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

3+0 Answers


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 (surprise), 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+ b= c2 ) to find the length of the last line. A = 5 and C = 10. 5= 25 and 102 = 100. Therefore, 25 + b2 = 100. 100 - 25 = 75, so b2 = 75. This means that b = the square root of 75. The area would be 10*sqrt(75), or about 86.6025.


Hope this helps!! smiley

falcon609  Aug 29, 2017

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



cool cool cool

CPhill  Aug 29, 2017
edited by CPhill  Aug 29, 2017

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.


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.



TheXSquaredFactor  Aug 29, 2017

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