+21 A quadrilateral with perimeter 52 has an inscribed circle with radius 5. What is the area of the quadrilateral?
+121 Let's denote the sides of the quadrilateral as \(a\), \(b\), \(c\), and \(d\). Given that the perimeter is 52, we have:
\[a + b + c + d = 52.\]
We're also given that the quadrilateral has an inscribed circle with a radius of 5. An inscribed circle touches the quadrilateral at its four sides, so the sum of the lengths of opposite sides of the quadrilateral is equal to its perimeter. That is:
\[a + c = b + d = 52/2 = 26.\]
We can use these equalities to express the sides of the quadrilateral in terms of \(a\) and \(b\):
\[c = 26 - a, \quad d = 26 - b.\]
The area of a quadrilateral inscribed in a circle can be calculated using the formula:
\[Area = \sqrt{(s - a)(s - b)(s - c)(s - d)},\]
where \(s\) is the semiperimeter (\(s = \frac{a + b + c + d}{2}\)).
Substituting the values we have:
\[s = \frac{a + b + c + d}{2} = \frac{52}{2} = 26.\]
\[Area = \sqrt{(26 - a)(26 - b)(26 - c)(26 - d)}.\]
Substituting the expressions for \(c\) and \(d\) in terms of \(a\) and \(b\):
\[Area = \sqrt{(26 - a)(26 - b)(a)(b)}.\]
We want to maximize the area, which occurs when \(a\) and \(b\) are equal. So, let's assume \(a = b\):
\[Area = \sqrt{(26 - a)(26 - a)(a)(a)} = \sqrt{(26 - a)^2 a^2} = (26 - a) a.\]
To maximize the area, we should choose \(a\) such that the expression \((26 - a) a\) is maximized. This occurs when \(a = 13\) (the midpoint between 0 and 26):
\[Area = (26 - 13) \cdot 13 = 13 \cdot 13 = 169.\]
So, the area of the quadrilateral is 169 square units.
