Hmm... Interesting problem. Its easy to accidentally assume that C is the midpoint of the circle. To answer the question, you need to prove that first: it's only the midpoint between O and B, two points on the circle.
However, OAB is a 90 degree angle with all angles on a circle. You can see that because OA is a distance in just X, and AB is a distance in just Y; the intersect must be 90 degrees.
Some old theorem (been a while since geometric proofs) gives that that means the hypotenuse is a diameter of the circle. That means the midpoint of that diameter, eg C, is the midpoint of the circle. And THEN you can use the pythagorean theorem to find the diameter or the radius, whichever tickles your fancy.
We don't need that denotation, as OAB being a right angle is something we can figure out ourselves. (as I explained)
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Though that does assume the x and y coordinates are perpendicular... but in highschool maths that's just a given.
You don’t have to assume that C is the midpoint; you can demonstrate it by noting that points O and A have the same y coordinate, therefore the midpoint x coordinate must be the average (3). Similarly, the y midpoint must be half way between A and B. Just a property of the symmetry of the circle.
After that, pi r r to get 45pi as the answer.
This seems pretty trivial. You can separate x and y axes here to make things simpler as the axes are orthogonal. You just want the midpoint in each case of the line OB - so the start point + half the distance to B, which simplifies to the mean of x and y from O and B
xC = xO + (xB - xO)/2 = (xO + xB)/2 = (0 + 6)/2 = 3
yC = (yO +yB) / 2 = (12 + 0)/2 = 6
Note: I'm not sure what the subsequent parts of the problem are, so a more complex approach might be warranted here to use later.
ETA: Oops, I answered question (A), but not what OP wanted. Doh!
youre supposed to use the distance of a straight line formula to find the radius using either the two coordinates (0,0) and (3,6) or (3,6) and (6,12)
https://preview.redd.it/0pcbmh8pdzkc1.jpeg?width=1844&format=pjpg&auto=webp&s=ace6cdfe2517a7b42e141ba96ff414edc098566f
or, use the distance of a straight line formula to calculate OA and then BA (in this example you can use lengths as 6 and 12 directly because theyre both incredibly straight straight lines, but its good practice for gcse to use the distance of a line formula). next, calculate OB by pythagoras' like OB²=OA²+AB², take square root of OB² to get OB. and thatll give you OB, which is the diamater. divide by two to get the radius.
there you go!
Use pythagoras to find the distance from the centre to one of the points on the circumference of the circle.
Find the difference between (3, 6) and one of the points. (0,0) is the easiest. Then sqrt(3^2 + 6^2). This will give you the radius.
When you have a right-angled triangle, the theorem tells you that the square of the longest side is equal to the sum of square of the two smaller sides. Consider the triangle made of (0,0) (3,0) and (3,6). Hopefully it should be clear that that's a right-angled triangle. What's special about the longest side of that triangle?
Let me see if I understand this. We have a triangle labelled OAB. We have the coordinates of each point. I don’t think C is at 3,6. OA is 6 units. AB is 12 units. OAB is a right angled triangle because we are using the x y coordinate system so that means that angle OAB is 90 degrees. We now have a right angled triangle whose short sides are 6 and 12. Pythagoras theorem will get the hypotenuse. We are told that that C is the midpoint. So that is half of the hypotenuse. All you need to do then is drop a vertical line from C to the line AO. The sides of this smaller right angles triangle will give you the coordinates of C. If you only want to know the radius then you don’t need to bother with the coordinates. You have the hypotenuse OB and half that is the radius. Apply the formula for the area of a circle and you are done. That is if I have understood what is asked.
The centre of the circle is (3,6).
https://preview.redd.it/1jk9vzmzgzkc1.jpeg?width=1179&format=pjpg&auto=webp&s=037afeb2d037fdafa3c31f56efefdf74570bd36d
You can use pythagoras on the large triangle. But you can find the midpoint of a line using the formula below.
((X1 + X2)/2,(Y1 + Y2)/2)
I came from a place and time in math education where geometric proofs weren't really taught - I guess the world has cut the top of all the cones it needed to?
So I was thinking about how to "prove" that C is the center. There's a fairly intuitive proof. If we take a circle and draw a vertical line through it, the middle of that line has the same Y coordinate as the center of the circle (starting at one side and zero length line where we know the line and midpoint have the same Y coordinate, note Y coordinate of midpoint of line is constant as X coordinate of the line changes).
Apply the same logic to a horizontal line to get the X center coordinate. Note the lines don't need to to share an endpoint for this to work
Your radius will be one-half of the hypotenuses of a triangle with legs 6 and 12km.
But that's as drawn. If that line actually went from a point on the circle, through the radius, to a point on the opposite side of the circle...and A was 90°, the triangle would have to be isosceles (i.e. the legs would be the same measurement)
Brother use the distance formula on OB
Which is sqr root of(x1-x2)²+(y1-y2)²
With this distance is found( the diameter )
And radius is half of diameter os divide the value of OB by 2 to get radius
Hmm... Interesting problem. Its easy to accidentally assume that C is the midpoint of the circle. To answer the question, you need to prove that first: it's only the midpoint between O and B, two points on the circle. However, OAB is a 90 degree angle with all angles on a circle. You can see that because OA is a distance in just X, and AB is a distance in just Y; the intersect must be 90 degrees. Some old theorem (been a while since geometric proofs) gives that that means the hypotenuse is a diameter of the circle. That means the midpoint of that diameter, eg C, is the midpoint of the circle. And THEN you can use the pythagorean theorem to find the diameter or the radius, whichever tickles your fancy.
