Area Of Regular Polygons

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January 6th 2007, 06:52 PM
Blayde

Area Of Regular Polygons

Ok, I need help with my geometry homework. If anyone can thoroughly explain or something I'd appreciate it much.

"If the area of a trapezoid is 4x^2+4x+1 and one base is 2x+3 and the height is 2x+1, what is the length of the other base? Answers is in terms of x. "

Also I know that the area of a regular hexagon with a radius of 7 has an area of 127, but I've gotten down to 147 x sqrt(3) x 1/2. I know for sure that it's 127, I just don't know how to get 127 from there and have it in simplest radical form.

And... to find the area of an equilateral triangle with the sides of 5 cms.

If anyone can help.. i'd appreciate it.
January 6th 2007, 07:10 PM
ThePerfectHacker

Quote:

Originally Posted by Blayde

Ok, I need help with my geometry homework. If anyone can thoroughly explain or something I'd appreciate it much.

"If the area of a trapezoid is 4x^2+4x+1 and one base is 2x+3 and the height is 2x+1, what is the length of the other base? Answers is in terms of x. "
.

$A=\frac{1}{2}(b_1+b_2)h$

$4x^2+4x+1=\frac{1}{2}(2x+3+b)(2x+1)$
$8x^2+8x+2=(2x+3+b)(2x+1)$
$(2x+1)(4x+2)=(2x+3+b)(2x+1)$
$4x+2=2x+3+b$
$4x-1=2x+b$
$2x-1=b$
January 6th 2007, 11:49 PM
earboth

Quote:

Originally Posted by Blayde

Ok, I need help with my geometry homework. ...

Also I know that the area of a regular hexagon with a radius of 7 has an area of 127, but I've gotten down to 147 x sqrt(3) x 1/2. I know for sure that it's 127, I just don't know how to get 127 from there and have it in simplest radical form.

...

Hello,

the area of a regular hexagon consists of 6 equilateral triangles. The side of this triangle is equal to r, in your case r = 7 cm.

$A_{6-gon}=6 \cdot \frac{1}{2} \underbrace{\cdot \underbrace{r}_{base} \cdot \underbrace{\frac{1}{2} \cdot r \cdot \sqrt{3}}_{height}}_{of \ triangle}=\frac{3}{2} r^2 \cdot \sqrt{3}$

Plug in r = 7 and you'll get: A = 127.3057...cm²

From the formula above you see that the area of an equilateral triangle with the side a is:

$A_{triangle}=\frac{1}{2} \cdot a \cdot \frac{1}{2} \cdot a \cdot \sqrt{3}=\frac{1}{4} \cdot a^2 \cdot \sqrt{3}$

With a = 5 cm you'll get: A = 10.825... cm²

EB
January 7th 2007, 05:10 AM
Quick

The area of a regular polygon with "radius" $r$ (meaning the distance from the center to a vertex) that has $n$ number of sides is:

$r^2n\left(\sin\left(\frac{360}{2n}\right)\times\co s\left(\frac{360}{2n}\right)\right)$

At least I think....

You should check that first.
January 7th 2007, 09:42 AM
Blayde

Quote:

Originally Posted by earboth

Hello,

the area of a regular hexagon consists of 6 equilateral triangles. The side of this triangle is equal to r, in your case r = 7 cm.

$A_{6-gon}=6 \cdot \frac{1}{2} \underbrace{\cdot \underbrace{r}_{base} \cdot \underbrace{\frac{1}{2} \cdot r \cdot \sqrt{3}}_{height}}_{of \ triangle}=\frac{3}{2} r^2 \cdot \sqrt{3}$

Plug in r = 7 and you'll get: A = 127.3057...cm²

From the formula above you see that the area of an equilateral triangle with the side a is:

$A_{triangle}=\frac{1}{2} \cdot a \cdot \frac{1}{2} \cdot a \cdot \sqrt{3}=\frac{1}{4} \cdot a^2 \cdot \sqrt{3}$

With a = 5 cm you'll get: A = 10.825... cm²

EB

Thanks everyone.

Thank you guys so much.