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Mathematics 14 Online
OpenStudy (anonymous):

What is the perimeter of the following polygon?

OpenStudy (anonymous):

OpenStudy (anonymous):

Use Pythagorus theorem to solve. Do you know how to do that?

OpenStudy (anonymous):

In this case, you can also use pythagorus triples.

OpenStudy (anonymous):

what are the numbers? I cant read them

OpenStudy (anonymous):

A2+b2=c2? I keep trying to plug it in but I cant get the right answer.

OpenStudy (anonymous):

you need to divide 6 by 2.

OpenStudy (anonymous):

The hight is 4 and the base is 6

OpenStudy (anonymous):

the base is 3, the height is 4. What is the hypotenuse?

OpenStudy (anonymous):

base is 3 because pythagorus theorem works only on 90 degree triangles.

OpenStudy (anonymous):

so make it two right angle triangles. Now you have 3,4 and...?

OpenStudy (anonymous):

Maybe, Am I suppost to add 3 an for to get x? or?

OpenStudy (anonymous):

four*

OpenStudy (anonymous):

\[3^2+4^2=x^2\]

OpenStudy (anonymous):

you add the square of the base and the square of the height to get the hypotenuse so yes, you add them.

OpenStudy (anonymous):

after you find x, this is the equation for the perimeter. 2x+6

OpenStudy (anonymous):

Okay I get that, now how do I solve the equation, this is where I get stuck.

OpenStudy (anonymous):

have you found x?

OpenStudy (anonymous):

I thought it was 7, but i'm not very sure.

OpenStudy (anonymous):

what is (3*3)+(4*4)=?

OpenStudy (anonymous):

9 + 16 = 25

OpenStudy (anonymous):

correct. now, 25 = x^2 how to find x? (I'll rewrite the equation) x^2=25

OpenStudy (anonymous):

12.5?

OpenStudy (anonymous):

\[x^2=25\]

OpenStudy (anonymous):

\[x=\sqrt{25}\]

OpenStudy (anonymous):

when you transfer square over to the other side, it becomes a square root.

OpenStudy (anonymous):

do you know what the square root of 25 is?

OpenStudy (anonymous):

it's 5

OpenStudy (anonymous):

yes. so now we know that x is 5. Now, there are two hypotenuse. So the equation would be 2x+6 or x+x+6 Since now we know that x=5, plug it into either one of the equations.

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