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Mathematics 13 Online
OpenStudy (anonymous):

What is the perimeter of the trapezoid below?

OpenStudy (anonymous):

OpenStudy (anonymous):

cant see the picture, but perimeter is just adding all of the sides

OpenStudy (anonymous):

do you know pythagorean theorem?

OpenStudy (anonymous):

I do

OpenStudy (anonymous):

so can you use that to find the missing length in your picture?

OpenStudy (anonymous):

I could but what numbers would I put in? Would I put in the top and bottom numbers? Or the one on the side with either the top or bottom number?

OpenStudy (anonymous):

well we know that pythagorean thm only works on right triangles, so in your figure the right triangle is |dw:1370619638197:dw|

OpenStudy (anonymous):

can you solve for the missing length?

OpenStudy (anonymous):

Yeah it would be \[\sqrt{136}\]

OpenStudy (anonymous):

not quite, what you did was solve for the hypotenuse no one of the sides, try again

OpenStudy (anonymous):

So how would I solve for one of the sides?

OpenStudy (anonymous):

if the thm says \[c^2=a^2+b^2\] and c is the hypotenuse, a and b are the sides, you can solve that equation for either a or b to get you the other side

OpenStudy (anonymous):

when i solve for b i get \[b=\sqrt{c^2-a^2}\]

OpenStudy (anonymous):

I got 10 for B

OpenStudy (anonymous):

\[b=\sqrt{100-36}=8\]

OpenStudy (anonymous):

dont know how you got 10

OpenStudy (anonymous):

perimeter =40

OpenStudy (anonymous):

I honestly don't know how either. So would the height be 10?

OpenStudy (anonymous):

you dont know how you goy 10?

OpenStudy (anonymous):

Not really but I squared the square root of 136 and then subtracted 36 from 136 and then square rooted 100. If I added 10+10+14+8 it equals 42 not 40

OpenStudy (anonymous):

the addition should be 10+8+8+14

OpenStudy (anonymous):

that height should be 8, the equation you should solve is \[b=\sqrt{100−36}=8\]

OpenStudy (anonymous):

So it would really be 40?

OpenStudy (anonymous):

yep, do you see why?

OpenStudy (anonymous):

I do and thanks :)

OpenStudy (anonymous):

you are very welcome :)

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