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# Salaam to anybody who can solve this

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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 Aug 29, 2017

### 3+0 Answers

#1
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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+ 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!! Aug 29, 2017
#2
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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   Aug 29, 2017
edited by CPhill  Aug 29, 2017
#3
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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.

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

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Aug 29, 2017