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Mathematics 11 Online
OpenStudy (anonymous):

Help please! I dont just want the answer. I need someone to help me through it! A square picture frame occupies an area of 112 ft2. What is the length of each side of the picture in simplified radical form?

OpenStudy (nory):

The area is 112. Let the length of a side be x. So what is our equation that we have about the picture frame?

OpenStudy (anonymous):

Im really not sure @Nory

OpenStudy (anonymous):

would 112 ft. 2 be 112^2? @Nory

OpenStudy (nory):

The area of the frame is 112, but it is also x*x (because of the area of the rectangle.) So our equation would be x^2 = 112. Do you know how to solve this equation?

OpenStudy (anonymous):

wouldnt that equation turn into x^2 - 112? @Nory

OpenStudy (nory):

Yes, but there's an easier way to solve it. Go back to x^2 = 112 and take square roots of both sides. What do you get?

OpenStudy (anonymous):

square root of 112 is 10.583? and x^2 is 2.718

OpenStudy (anonymous):

@Nory

OpenStudy (nory):

Write it as a radical. x = √122 x is the length of a side.

OpenStudy (nory):

Oh wait, 112. Not 122. What am I thinking?

OpenStudy (anonymous):

haha its okay. so x = the square root of 112? which is 10.583?

OpenStudy (anonymous):

@Nory

OpenStudy (anonymous):

@jim_thompson5910 could you help me?

jimthompson5910 (jim_thompson5910):

can you factor 112 where one factor is a perfect square

OpenStudy (anonymous):

yes?

jimthompson5910 (jim_thompson5910):

how so

OpenStudy (anonymous):

you have to find a number that can be multiplied with another to get 112

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

ex: 112 = 2*56 but neither factor is a perfect square

OpenStudy (anonymous):

i dont get how you find out if a number is a perfect square.

jimthompson5910 (jim_thompson5910):

it might help to list out the perfect squares (up to 112) 1, 4, 9, 16, 25, 36, 49, 64, 81, 100

OpenStudy (anonymous):

so i would find the best perfect square you can divide out of 112?

jimthompson5910 (jim_thompson5910):

so for instance, 9 is NOT a factor since 112/9 = 12.4444

jimthompson5910 (jim_thompson5910):

it would be a factor if you got a whole number when you divided

OpenStudy (anonymous):

so the number it equals would have to be a whole number?

jimthompson5910 (jim_thompson5910):

yes if you divided and got a whole number, then the number you divided by is a factor

jimthompson5910 (jim_thompson5910):

ex: 112/2 = 56, so this means 2 is a factor of 112 it also means 56 is a factor as well

jimthompson5910 (jim_thompson5910):

but neither are perfect squares

OpenStudy (anonymous):

im so confused

jimthompson5910 (jim_thompson5910):

do you see how 2 is a factor of 112?

OpenStudy (anonymous):

so 56 and 2 are not perfect squares because their square roots are not whole numbers?

jimthompson5910 (jim_thompson5910):

exactly

jimthompson5910 (jim_thompson5910):

25 is a perfect square because the square root of 25 is 5 (a whole number)

OpenStudy (anonymous):

so i have to find a number that is a factor of 112 and is also a perfect square?

jimthompson5910 (jim_thompson5910):

exactly

jimthompson5910 (jim_thompson5910):

and you want to find the largest possible factor that's a perfect square

OpenStudy (anonymous):

okay so the largest number is 16?

jimthompson5910 (jim_thompson5910):

yep good

jimthompson5910 (jim_thompson5910):

so 112/16 = ???

OpenStudy (anonymous):

7

jimthompson5910 (jim_thompson5910):

this means that 112 = 16*7

OpenStudy (anonymous):

so each side is 7?

jimthompson5910 (jim_thompson5910):

and this further means.... \[\large \sqrt{112} = \sqrt{16*7}\] \[\large \sqrt{112} = \sqrt{16}*\sqrt{7}\] \[\large \sqrt{112} = 4\sqrt{7}\]

jimthompson5910 (jim_thompson5910):

notice how I simplified by exploiting the fact that you can factor where one factor is a perfect square (the largest perfect square possible)

OpenStudy (anonymous):

yea you got 4. so i have to simplify 16 to finish the equation?

jimthompson5910 (jim_thompson5910):

yes you have to take the square root of 16 to get 4

OpenStudy (anonymous):

this was easier then I thought! thank you once again!

jimthompson5910 (jim_thompson5910):

yw

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