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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
- a)16(π – 1)
- b)32π
- c)16π
- d)32(π – 1)
Correct answer is option 'D'. Can you explain this answer?
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In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter ...
To find the area of the overlapping region between the two circles, we need to find the radius of each circle first.
Since AB = AC = 8 cm, triangle ABC is an isosceles triangle. Therefore, the altitude from A to BC is also the perpendicular bisector of BC. This means that the circle with BC as diameter is actually the circumcircle of triangle ABC.
Let's call the center of the circumcircle O. Since OA is the radius of the circumcircle, it is also the perpendicular bisector of BC. Therefore, OA is also the altitude from A to BC.
Let's call the intersection point of OA and BC as D. Triangle AOD is a right triangle, and since AO is the radius of the circumcircle, it is also the hypotenuse of triangle AOD. Therefore, AD = 4 cm.
Since triangle ABC is isosceles, the altitude from A to BC (AD) is also the median and angle bisector of angle BAC. Therefore, angle BAD is a right angle.
Let's call the center of the circle with center at A as P. Since AP is the radius of this circle, we need to find AP.
Triangle ABP is a right triangle, and since AB = AC = 8 cm, it is an isosceles right triangle. Therefore, angle BAP is a right angle.
Since angle BAD is a right angle and angle BAP is a right angle, angles BAD and BAP are congruent. Therefore, triangles BAD and BAP are similar.
Using the property of similar triangles, we can set up the following proportion:
BD / BA = BA / BP
Since BD = AD = 4 cm and BA = 8 cm, we can substitute these values into the proportion:
4 / 8 = 8 / BP
Simplifying the proportion:
1/2 = 8 / BP
Cross multiplying:
BP = 16 cm
Therefore, the radius of the circle with center at A is 16 cm.
Now, we can find the area of the overlapping region between the two circles.
The area of a circle is given by the formula A = πr^2.
The area of the overlapping region between the two circles is the difference between the areas of the two circles.
The area of the circumcircle is A1 = π(8 cm)^2 = 64π cm^2.
The area of the circle with center at A is A2 = π(16 cm)^2 = 256π cm^2.
The area of the overlapping region is A1 - A2 = 64π cm^2 - 256π cm^2 = -192π cm^2.
Therefore, the area of the overlapping region between the two circles is -192π cm^2.
Since AB = AC = 8 cm, triangle ABC is an isosceles triangle. Therefore, the altitude from A to BC is also the perpendicular bisector of BC. This means that the circle with BC as diameter is actually the circumcircle of triangle ABC.
Let's call the center of the circumcircle O. Since OA is the radius of the circumcircle, it is also the perpendicular bisector of BC. Therefore, OA is also the altitude from A to BC.
Let's call the intersection point of OA and BC as D. Triangle AOD is a right triangle, and since AO is the radius of the circumcircle, it is also the hypotenuse of triangle AOD. Therefore, AD = 4 cm.
Since triangle ABC is isosceles, the altitude from A to BC (AD) is also the median and angle bisector of angle BAC. Therefore, angle BAD is a right angle.
Let's call the center of the circle with center at A as P. Since AP is the radius of this circle, we need to find AP.
Triangle ABP is a right triangle, and since AB = AC = 8 cm, it is an isosceles right triangle. Therefore, angle BAP is a right angle.
Since angle BAD is a right angle and angle BAP is a right angle, angles BAD and BAP are congruent. Therefore, triangles BAD and BAP are similar.
Using the property of similar triangles, we can set up the following proportion:
BD / BA = BA / BP
Since BD = AD = 4 cm and BA = 8 cm, we can substitute these values into the proportion:
4 / 8 = 8 / BP
Simplifying the proportion:
1/2 = 8 / BP
Cross multiplying:
BP = 16 cm
Therefore, the radius of the circle with center at A is 16 cm.
Now, we can find the area of the overlapping region between the two circles.
The area of a circle is given by the formula A = πr^2.
The area of the overlapping region between the two circles is the difference between the areas of the two circles.
The area of the circumcircle is A1 = π(8 cm)^2 = 64π cm^2.
The area of the circle with center at A is A2 = π(16 cm)^2 = 256π cm^2.
The area of the overlapping region is A1 - A2 = 64π cm^2 - 256π cm^2 = -192π cm^2.
Therefore, the area of the overlapping region between the two circles is -192π cm^2.
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