I’m really stuck on this, I can’t even work out one side

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malley

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Help plz
 

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You have shown no work at all. You CANNOT be stuck at the very beginning. Try something.

Have you considered drawing the diagonals of the rectangle and pondering how they relate to the circle?
 
tkhunny said:
You have shown no work at all. You CANNOT be stuck at the very beginning. Try something.

Have you considered drawing the diagonals of the rectangle and pondering how they relate to the circle?
Click to expand...
I have done working lol I just didn’t on the sheet. I got the diagonal to be 12 but that can’t be a hypotenuse for a triangle with al sides as integers and the perimeter is 28 so the other two sides add up to 14? I just don’t get it
 
Who said the sides are integers?

Define one side as x; what is the other side, and what equation do you get?
 
lets the sides of the rectangle to be a and b.
you have one equation as you mentioned: a+b=14
can you look at the diagonal of rectangle and find any other relationship? how about if you draw like half of the diagonal

let me ask you something about rectangle, do you know where two diagonal of a rectangle intersect?where is this position on the diagonal?
 
Jagulba said:
lets the sides of the rectangle to be a and b.
you have one equation as you mentioned: a+b=14
can you look at the diagonal of rectangle and find any other relationship? how about if you draw like half of the diagonal

let me ask you something about rectangle, do you know where two diagonal of a rectangle intersect?where is this position on the diagonal?
Click to expand...
The diagonal is the diameter? I’m sorry if I’m being dumb
 
malley said:
I made the other side 14-x and did some stuff and got x=2 and y=12 which makes no sense
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Please show the work you did, so we can find the error. Clearly it is not true that 2^2 + 12^2 = 12^2, so something went wrong in the solving process (assuming the equation itself was correct).
 
oh there you go, well done. you got it. the rest is simple algebra
divide all of them by 2
x^2-14x+26

x^2-14x+49-49+26=0
(x-7)^2=49-26=23
 
no
its like i write:
(15-12)^2=9
and then you say so 15 is 12? what happened to 9 on the other side of the equation
or in this case:
(x-7)^2=23
if you say x=7 then what happened to 23?
 
Jagulba said:
no
its like i write:
(15-12)^2=9
and then you say so 15 is 12? what happened to 9 on the other side of the equation
or in this case:
(x-7)^2=23
if you say x=7 then what happened to 23?
Click to expand...
Omg thanks so much I actually did it the area is 26 thanks so much
 
That's right.

Since you didn't show which way you finished, I'll show two (for the sake of others who might be reading this).

One, by either completing the square or using the quadratic formula, you find that x = 7 ± sqrt(23); then y = 14 - x = 7 ∓ sqrt(23), that is, the opposite sign. When you multiply (7 + sqrt(23))(7 - sqrt(23)), you get the difference of squares, 49 - 23 = 26. You could also do this with decimals, but you might not be sure the answer is exact.

Or, realizing that x could have been either of the sides, you see that x and y are the two solutions of x^2-14x+26 = 0. The constant term is the product of the two solutions, so the area, xy, is 26!
 
Dr.Peterson said:
That's right.

Since you didn't show which way you finished, I'll show two (for the sake of others who might be reading this).

One, by either completing the square or using the quadratic formula, you find that x = 7 ± sqrt(23); then y = 14 - x = 7 ∓ sqrt(23), that is, the opposite sign. When you multiply (7 + sqrt(23))(7 - sqrt(23)), you get the difference of squares, 49 - 23 = 26. You could also do this with decimals, but you might not be sure the answer is exact.

Or, realizing that x could have been either of the sides, you see that x and y are the two solutions of x^2-14x+26 = 0. The constant term is the product of the two solutions, so the area, xy, is 26!
Click to expand...
Thanks yea I did the first way
 
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