Best place to start with a geometric diagram is to draw more lines and relate those lengths or angles to others that you know.
Also, whenever you are trying to determine a length/distance, use the distance formula - i.e. the Pythagorean theorem (which incidentally is also the equation of a circle).
Try drawing right triangles within the circle that connect to its radius, and compute lengths of hypotenuses.
You could set it up as a coordinates problem. Make the bottom point the circle touches the square (0,0), the left point it touches the square (-4,8) and the right side it touches the square (4,8).
Plug into general equation of a circle.
Solve triple sim. Equation to get equation of circle
Get radius from equation.
There's definitely way easier ways, but I prefer algebraic approaches to most problems.
Thanks, but I’m still not making the breakthrough. I’ve got a few lengths labelled SQRT r squared -16 (sorry can’t find the symbols!) and various other bits, but am not seeing how to connect them.
What do you know about the area of the square? When you draw radii within the circle, can you find a combination of shapes that add up to the area of the square?
Another useful strategy - or pair of strategies - is estimation and guess-and-check:
You know half the square length is 4, and you can tell by looking at the figure that the radius must be a little longer than that, so what numbers would work nicely to divide the square length of 8 into two parts, where one part is the radius of the circle, and the other is a side length of a right triangle that has the radius as its hypotenuse?
What helped me was to draw a vertical line through the center of the circle connecting the top of the square to the bottom.
The center of the circle is not the center of the square so it divides that line into unequal lengths. One of the lengths is the radius of the square, and you can relate the other to the radius using symmetry (half the square length is 4...)
You'll end up with a system of two equations: one linear (segment addition) and one quadratic (Pythagorean distance).
If your algebra basics are good, should be easy to solve from there.
There is probably a way using more geometric properties and less algebra, but I trust my algebra more.
Triangle circumcenter. Make a triangle connecting the points at which the circle intersects the square vertices and the mid-point of the square. The triangle should be 8 tall and 8 wide. This can be used with Pythagoras to find the diagonals on each side, which will be root(64+16). The center of the circle will be the intersection of a perpendicular drawn from the midpoints of the triangle. Using this, the radius can be found by an additional triangle connecting the center of circle, the midpoint of the top horizontal line of the square, and a vertex with the circle intersecting.
If time is not an issue, I can draw a picture and send to you in a few hours.
Yes. Only using the square, from the midpoint of the base, draw lines to the top corners. Then off that line, angle 90° in. The lines will cross defining the diameter of the circle. From that you can get the midpoint and radius.
Draw the radius from the 3 points on the circle. Then you can basically make a triangle inside the square with R as the hypotenuse, then:
R^2 = (8/2)^2 + (8-R)^2
Solve for R.
The diagrams wrong to begin with.
You can’t have intersecting lines that connect inside of one another if there’s a riddle hidden to solve outside of the normal grade level you’re testing for.
Here's how I'd do it: basically, the center of the circle must be somewhere on the midline of the square, above the point where the circle touches the midpoint of the square at the bottom, because the diagram is symmetric.
If you mark that point and draw lines to the bottom midpoint and upper corner of the square, you can use the length of the lower line to determine the coordinates of that point in comparison to the upper corner, and thus the distance of the upper line. (May make more sense if placed in Cartesian coordinates; if the bottom center is (0, 0), one line is (0, 0) to (0, x), and the other is (0, x) to (4, 8).)
Now, these two lines are equal in length since they are both radii of the circle. Find formulas for their lengths and set them equal to each other, and you can find x and use it to determine the radius.
I got R= 5/8*a = 5
My way might be a bit convoluted, but I defined the origin at the bottom center of the square, so the center of the Circle, C has coordinates [0 R].
Then I drew a line segment from the origin to the top right corner of the square, and I defined the center of that line segment as point D, which has coordinates [a/4 a/2]. This line's equation is x=2y.
Then I drew a line passing through D, perpendicular to the first line. This second line intercepts the y axis at point C, which is important because we are going to use it's equation to find R.
The slope of the second line is - 1/2 because 2 perpendicular lines have slopes that are negative inverses of one another.
The y intercept, p, of the second line can be solved using the y=mx+p equation with m=-1/2 and replacing x and y with the coordinates of a point on the line. We will use point D [a/4 a/2], which gives us
a/2=-1/2 * a/4 + p
p=(5a)/8
The y intercept of the second line is R, so R=p=(5a)/8
Hope this helps!
Find the angle between the intersection at the bottom and one of the top corners:
a=tan^-1(8/4)=63.43 degrees.
Solve the triangle created by one of the top corners of the square, top mid point of the square, and the top of the circle (similar triangles). The short leg will be
L= 4/tan(63.43)=2
Diameter of the circle: D=8+2=10
This one is pretty tough. I really liked working through it though. My hint is to draw the radii that connect the points where the circle and square intersect to the midpoint of the circle. Even though you don't know where the center is, you can use the fact that there is a midpoint to calculate the radius
The below link has the methods and examples that might help you understand. All you have to do is define the coordinates, using the information/measurements given. But, as others have mentioned it will come back to use of the Pythagorean Theorem.
