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 centred 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:

We have a right-angled isosceles triangle $ABC$ with the right angle at $A$, where $AB = AC = 6$ cm and $BC$ is the hypotenuse.

There's a semicircle $BQC$ with $BC$ as diameter on the opposite side of $A$, and an arc $BPC$ centered at $A$ between line $BC$ and the semicircle.

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First, let's find the length of hypotenuse $BC$ using Pythagoras theorem:

$BC^2 = AB^2 + AC^2 = 6^2 + 6^2 = 36 + 36 = 72$

$BC = \sqrt{72} = 6\sqrt{2}$ cm

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The semicircle $BQC$ has radius:

$r = \dfrac{BC}{2} = \dfrac{6\sqrt{2}}{2} = 3\sqrt{2}$ cm

Area of semicircle $BQC$:

$\text{Area} = \dfrac{1}{2} \times \pi \times (3\sqrt{2})^2 = \dfrac{1}{2} \times \pi \times 18 = 9\pi$ sq cm

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The sector $BPC$ is centered at $A$ with radius $AB = 6$ cm and angle $\angle BAC = 90°$.

This makes it a quarter circle:

$\text{Area of sector} = \dfrac{90°}{360°} \times \pi \times 6^2 = \dfrac{1}{4} \times \pi \times 36 = 9\pi$ sq cm

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The area of triangle $ABC$:

$\text{Area} = \dfrac{1}{2} \times AB \times AC = \dfrac{1}{2} \times 6 \times 6 = 18$ sq cm

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The region enclosed by arc $BPC$ and semicircle $BQC$ is:

$\text{Required Area} = \text{Area of semicircle} - \text{Area of sector} + \text{Area of triangle}$

$= 9\pi - 9\pi + 18 = 18$ sq cm

The answer is $18$ sq cm.