In a triangle $ABC, AB = AC = 8$cm. $A$ circle drawn with $BC$ as diameter passes through $A$. Another circle drawn with center at $A$ passes through $B$ and $C$. Then the area, in sq. cm, of the overlapping region between the two circles is

We have triangle ABC where AB = AC = 8 cm, making it an isosceles triangle.

Here's the key insight: Since a circle with BC as diameter passes through point A, we can use Thales' theorem.

Thales' Theorem: If a point lies on a circle and the line segment connecting it to the endpoints of a diameter, then the angle at that point is 90°.

Since A lies on the circle with BC as diameter, angle BAC = 90°.

This means triangle ABC is a right-angled isosceles triangle with the right angle at A.

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Using the Pythagorean theorem in right triangle ABC:

$BC^2 = AB^2 + AC^2 = 8^2 + 8^2 = 64 + 64 = 128$

$BC = \sqrt{128} = \sqrt{64 \times 2} = 8\sqrt{2}$ cm

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Circle 1: Center at midpoint of BC, radius = BC/2 = 4√2 cm

Circle 2: Center at A, radius = AB = AC = 8 cm

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The overlapping region consists of two parts:

Since the circle with BC as diameter passes through A, point A lies exactly on this circle. The overlapping region includes the entire semicircle on one side of BC.

Area of semicircle = $\frac{1}{2} \times \pi r^2 = \frac{1}{2} \times \pi \times (4\sqrt{2})^2 = \frac{1}{2} \times \pi \times 32 = 16\pi$

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From the circle centered at A, we need the segment that extends beyond triangle ABC.

This segment = Area of sector - Area of triangle ABC

Since angle BAC = 90°, the sector is a quarter circle.

Area of sector = $\frac{90°}{360°} \times \pi \times 8^2 = \frac{1}{4} \times \pi \times 64 = 16\pi$

Area of triangle ABC = $\frac{1}{2} \times AB \times AC = \frac{1}{2} \times 8 \times 8 = 32$

Area of segment = 16π - 32

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Total area = Area of semicircle + Area of segment

$= 16\pi + (16\pi - 32) = 32\pi - 32 = 32(\pi - 1)$

Therefore, the area of the overlapping region is 32(π - 1) sq. cm.

Key Takeaway: When you see a circle with a chord as diameter passing through another point, immediately think of Thales' theorem - it creates a right angle that often simplifies the entire problem!