AB is a diameter of a circle and C is any point on the circumference o...
Yes A is correct coz suppose O be the centre of circle. When the triangle ABC is isosceles then CO will be the radius of the the circle and also the altitude and the base is diameter. These 2 are maximum in case of the triangle inscribed in a semicircle. so area will be maximum.
AB is a diameter of a circle and C is any point on the circumference o...
Analysis:
Let's analyze each option one by one.
a) The area of triangle ABC is maximum when it is isosceles.
To prove this, let's consider the triangle ABC as shown below:
A
/ \
C------B
Since AB is a diameter of the circle, angle ACB is a right angle. Let's consider angle CAB (angle A) and angle ABC (angle B) to be x.
Since the sum of angles in a triangle is 180 degrees, we have:
x + x + 90 = 180
2x = 90
x = 45
Now, let's draw the altitude from point C to the base AB and call it CD. Since triangle ABC is isosceles, the altitude will bisect the base AB. Let's call the point of intersection of CD and AB as E.
A
/ | \
C--D--B
|
E
Since CD bisects AB, CE = ED. Let's consider CE = ED = h.
Now, let's calculate the area of triangle ABC.
Area of triangle ABC = (1/2) * AB * CD
= (1/2) * AB * h
Since AB is fixed (as it is the diameter of the circle), the area of triangle ABC is directly proportional to the height h. And since CE = ED, the height h is maximum when triangle ABC is isosceles.
Therefore, the area of triangle ABC is maximum when it is isosceles. Hence, option a) is correct.
b) The area of triangle ABC is minimum when it is isosceles.
This statement is incorrect as we have just proved in option a) that the area of triangle ABC is maximum when it is isosceles.
c) The perimeter of triangle ABC is minimum when it is isosceles.
This statement is not mentioned in the question, so we cannot conclude anything about the perimeter of triangle ABC.
d) None of these.
Since option a) is correct, option d) is incorrect.
Hence, the correct answer is option a) - the area of triangle ABC is maximum when it is isosceles.
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