sanju09 wrote:A triangle is inscribed in a circle whose radius is 2 centimeters. Each vertex of the triangle is on the circle and one side of the triangle is a diameter of the circle. What is the largest possible area of the triangle in square centimeters?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12

From the diagram in the attachment, we can say that area of the triangle(in black) is (1/2)*a*b. So for (1/2)*a*b be be maximum, the value of a should be equal to the value of b.

According to Pythagorean Theorem, a^2 + b^2 = 4^2. Since a=b, a^2 + a^2 = 4^2.
2*a^2 = 4^2
a^2 = 4*4/2
a = 2√2

Area of the triangle = (1/2)*a*b = (1/2)*2√2*2√2 = 4 sq cms

neelgandham wrote:

sanju09 wrote:A triangle is inscribed in a circle whose radius is 2 centimeters. Each vertex of the triangle is on the circle and one side of the triangle is a diameter of the circle. What is the largest possible area of the triangle in square centimeters?
(A) 2
(B) 4
(C) 6
(D) 8
(E) 12

From the diagram in the attachment, we can say that area of the triangle(in black) is (1/2)*a*b. So for (1/2)*a*b be be maximum, the value of a should be equal to the value of b.

According to Pythagorean Theorem, a^2 + b^2 = 4^2. Since a=b, a^2 + a^2 = 4^2.
2*a^2 = 4^2
a^2 = 4*4/2
a = 2√2

Area of the triangle = (1/2)*a*b = (1/2)*2√2*2√2 = 4 sq cms

The other way round, we can take diameter (4) as fixed base, such that the radius (2) is the maximum possible height for the inscribed triangle, and hence the largest possible area of the triangle

= ½ (4) (2) = [spoiler]4.


Pick B[/spoiler]