Question

In a triangle ABC,AB=AC=8cm. 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

• A. \(16(\pi−1)\)
• B. \(32(π−1)\)
• C. \(32π\)
• D. \(16π\)

Answer 1

The Correct Option is B

The correct answer is B:(\(32(\pi-1)\)
Given: 
Triangle ABC with AB=AC=8cm. 
Circle with BC as diameter passes through A (let's call this Circle 1). 
Circle with center at A passes through B and C (let's call this Circle 2). 

First, let's find the radius of Circle 1: 
The diameter of Circle 1 is BC,which is also the side AC of triangle ABC.Since AB=AC=8cm,BC=8cm. 
So, the radius of \(Circle\space 1=\frac{BC}{2}=\frac{8}{2}=4cm\). 
Next,let's find the radius of Circle 2: 
Since Circle 2 has center at A and passes through B and C, the radius of Circle 2 is AB=8cm. 
Now,let's find the distance between the centers of Circle 1 and Circle 2: 
The centers of Circle 1 and Circle 2 are A and B,respectively.The distance between A and B is AB=8cm. 
The overlapping region between two circles occurs when the distance between their centers is less than the sum of their radii.In this case, the condition for overlap is: 
Distance between centers<Radius of Circle 1+Radius of Circle 2 
8 cm< 4cm+8cm 
8 cm< 12cm 
Since 8cm is indeed less than 12cm, there is an overlapping region between the two circles. 
To find the area of the overlapping region, we can use the formula for the area of the intersection of two circles: 
Area of overlap=Circle 1 area sector-Triangle area+Circle 2 area sector 
Circle 1 area sector=\((\frac{θ}{360}) \times π \times r^2\), where θ is the angle of the sector formed by the overlapping region (in degrees), and r is the radius of Circle 1. 
Triangle area=\(0.5\times{base}\times{height}\), where the base is AB=8cm, and the height is the distance between the centers of the circles (8 cm). 
Circle 2 area sector=\((\frac{θ}{360}) \times π \times r^2\), where θ is the angle of the sector formed by the overlapping region (in degrees), and r is the radius of Circle 2. 
Substituting the values: 
\(θ\)=360° (since the overlapping region is essentially the whole circle) 
r (Circle 1)=4 cm 
r (Circle 2)=8 cm 
Circle 1 area sector=\((\frac{360}{360}) \times π \times (4^2) = 16π\) 
Triangle area=\(0.5 \times 8 \times 8=32\)
Circle 2 area sector=\((\frac{360}{360}) \times π \times (8^2) = 64π\)
Area of overlap=\(16π-32+64π=80π-32=48π\)
So, the correct option is: B. 32(π−1) 

Answer 2

Since BC is circle C2's diameter, we may state that \(∠BAC=90 ^∘\) because the semicircle's angle is \(90^ ∘\). 
Thus, the overlapping area is equal to \(\frac{1}{2}\) (the circle's C2) plus the area of the minor sector formed by BC in C1.

Since AB = AC = 8 cm and \(∠BAC = 90^∘\), we may determine that BC =\(8\sqrt{2}\) cm.
Half of BC's length is the radius of C2. 
Area of C2 =\(\pi (4\sqrt{2})^2=32\pi \text{cm}^2\)

Since C1 travels through B and has A as its center, its radius, or AB, is equal to 8 cm. 
This corresponds to the area of the minor sector formed by BC in C1 = \(\frac{1}{4}\)(Area of circle C1) - Area of triangle ABC =\(\frac{1}{4}\pi(8)^2-(\frac{1}{2}\times8\times8)\)
\(=16\pi-32\text{cm}^2\)
Thus, the area that overlaps between the two circles =\(\frac{1}{2}\)(Area of circle C2) + Area of the minor sector made be BC in C1
\(=\frac{1}{2}(32\pi)+(16\pi-32)\)
\(=32(\pi-1) cm^2\)