What is the greatest possible area of a triangular region with one vertex at the center of the circle of radius 1 and other 2 vertices on the circle.
The ans I have is 1/2 . Pls can anyone solve this.
2007-02-05 17:51:54 · 3 answers · asked by rajagopal r 1 in Science & Mathematics ➔ Mathematics
Let x be the altitude of the triangle, as measured from the center of the circle. Half of the base of the triangle is then √ (1-x²), and the area is x√ (1-x²).
Differentiate for x, set it equal to 0, and solve for x, getting x = 1/√ 2. Area will then be 1/2.
2007-02-05 18:01:24 · answer #1 · answered by Scythian1950 7 · 1⤊ 0⤋
A = sinθcosθ, 0 < θ < 90 (0 < 2θ < 180)
dA/dθ = -sin^2θ + cos^2θ = 0 for max area
tanθ = 1
θ = Ï/4
sin(Ï/4)cos(Ï/4) = â2â2/(2*2) = 2/4 = 1/2
2007-02-06 02:05:41 · answer #2 · answered by Helmut 7 · 0⤊ 0⤋
area of triangle = 1/2r^2 sin(theta)
=1/2 sin(theta)
now,for area to be maximum sin(theta) should be maximum=1
therefore, maximum area is 1/2
[theta is the central angle]
2007-02-08 02:54:54 · answer #3 · answered by Apoorv g 2 · 0⤊ 0⤋