CAT 2025Slot 3QAQuestion & SolutionQues & Sol
Question
ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm, then the area, in sq. cm, of the trapezium is
Options
Solution
1. Concept Used
-
Topic: Geometry — Trapezium with an inscribed circle (incircle), properties of similar triangles, and incircle radius formula for right-angled triangles.
-
Formula: For a right-angled triangle with legs $a$ and $b$ and hypotenuse $c$:
$$ r=\frac{a+b-c}{2} $$
Area of a trapezium:
$$ \text{Area}=\frac{1}{2}\times(\text{sum of parallel sides})\times\text{height} $$
2. Calculation
Let $DC=x$, so $AB=3x$ since $AB=3DC$.
Since $AD\perp AB$, angle $A$ is $90^\circ$. The inscribed circle has radius $r=3$ cm, so its diameter equals the height:
$$ AD=2r=6\text{ cm} $$
Now extend $BC$ and $AD$ to meet at point $E$. Since $AB\parallel DC$ and $\angle A=90^\circ$, triangles $ABE$ and $DCE$ are similar right-angled triangles with ratio $AB:DC=3:1$.
Therefore,
$$ AE=3\times DE $$
Since $AE=DE+AD$,
$$ 3DE=DE+6 $$
$$ 2DE=6 $$
$$ DE=3\text{ cm},\qquad AE=9\text{ cm} $$
In right-angled triangle $DCE$:
$$ CE=\sqrt{x^2+9} $$
Drop a perpendicular from $C$ to $AB$ at point $T$. Since $DC=x$ and $AB=3x$, we have:
$$ AT=x,\qquad TB=2x $$
The height $CT=AD=6$, so:
$$ BC=\sqrt{(2x)^2+6^2} $$
$$ BC=\sqrt{4x^2+36} $$
$$ BC=2\sqrt{x^2+9} $$
Also,
$$ BE=BC+CE $$
$$ BE=2\sqrt{x^2+9}+\sqrt{x^2+9} $$
$$ BE=3\sqrt{x^2+9} $$
Now apply the incircle radius formula to right-angled triangle $ABE$:
$$ r=\frac{AB+AE-BE}{2} $$
Therefore,
$$ r=\frac{3x+9-3\sqrt{x^2+9}}{2} $$
Setting $r=3$:
$$ 3=\frac{3x+9-3\sqrt{x^2+9}}{2} $$
$$ 6=3x+9-3\sqrt{x^2+9} $$
$$ 3\sqrt{x^2+9}=3x+3 $$
$$ \sqrt{x^2+9}=x+1 $$
Squaring both sides:
$$ x^2+9=x^2+2x+1 $$
$$ 9-1=2x $$
$$ x=4 $$
Therefore,
$$ DC=4\text{ cm},\qquad AB=12\text{ cm},\qquad AD=6\text{ cm} $$
The area of the trapezium is:
$$ \text{Area}=\frac{1}{2}\times(AB+DC)\times AD $$
$$ =\frac{1}{2}\times(12+4)\times6 $$
$$ =\frac{1}{2}\times16\times6 $$
$$ =48\text{ sq. cm} $$
3. Solution
Answer = Option 1 ✅
The area of the trapezium is 48 sq. cm.