Let ABC be a right-angled isosceles triangle with hypotenuse BC. Let BQC be a semi-circle, away from A, with diameter BC. Let BPC be an arc of a circle centered at A and lying between BC and BQC. If AB has length 6 cm then the area, in sq cm, of the region enclosed by BPC and BQC is

The correct option is (B).

Let $△ABC$ be a right-angled isosceles triangle with $\angle A=90{\circ }^{}$ and equal sides $AB=AC=6$ cm.
Using Pythagoras' theorem, the hypotenuse $BC$ is:
$BC=\sqrt{A{B}^2+A{C}^2}=\sqrt{6^2+6^2}=6\sqrt{2}$ cm.

We are given:
1. $BQC$ is a semi-circle with diameter $BC$. Its radius is:
$r=\frac{BC}{2}=3\sqrt{2}$ cm.
The area of this semi-circle is:
$\text{Area}\left(BQC\right)=\frac{1}{2}\pi r^2=\frac{1}{2}\pi {\left(3\sqrt{2}\right)}^2=9\pi$ sq cm.

2. $BPC$ is an arc of a circle centered at $A$ with radius $R=AB=6$ cm. Since $\angle BAC=90{\circ }^{}$, the region $ABPC$ is a quadrant (quarter circle) of radius 6 cm.
The area of this sector is:
$\text{Area of sector}ABPC=\frac{1}{4}\pi R^2=\frac{1}{4}\pi \left(6^2\right)=9\pi$ sq cm.

3. Let's find the area of the region between segment $BC$ and the arc $BPC$ (which is a circular segment):
$\text{Area of segment}BPC=\text{Area of sector}ABPC-\text{Area}(△ABC)$
The area of the right-angled triangle $△ABC$ is:
$\text{Area}(△ABC)=\frac{1}{2}\times 6\times 6=18$ sq cm.
Thus, $\text{Area of segment}BPC=9\pi -18$.

4. The region enclosed between the arc $BPC$ and the semi-circle $BQC$ (the lune of Hippocrates) is calculated by subtracting the area of segment $BPC$ from the area of the semi-circle $BQC$:
$\text{Required Area}=\text{Area of semi-circle}BQC-\text{Area of segment}BPC$
$\text{Required Area}=9\pi -(9\pi -18)=18$ sq cm.