Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

Good explaing on how to find X :attached file.

OpenStudy (anonymous):

OpenStudy (anonymous):

x^2=45^2+45^2 x=sqrt(2)*45 This is according to pythagoras theorem. Also the figure is a square so all sides are equal x=63.64 approx

OpenStudy (anonymous):

so basically ur timesing the givin number twice and adding them to gether to get the answer?

OpenStudy (anonymous):

according to pythagoras theorem in a right angled triangle base^2+height^2=hypotenuse^2 used this to find x

OpenStudy (anonymous):

the given figure is a square with side 45. diagonal of a square = side*sqrt(2)

OpenStudy (anonymous):

i got 4050?

OpenStudy (anonymous):

u should get 63.64

OpenStudy (anonymous):

do i 4050^2 ?

OpenStudy (anonymous):

you have to square root your answer to get x. you have found value of x^2

OpenStudy (anonymous):

how do i put tht in the calcutaor?

OpenStudy (anonymous):

just write 4050 and press the button that looks like

OpenStudy (anonymous):

\[\sqrt{}\]

OpenStudy (anonymous):

u should have a root signed button. type in 4050 and press it.

OpenStudy (anonymous):

ohhh okay see i thought you ment exponent

OpenStudy (anonymous):

who do i give the medal to . you guys are really helpful thank you very much :)

OpenStudy (anonymous):

oh and my name is alyssa lol thank you on the help on this one @Vaidehi09 & @ProgramGuru

OpenStudy (anonymous):

the medal dosent matter. What matters is that you have understood the problem and successfully solved it!

OpenStudy (anonymous):

and also that u'll be able to solve such probs in future. well said @ProgramGuru.

OpenStudy (anonymous):

since the bisect of square produce two triangle with 45 degree you can use the formula sin theta= perpendicular/ hypotenuse so sin 45=45/ hypotenuse sin 45= 0.71 so 0.71=45/ hypotenuse or hypotenuse= 45/0.71 hypotenuse = 63.38

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!