Multiple choice

A triangle has vertices $A,\ B$ and $C$ and the respective opposite sides have lengths $a,b$ and $c$. This triangle is inscribed in a circle of radius $R$. If $b=c=1$ and the altitude from $A$ to side $BC$ has length $\sqrt{\dfrac{2}{3}}$, then $R$ equals.

  1. $\dfrac{1}{\sqrt{3}}$
  2. $\dfrac{2}{\sqrt{3}}$
  3. $\dfrac{\sqrt{3}}{2}$
  4. $\dfrac{\sqrt{3}}{2\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In triangle ABC, b = c = 1. Let the base BC = a. The altitude h = sqrt(2/3). Area = 0.5 * a * h = 0.5 * a * sqrt(2/3). Also, Area = (abc) / (4R) = (1 * 1 * a) / (4R) = a / (4R). Equating: 0.5 * a * sqrt(2/3) = a / (4R). 0.5 * sqrt(2/3) = 1 / (4R). R = 1 / (2 * sqrt(2/3)) = sqrt(3) / (2 * sqrt(2)).

AI explanation

In triangle ABC, since b = c = 1, the triangle is isosceles with AB = AC = 1. Using the Pythagorean theorem in the right triangle formed by the altitude from A to BC, half of the base BC is sqrt(1^2 - (sqrt(2/3))^2) = sqrt(1 - 2/3) = sqrt(1/3) = 1/sqrt(3), making the full base a = 2/sqrt(3). Applying the sine rule, 2R = a / sin A; first find cos A using the cosine rule cos A = (1^2 + 1^2 - (2/sqrt(3))^2) / 2 = (2 - 4/3) / 2 = 1/3, so sin A = sqrt(1 - (1/3)^2) = 2sqrt(2)/3. Thus, R = a / (2 sin A) = (2/sqrt(3)) / (2 * 2sqrt(2)/3) = (2/sqrt(3)) * (3 / (4sqrt(2))) = sqrt(3) / (2sqrt(2)). The value of R is sqrt(3) / (2sqrt(2)).

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