Question

Geometry.

Answer 1

(Area of triangle-area of circle) * $10=
A(triangle) = \[1/2(b*h)\]
A(circle) =\[\pi*r ^{2}\]

Answer 2

make sense?

Answer 3

Yeah I know that, I just don't know how to find the area of a triangle without knowing the height.

Answer 4

cut the triangle in half. now you have a triangle & know 2 sides.

Answer 5

the 3rd side will be the hieght of the original

Answer 6

Are oyu meaning the middle? As in the line you draw will be the third side? because even then I wouldnt know how tall that is.

Answer 7

u can also find the area of an equilateral triangle by this formula\[\sqrt{3}/4timesa ^{2}\]

Answer 8

\[\sqrt{3}/4*a ^{2}\]

Answer 9

If you cut the triangle down the middle and let x be your new side.
use the Pythagorean thereom:
\[36^{2}=18^{2}+x ^{2}\]

Answer 10

Dude why did you delete that, I was trying to figure it out!!!:)

Answer 11

Oh ok. I gotcha now @red0801

Answer 12

there u go!

Answer 13

now you can take the area of the triangle subtract the area of the circle. multiply by cost of material! :-)

& wallah..ur an engineer!

Answer 14

@DarthTony thank you for helping! and red, lol not quite.

Answer 15

\[\frac{\sqrt{3}}{4}\times36^{2}-\pi\times6^{2}\]
or in my pic
\[(2\times(\frac{(\sqrt{18^{2}+36^{2})}\times18}{2}))-(\pi\times6^{2})\]

Answer 16

um actually
\[(2\times(\frac{\sqrt{36^{2}-18^{2}}\times18}{2}))-(\pi\times6^{2})\]
not18^2+36^

Answer 17

Thanks so much bro!

Answer 18

is that right i'm very blur

Answer 19

??

Answer 20

nothing

Answer 21

mmk whatever you say.

Answer 22

Thanks @red0801

Answer 23

Tony..u are only accounting for 1/2 of the base.

Answer 24

no prob. good luck

Answer 25

@Ahmad_Shadab thanks for your help too. And I got the right answer, so what do you mean only accounting for half?

Answer 26

what did you get for the height?

Answer 27

my second post
and \[(\frac{\sqrt{3}}{4}\times36^{2})-\pi6^{2}\]
are correct solution :)

Answer 28

Sry..only saw this 1

(2×(362−182−−−−−−−−√×182))−(π×62)

Answer 29

:-)

Answer 30

should we close this question?
we do discussing about?? @Emah already got an answer.