezium such that \(AB\) is parallel to \(CD\). \(\angle ABC=90^\circ\) and \(\angle BAD=45^\circ\). Lengths: \(CD=5\ \mathrm{cm}\) and \(BC=4\ \mathrm{cm}\). Find the area of the trapezium.

• A. \(20\ \mathrm{cm}^2\)
• B. None of these
• C. \(28\ \mathrm{cm}^2\)
• D. \(24\ \mathrm{cm}^2\)

Hint:

When one leg is perpendicular to the bases in a trapezium, that leg is the height}. Use coordinates or projections with the given angle to recover the unknown base.

Answer 1

The Correct Option is C

Step 1: Determine the trapezium's height
Since \(\angle ABC=90^\circ\) and \(AB\parallel CD\), \(BC\) is perpendicular to both bases, thus the height \(h=BC=4\ \mathrm{cm}\).
Step 2: Calculate the length of base \(AB\)
Let \(A\) be at \((0,0)\), and \(AB=x\), so \(B\) is at \((x,0)\). Given \(CD\parallel AB\), \(BC=4\), then \(C\) is at \((x,-4)\) and \(D\) is at \((x-5,-4)\) (because \(CD=5\)).
Vector \(\overrightarrow{AD}=(x-5,-4)\). As \(\angle BAD=45^\circ\), the slope magnitude of \(AD\) is \(|-4/(x-5)|=\tan 45^\circ=1\), which means \(|x-5|=4\).
Choosing \(x-5=4\) (to maintain vertex order) yields \(x=9\), therefore \(AB=9\ \mathrm{cm}\).
Step 3: Compute the Area
\(\displaystyle \text{Area}=\frac{(AB+CD)}{2}\times h=\frac{(9+5)}{2}\times 4=7\times 4=28\ \mathrm{cm}^2.\)