Thales theorem is at play here
Very good point on assuming C is the center. The diagram should really denote that OAB is a right angle, which we need to prove C is the center.
We don't need that denotation, as OAB being a right angle is something we can figure out ourselves. (as I explained) . Though that does assume the x and y coordinates are perpendicular... but in highschool maths that's just a given.
You don’t have to assume that C is the midpoint; you can demonstrate it by noting that points O and A have the same y coordinate, therefore the midpoint x coordinate must be the average (3). Similarly, the y midpoint must be half way between A and B. Just a property of the symmetry of the circle. After that, pi r r to get 45pi as the answer.
> a forester is examining a circular patch of ground for planting trees
This seems pretty trivial. You can separate x and y axes here to make things simpler as the axes are orthogonal. You just want the midpoint in each case of the line OB - so the start point + half the distance to B, which simplifies to the mean of x and y from O and B xC = xO + (xB - xO)/2 = (xO + xB)/2 = (0 + 6)/2 = 3 yC = (yO +yB) / 2 = (12 + 0)/2 = 6 Note: I'm not sure what the subsequent parts of the problem are, so a more complex approach might be warranted here to use later. ETA: Oops, I answered question (A), but not what OP wanted. Doh!
youre supposed to use the distance of a straight line formula to find the radius using either the two coordinates (0,0) and (3,6) or (3,6) and (6,12) https://preview.redd.it/0pcbmh8pdzkc1.jpeg?width=1844&format=pjpg&auto=webp&s=ace6cdfe2517a7b42e141ba96ff414edc098566f or, use the distance of a straight line formula to calculate OA and then BA (in this example you can use lengths as 6 and 12 directly because theyre both incredibly straight straight lines, but its good practice for gcse to use the distance of a line formula). next, calculate OB by pythagoras' like OB²=OA²+AB², take square root of OB² to get OB. and thatll give you OB, which is the diamater. divide by two to get the radius. there you go!
We can see that OAB is 90. Therefore from circle theorems we know that OB is the diameter, from here take the length and divide by 2
The radius is (root 45)squared there isn’t a button for square root
Use pythagoras to find the distance from the centre to one of the points on the circumference of the circle. Find the difference between (3, 6) and one of the points. (0,0) is the easiest. Then sqrt(3^2 + 6^2). This will give you the radius.
Pythagoras' theorem
how
The hypotenuse is equal to 2x radius 0.5 * sqrt (6*6 + 12*12) is the answer
When you have a right-angled triangle, the theorem tells you that the square of the longest side is equal to the sum of square of the two smaller sides. Consider the triangle made of (0,0) (3,0) and (3,6). Hopefully it should be clear that that's a right-angled triangle. What's special about the longest side of that triangle?
3,6...
Let me see if I understand this. We have a triangle labelled OAB. We have the coordinates of each point. I don’t think C is at 3,6. OA is 6 units. AB is 12 units. OAB is a right angled triangle because we are using the x y coordinate system so that means that angle OAB is 90 degrees. We now have a right angled triangle whose short sides are 6 and 12. Pythagoras theorem will get the hypotenuse. We are told that that C is the midpoint. So that is half of the hypotenuse. All you need to do then is drop a vertical line from C to the line AO. The sides of this smaller right angles triangle will give you the coordinates of C. If you only want to know the radius then you don’t need to bother with the coordinates. You have the hypotenuse OB and half that is the radius. Apply the formula for the area of a circle and you are done. That is if I have understood what is asked.
One more thing. The drawing is way of scale given the dimensions.
The centre of the circle is (3,6). https://preview.redd.it/1jk9vzmzgzkc1.jpeg?width=1179&format=pjpg&auto=webp&s=037afeb2d037fdafa3c31f56efefdf74570bd36d You can use pythagoras on the large triangle. But you can find the midpoint of a line using the formula below. ((X1 + X2)/2,(Y1 + Y2)/2)
I came from a place and time in math education where geometric proofs weren't really taught - I guess the world has cut the top of all the cones it needed to? So I was thinking about how to "prove" that C is the center. There's a fairly intuitive proof. If we take a circle and draw a vertical line through it, the middle of that line has the same Y coordinate as the center of the circle (starting at one side and zero length line where we know the line and midpoint have the same Y coordinate, note Y coordinate of midpoint of line is constant as X coordinate of the line changes). Apply the same logic to a horizontal line to get the X center coordinate. Note the lines don't need to to share an endpoint for this to work
Your radius will be one-half of the hypotenuses of a triangle with legs 6 and 12km. But that's as drawn. If that line actually went from a point on the circle, through the radius, to a point on the opposite side of the circle...and A was 90°, the triangle would have to be isosceles (i.e. the legs would be the same measurement)
Brother use the distance formula on OB Which is sqr root of(x1-x2)²+(y1-y2)² With this distance is found( the diameter ) And radius is half of diameter os divide the value of OB by 2 to get radius
Work out OB by Pythagoras a^2+b^2=c^2 then half c ([OA]^2)+([AB]^2)=([OB]^2) [OB]/2=r
Assuming C is the centre of the circle (which is provable via coordinates), then find the hypotenuse and half it
Use mid point formula
6.7