will someone please help me with these 9 problems that i have already did and got wrong... Please help

- anonymous

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- anonymous

these two i had to find the base area... and i got them wrong

##### 2 Attachments

- anonymous

What are the shapes of bases on each picture? What are the formulas to calculate areas of these figures?

- anonymous

traingle and a circle

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- anonymous

Correct. What kind of triangle? It's a *regular* pyramid.

- anonymous

equallaterial

- anonymous

Yup. So you have to find an area of a circle given it's radius and an area of an equilateral triangle given the length of this side. Both formulas can be found easily, e.g. on wikipedia.
If you want an additional challenge you can devise the formula for triangle using the Pythagorean theorem (if you draw its height you'll get a right triangle).

- anonymous

for the triangle wouldnt it be 1/2*4*4=8cm

- anonymous

Nope, its height does not equal 4.

- anonymous

what is the height

- anonymous

"Altitude" is another word. http://en.wikipedia.org/wiki/Altitude_%28triangle%29
I assume you used the formula for triangle that's given as area = 1/2 * base * height.

- anonymous

yea...
for the cone the base area would be 113.0976 or 113.1in

- anonymous

\[113.1 in^{2} \]
That is square inches. Correct.

- anonymous

i still dont know how to find the base area of the pyramid

- anonymous

Area of base: ½b×h

- anonymous

1/24*7=14cm

- anonymous

1/2*4*7=14cm

- anonymous

The wikipedia page about equilateral triangles has the formula that uses only the side length.

- anonymous

1/2*4*7=14cm so this isnt right

- anonymous

No, it's not right.

- anonymous

i give up

- anonymous

Too early. You're given an equilateral triangle with side length of 4cm. Apply the formula from: http://en.wikipedia.org/wiki/Equilateral_triangle

- anonymous

A=\frac{\sqrt{3}}{4} a^2

- anonymous

Yes, this one.

- anonymous

6.93

- anonymous

Correct!

- anonymous

okay.. how are these two wrong.. finding the volume

##### 2 Attachments

- anonymous

In both cases you got the base area wrong. First case is the same as the problem you've just solved. Second one has a base that can be thought of as consisting of multiple triangles.

- anonymous

the first one the base area would be 10.83

- anonymous

Indeed.

- anonymous

so the volume would be 32.49in cubed

- anonymous

True.

- anonymous

what formula would i use to get the base area for the hexagon

- anonymous

It consists of six equilateral triangles.

- anonymous

\[\sqrt{6}\div4\times3^{3}\]

- anonymous

\[3^{3}?\]

- anonymous

7^3

- anonymous

Let's do it more slowly. What's the area of one of the triangles that form that hexagon?

- anonymous

\[\sqrt{3}\div4\times7^{3}\]

- anonymous

7^2

- anonymous

Yes. And there's six of them. So the area of the whole hexagon is..?

- anonymous

21.21*5

- anonymous

5?

- anonymous

21.22*6

- anonymous

Much better. You can get the correct volume now.

- anonymous

127.31

- anonymous

Looks good to me.

- anonymous

381.92 would be the volume?

- anonymous

No, you must have missed something in the formula for volume.

- anonymous

127.31 this is the area of one side base or all

- anonymous

There's only one base in this case.

- anonymous

127.31 is the base area right

- anonymous

Yes.

- anonymous

than to get the volume dont i * by 3

- anonymous

You do, but there's a division there too.

- anonymous

190.97

- anonymous

Pyramid (1/3) * (base area) * height

- anonymous

That's the correct formula, yes. Height is 3cm if I read the picture correctly.

- anonymous

yeah

- anonymous

(1/3) * 3 * 127.31 = ..?

- anonymous

190.97cm^3

- anonymous

127.32 * 3 / 3 = ..?

- anonymous

127.32

- anonymous

Aye.

- anonymous

(127.32 * 3) / 3 = 127.32

- anonymous

Multiply by the numerator, divide by denominator. That's how you multiply a number by fraction. Surprisingly 127.32 cubic centimeters is the answer. It's the same number as the base area, but unit differs (cubed vs squared).

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