A Triangle Is Inscribed in a Semicircle of Diameter 6R $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$

question 18

Multiple Choice

A triangle is inscribed in a semicircle of diameter 6R. Show that the smallest possible value for the area of the shaded region is $\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }$ . Hint: The area of the shaded region is a minimum when the area of the triangle is a maximum. Find the value of x that maximizes the square of the area of the triangle. This will be the same x that maximizes the area of the triangle.

• A) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to \(A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }\) . The square of the area of the triangle is equal to \(( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }\) , and the substitution \(x ^ { 2 } = t\) will transform this expression into the quadratic function \(- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )\) Since we want to find the maximum value of t, we will substitute the value \(t = 18 R ^ { 2 } = x ^ { 2 }\) into the equation. Solving for t gives us the following minimum area of the shaded region: \(t = \frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }\) .
• B) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The area of the triangle is equal to \(A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }\) . The square of the area of the triangle is equal to \(( A ( x ) ) ^ { 2 } = 9 R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }\) The substitution \(x ^ { 2 } = t\) will transform this expression into the quadratic function \(- \frac { 1 } { 4 } t ^ { 2 } + 9 R ^ { 2 } t ( 1 )\) Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is \(t = \frac { - b } { 2 a } = 18 R ^ { 2 }\) Substituting the value \(t = 18 R ^ { 2 }\) into the equation \(t = x ^ { 2 }\) gives us \(x ^ { 2 } = 18 R ^ { 2 }\) and consequently \(x = 3 R \sqrt { 2 }\) (The negative root can be rejected since the side of a triangle can't be negative). With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \(\frac { 9 \pi R ^ { 2 } } { 2 } - \frac { 1 } { 2 } x \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }\) Substituting the value \(x = 3 R \sqrt { 2 }\) in the equation (2) gives us that the minimum value of the shaded region is equal to \(\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }\) .
• C) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to \(\frac { 1 } { 4 } 9 R ^ { 2 } x ^ { 2 } + \frac { 1 } { 4 } x ^ { 4 }\) . The substitution \(x ^ { 2 } = t\) will transform this into the quadratic function \(\frac { 1 } { 4 } 9 R ^ { 2 } t + \frac { 1 } { 4 } t ^ { 2 }\) Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is \(t = \frac { - b } { 2 a } = 9 R ^ { 2 }\) Substituting the value \(t = 9 R ^ { 2 }\) into the equation \(t = x ^ { 2 }\) gives us \(x ^ { 2 } = 9 R ^ { 2 }\) and consequently \(x = 3 R\) . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \(\pi 9 R ^ { 2 } - \frac { 1 } { 2 } x \sqrt { 6 R ^ { 2 } - x ^ { 2 } }\) Substituting the value \(x = 3 R\) into the equation (2), we find that the minimum value of the shaded region is equal to \(\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }\) .
• D) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to \(A ( x ) = \frac { x } { 2 } \sqrt { ( 6 R ) ^ { 2 } - x ^ { 2 } }\) , which is a quadratic function. The graph of this function will be a parabola opening downward, so we can write the maximum value of this function as: \(x ^ { 2 } = \frac { - b } { 2 a } = 18 R ^ { 2 }\) We can then write \(x ^ { 2 } = 18 R ^ { 2 }\) as \(x ^ { 2 } = 18 R ^ { 2 }\) and calculate the minimum area of the shaded region. Substituting this value into the area equation, we find its minimum area: \(\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }\) .
• E) The hint tells us that the area of the region is a minimum when the area of the triangle is a maximum. We first find the value of x that maximizes the square of the area of the triangle. The square of the area of the triangle is equal to \(\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 4 }\) . The substitution \(x ^ { 2 } = t\) will transform this into the quadratic function \(\frac { 1 } { 2 } 9 ^ { 2 } R ^ { 2 } t - \frac { 1 } { 4 } t ^ { 2 }\) Since the graph of equation (1) will be a parabola opening downward, the input t that yields a maximum value for this function is \(t = \frac { - b } { 2 a } = 3 R ^ { 2 }\) Substituting the value \(t = 3 R ^ { 2 }\) into the equation \(t = x ^ { 2 }\) gives us \(x ^ { 2 } = 3 R ^ { 2 }\) and consequently \(x = R \sqrt { 7 }\) . With this value of x, we can calculate the minimum area of the shaded region. The minimum area of the shaded region is equal to \(\frac { \pi 9 R ^ { 2 } } { 4 } - \frac { 1 } { 2 } x \sqrt { 9 R ^ { 2 } - x ^ { 2 } }\) Substituting the value \(x = R \sqrt { 7 }\) into the equation (2), we find that the minimum value of the shaded region is equal to \(\frac { 9 ( \pi - 2 ) R ^ { 2 } } { 2 }\) .