In the figure below, triangle ZAX is equilateral and has perimeter 30.

What is the exact area of the quadrilateral FLAX? Explain how you determined your answer

Guest Aug 29, 2017

#1**+2 **

The area of quadrilateral FLAX is 10 x sqrt(75).

Triangle ZAX is equilateral and has a perimeter of 30. Each side is equal and there are 3 sides (), so 30/3=10. Therefore, line XA and FL are 10 units long. Since triangle ZAX is equilateral, point Z is in the middle of line FL. Line FL is 10, so lines FZ is 5 units long. Line ZX (part of the triangle) is 10 units long. We can use the Pythagorean theorem (a^{2 }+ b^{2 }= c^{2} ) to find the length of the last line. A = 5 and C = 10. 5^{2 }= 25 and 10^{2} = 100. Therefore, 25 + b^{2} = 100. 100 - 25 = 75, so b^{2} = 75. This means that b = the square root of 75. The area would be 10*sqrt(75), or about 86.6025.

Hope this helps!!

falcon609
Aug 29, 2017

#2**+1 **

Since ZAX is equilateral.....it must have a side = 10

And the height of the quadrilateral, h, is given by :

h = 10 * sin (60 degrees) = 10√3/2 = 5/√3 units

And the area of the quadrilateral = side of ZAX * height of quadrilateral = 10 * 5√3 =

50 √3 square units

CPhill
Aug 29, 2017

#3**+1 **

Coincidentally, trigonometry was unnecessary in this problem (although use whatever solving method you'd like!).

The height of the rectangle (we know it is a rectangle because it a quadrilateral with 4 right angles) \(FLAX\) is of equal length to the height of\(\triangle ZAX\).

Let's draw in \(\overline{ZB}\) such that the segment is a perpendicular bisector of \(\overline{FL}\) and \(B\) lies on the midpoint of \(\overline{XA}\).

Inserting the altitude of an isosceles (or equilateral, in this case) triangle has 2 lesser-known properties.

1) The altitude bisects the intersected angle.

2) The altitude bisects the intersected segment.

Put all of this information together \(m\angle AZB=30^{\circ}\hspace{1mm}\text{and}\hspace{1mm}m\angle ZBA=90^{\circ}\hspace{1mm}\text{and}\hspace{1mm} m\angle BAZ=60^{\circ}\). \(AB=5\hspace{1mm}\text{and}\hspace{1mm} AZ=10\)

It's a 30-60-90 triangle. Yay! By definition, the ratio of the side lengths is \(1:\sqrt{3}:2\). Knowing this, we can calculate the length of the altitude.

\(\frac{ZB}{\sqrt{3}}=\frac{5}{1}\) | Multiply by the square root of 3 on both sides of the equation. |

\(ZB=5\sqrt{3}\) | |

It has already been established that ZB is the height of the rectangle. Therefore, multiply the length and the width together to get the total area.

\(A=10*5\sqrt{3}=50\sqrt{3}units^2\)

TheXSquaredFactor
Aug 29, 2017