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24 Mar 2020, 07:49
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An isosceles triangle is inscribed in a circle such that one of the triangle’s sides coincides with the circle’s longest chord. If the measure of another side of the triangle is 5, what is the difference between the area of the circle and the area of the triangle?
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
24 Mar 2020, 10:05
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the given ∆ is isoscele ; 45;45;90 ; 5;5;5√2
so dia = 5√2 ; radius ; 5/√2
area of circle ; 25/2 * pi and area of ∆ ; 1/2 *25
so ∆ of area ;
\(\frac{25}{2}(\pi - 1)\)
IMO D
so dia = 5√2 ; radius ; 5/√2
area of circle ; 25/2 * pi and area of ∆ ; 1/2 *25
so ∆ of area ;
\(\frac{25}{2}(\pi - 1)\)
IMO D
Bunuel
An isosceles triangle is inscribed in a circle such that one of the triangle’s sides coincides with the circle’s longest chord. If the measure of another side of the triangle is 5, what is the difference between the area of the circle and the area of the triangle?
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
26 Mar 2020, 14:25
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Bunuel
An isosceles triangle is inscribed in a circle such that one of the triangle’s sides coincides with the circle’s longest chord. If the measure of another side of the triangle is 5, what is the difference between the area of the circle and the area of the triangle?
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
A. \(\frac{25}{2}(1 - \pi)\)
B. \(25(1 - \pi)\)
C. \(\frac{25}{2}(\pi + 1)\)
D. \(\frac{25}{2}(\pi - 1)\)
E. \(\frac{25}{4}(\pi - 1)\)
The longest chord of a circle is its diameter. So the isosceles triangle that is inscribed in the circle must have a side that is the diameter of the circle. Recall that every triangle inscribed in a circle that has a side that is the diameter of the circle is a right triangle. Therefore, the isosceles triangle must be an isosceles right triangle, i.e., a 45-45-90 triangle. Since we are given that a leg of this isosceles right triangle is 5, its hypotenuse is 5√2. Since the hypotenuse of the triangle is the diameter of the circle, the diameter is also 5√2.
The area of the circle is π(5√2/2)^2 = π(25*2/4) = 25π/2. The area of the triangle is ½ * 5^2 = 25/2. Therefore, the difference between the two areas is 25π/2 - 25/2 = 25/2 * (π - 1).
Answer: D
