# Thread: Urgent help with geometry problem

1. ## Urgent help with geometry problem

Hi, I'm new to the forum and I need help with this geometry problem.

It is meant to be a layout of a car park constructed out of a normal hexagon and a normal rectangle. So all sides of the hexagon are 9m.

I need to find the following:

1.
Find the perimeter. This one is simple enough.

2.
If A is the midpoint of DC, find the size of the angle y.

3.
Find the length of side x.

4.
Calculate the area of the triangle ABC.

5.
Calculate the area of the whole shape.

I'm mostly just unsure of where to start and what methods to use. If anyone could show me how to work through these questions or point me in the right direction that would be greatly appreciated.

EDIT: I guess this should probably have gone in the Trig forum.

2. Originally Posted by Jahmez
Hi, I'm new to the forum and I need help with this geometry problem.

It is meant to be a layout of a car park constructed out of a normal hexagon and a normal rectangle. So all sides of the hexagon are 9m.

I need to find the following:
1.
Find the perimeter. This one is simple enough.
2.
If A is the midpoint of DC, find the size of the angle y.
3.
Find the length of side x.
4.
Calculate the area of the triangle ABC.
5.
Calculate the area of the whole shape.

I'm mostly just unsure of where to start and what methods to use. If anyone could show me how to work through these questions or point me in the right direction that would be greatly appreciated.

EDIT: I guess this should probably have gone in the Trig forum.
Hello,

to # 2.:
since the 2 trapezoids are the halves of a regular hexagon you get:
AB = BC = AC = 9. That means triangle ABC is an equilateral triangle. Therefore the angle y = 60°.

to #3:
x = AB = 9 m. See above.

to #4.:
If s is the side of an equilateral triangle it's area is calculated by:

$\displaystyle A_{eqi.~triangle} = \frac{1}{4} \cdot s^2 \cdot \sqrt{3}$. Plug in the value you know (s = 9 m).

$\displaystyle A_{eqi.~triangle} = \frac{1}{4} \cdot 9^2 \cdot \sqrt{3} = \frac{81}{4} \cdot \sqrt{3}$

to #5.:

The complete area consists of 6 triangles and 1 rectangle. Since AC = 9 m the width of the rectangle is 18 m. Thus the complete area is:

$\displaystyle A_{complete}=6 \cdot \frac{81}{4} \cdot \sqrt{3} + 18 \cdot 30= \frac{243}{2} \cdot \sqrt{3} + 540 \approx 750.444~m^2$

3. Thanks earboth