CAT 2022 Slot 1 · Quant · Question 11

Since \angle BAD={90}^{∘}, the side AB is perpendicular to AD.

AD is parallel to BC, so AB is perpendicular to BC as well, making AB the height of the trapezium.

The perimeter gives AB+BC+CD+DA=36.

Substituting the known sides, AB+3+CD+8=36, so AB+CD=25.

Write AB=h, so CD=25-h.

Drop a perpendicular from C to AD, meeting it at P.

Then ABCP is a rectangle, so AP=BC=3 and CP=AB=h.

The leftover piece is PD=AD-AP=8-3=5.

In the right triangle CPD, C{D}^{2}=C{P}^{2}+P{D}^{2}, that is {\left( 25-h \right)}^{2}={h}^{2}+25.

Expanding, 625-50h+{h}^{2}={h}^{2}+25.

The {h}^{2} terms cancel, leaving 600=50h, so h=12.

The area of a trapezium is half the sum of the parallel sides times the height: \frac{1}{2}\left( 3+8 \right)\times 12.

That is \frac{11\times 12}{2}=66 sq cm.

Shortcut. The right-angle corner makes AB the height outright, so the only unknown is h — and squaring 25-h cancels {h}^{2}, leaving a linear equation. The slant side turns out to be 13, the familiar 5-12-13 triple.