Circles
Circle
Allen Star Batch
Grade 11
Question:
Let $ABCD$ be a quadrilateral in which $AB \parallel CD$, $AB \perp AD$ and $AB = 3CD$. If the area of the quadrilateral $ABCD$ is 4, then the radius of the circle touching all the four sides of the quadrilateral is:
$\sin\frac{\pi}{12}$
$\sin\frac{\pi}{3}$
$\sin\frac{\pi}{4}$
$\sin\frac{\pi}{6}$
Step-by-Step Solution
Key Concept: Tangency condition combined with trapezium area constraint determines the radius uniquely.
For trapezium $ABCD$ with area $\frac{1}{2}(a + 3a)(2r) = 4$, we get $ar = 1$. The equation of line $BC$ is $y = -r^2(x - \frac{3}{r})$, giving $y + r^2x - 3r = 0$. For $BC$ to be tangent to the circle: $\frac{|r + r^3 - 3r|}{\sqrt{1+r^4}} = r$. Simplifying: $r^4 + 4 - 4r^2 = 1 + r^4$, so $r = \frac{\sqrt{3}}{2}$.
Correct Answer: 2