Ask your own question, for FREE!
Mathematics 11 Online
OpenStudy (anonymous):

PLEASE HELP WITH HW!

OpenStudy (anonymous):

OpenStudy (anonymous):

OpenStudy (anonymous):

Do you know the formula for a triangle?

OpenStudy (anonymous):

30-60-90?

OpenStudy (anonymous):

Area=1/2 X B X H

OpenStudy (anonymous):

so is it: .5 x 12 x 5?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

and for the second problem?

OpenStudy (anonymous):

That is a rectangular. So u will use the formula for that

OpenStudy (anonymous):

what formula?

OpenStudy (anonymous):

A=L*W

OpenStudy (anonymous):

Idk how to do that one..

OpenStudy (anonymous):

and btw, .5 x 12 x 5 is 30. that's not an option..

OpenStudy (anonymous):

@KE8717504 ?

OpenStudy (anonymous):

@sabeelio do you know how to do this?

OpenStudy (anonymous):

okay so you must use pythagoren theorem since this is a right triangle which is a^2 + b^2 = c^2

OpenStudy (anonymous):

a and b are the legs and c is the hypotenus so plug in your values into the equationa nd you'll ge tyour answer

OpenStudy (anonymous):

the same theorom should help with your answer for number 2

OpenStudy (anonymous):

so what do you think your equation would be?

OpenStudy (anonymous):

im confused..

OpenStudy (anonymous):

are you saying it'd be: 12^2+ 6^2 = c^2?

OpenStudy (anonymous):

@sabeelio

OpenStudy (anonymous):

yes exactly then just solve for c and that is your asnwer

OpenStudy (anonymous):

so uoy have 144 +25 = 169 and the square root of 169 is 13 so that is your answer

OpenStudy (anonymous):

now if you apply that to the 2nd question and find the value of c you can get the right answer there as well, it is a rectangle so it is 16 on top and bopttom and 12 on left and right and split into 2 triangleks, so you just find the hypotenus (c) of one of the triangles and you will get your answer

OpenStudy (anonymous):

@sabeelio I don't understand what you are saying?

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!