Question

a napkin is folded into an isosceles triangle, triangle abc, and placed on a plate, as shown. the napkin has a perimeter of 38 centimeters. to the nearest square centimeter, how many square centimeters of the plate are covered by the napkin? 16 square centimeters 30 square centimeters 56 square centimeters 60 square centimeters

Explanation:

Step1: Analyze the isosceles triangle

The napkin is an isosceles triangle \(ABC\) with \(AB = 8\) cm, and the perimeter is 38 cm. Let the equal sides ( \(AC = BC\) ) be \(x\). So, \(2x + 8=38\). Solving for \(x\): \(2x = 38 - 8 = 30\), so \(x = 15\) cm. So \(AC = BC = 15\) cm.

Step2: Use the formula for the area of a triangle

The area of a triangle is \(\frac{1}{2}ab\sin C\), where \(a\) and \(b\) are two sides and \(C\) is the included angle. Here, \(a = 15\), \(b = 15\), and the included angle (at \(C\))? Wait, no, looking at the diagram, the angle at \(O\) (wait, maybe the triangle is \(OAC\) or \(OAB\)? Wait, maybe the triangle is isosceles with two sides as 15 (from perimeter) and base 8, and the angle between the two equal sides? Wait, maybe the height? Wait, alternatively, using the formula \(\frac{1}{2} \times \text{base} \times \text{height}\). Wait, maybe the triangle has sides \(OA = OC\) (since it's isosceles), and \(AC = 8\) cm, and the angle at \(O\) is \(30^\circ\). Wait, maybe the area of triangle \(OAC\) or \(OAB\)? Wait, the napkin is triangle \(ABC\), and we need the area of the plate not covered? Wait, no, the question is how many square centimeters of the plate are covered by the napkin? Wait, maybe the triangle has sides: let's re - evaluate.

Wait, the perimeter of the napkin (triangle \(ABC\)) is 38 cm, \(AB = 8\) cm, so the other two sides \(AC\) and \(BC\) are equal. So \(AC=BC=\frac{38 - 8}{2}=15\) cm. Now, if we consider the triangle with two sides \(a = 15\) cm, included angle \(30^\circ\)? Wait, no, maybe the triangle is \(OAC\) where \(OA = OC = 15\) cm, angle at \(O\) is \(30^\circ\), and \(AC = 8\) cm? Wait, no, let's use the formula for the area of a triangle with two sides and included angle: \(A=\frac{1}{2}ab\sin C\). If \(a = 15\), \(b = 15\), and \(C = 30^\circ\), then \(A=\frac{1}{2}\times15\times15\times\sin30^\circ\). \(\sin30^\circ=\frac{1}{2}\), so \(A=\frac{1}{2}\times15\times15\times\frac{1}{2}=\frac{225}{4} = 56.25\approx56\) square centimeters.

Answer:

56 square centimeters (corresponding to the option "56 square centimeters")