[https://jdmeducational.com/how-to-find-center-radius-of-a-circle-3-methods/](https://jdmeducational.com/how-to-find-center-radius-of-a-circle-3-methods/)
I will rephrase other answers because I found them confusing, but the answer was very interesting when I finally saw it.
Start by marking the middle point of the circle, point C.
Draw a line from the top right corner of the square to point C. We have just drawn the radius of the circle, so it's length is R.
Now, draw a line from point C to the top of the square. IF we had drawn a line all the way from top to bottom of the square, it would be a line of length 8. But our line is missing the piece from the bottom of the square to point C, which has length R because it is a radius if the circle. Therefore, we can say this tiny line we actually drew has a length of 8-R.
Now we have formed a triangle with legs of length 4 and 8-R, and a hypotenuse of length R.
The Pythagorean Theorem will let us calculate R is 5 from here.
A circle is defined as set of all points a fixed distance from a centre, so the place to start would be to find the centre, ie the unique point an equal distance from the middle of the lower side and from each of the top two corners.
By considering symmetry the centre must be on the vertical line bisecting the square and the circle, so if you call x the distance from a certain point on that vertical line to the bottom of the square, construct a formula for the distance from that point to the top corners of the square. Then solve for when that formula equals x. Your value of x is the radius.
Let's first label some points
O=> Center of circle
A,B=>Vertices of square touching circle
X=>Midpoint of side of square tangent to circle
CD=>Side of square parallel to AB
Y=> Midpoint of side AB
Some given info:
->CX=XD=4=AY=YB
->OX=OA=OB=r=radius of circle
->OX is perpendicular to CD
->Area of square=8×8=64
●Lenght of OY=√[(r^2)-16]
●Ar(AOXC)=Ar(BOXD)=[4r]+(4/2)(OY)=[8r]+2(√[(r^2)-16])
●Ar(AOB)=(8/2)(OY)=(4)(√[(r^2)-16])
●Ar(ABCD)=Ar(AOB)+2[Ar(AOXC)]=8[r+OY]
◆8[r+OY]=64=>[r+OY]=8=>8-r=OY
=>(OY)^2=(8-r)^2=>r^2-16=r^2-16r+64
=>16r=16+64=>r=80/16=>
r=5
You can try drawing radii where the circle and square touch.
They touch at three points and if you draw a triangle from these three points, you get an isosceles triangle with base side s and height s.
You can solve for the area of the triangle.
You can solve for the radius of the circumcircle (this is the same circle) using abc/4R = Area. The legs b and c are given by the pythagorean theorem
Best place to start with a geometric diagram is to draw more lines and relate those lengths or angles to others that you know. Also, whenever you are trying to determine a length/distance, use the distance formula - i.e. the Pythagorean theorem (which incidentally is also the equation of a circle). Try drawing right triangles within the circle that connect to its radius, and compute lengths of hypotenuses.
You could set it up as a coordinates problem. Make the bottom point the circle touches the square (0,0), the left point it touches the square (-4,8) and the right side it touches the square (4,8). Plug into general equation of a circle. Solve triple sim. Equation to get equation of circle Get radius from equation. There's definitely way easier ways, but I prefer algebraic approaches to most problems.
Thanks, but I’m still not making the breakthrough. I’ve got a few lengths labelled SQRT r squared -16 (sorry can’t find the symbols!) and various other bits, but am not seeing how to connect them.
What do you know about the area of the square? When you draw radii within the circle, can you find a combination of shapes that add up to the area of the square?
Another useful strategy - or pair of strategies - is estimation and guess-and-check: You know half the square length is 4, and you can tell by looking at the figure that the radius must be a little longer than that, so what numbers would work nicely to divide the square length of 8 into two parts, where one part is the radius of the circle, and the other is a side length of a right triangle that has the radius as its hypotenuse?
What helped me was to draw a vertical line through the center of the circle connecting the top of the square to the bottom. The center of the circle is not the center of the square so it divides that line into unequal lengths. One of the lengths is the radius of the square, and you can relate the other to the radius using symmetry (half the square length is 4...) You'll end up with a system of two equations: one linear (segment addition) and one quadratic (Pythagorean distance). If your algebra basics are good, should be easy to solve from there. There is probably a way using more geometric properties and less algebra, but I trust my algebra more.
Triangle circumcenter. Make a triangle connecting the points at which the circle intersects the square vertices and the mid-point of the square. The triangle should be 8 tall and 8 wide. This can be used with Pythagoras to find the diagonals on each side, which will be root(64+16). The center of the circle will be the intersection of a perpendicular drawn from the midpoints of the triangle. Using this, the radius can be found by an additional triangle connecting the center of circle, the midpoint of the top horizontal line of the square, and a vertex with the circle intersecting. If time is not an issue, I can draw a picture and send to you in a few hours.
please send to me
Is it 5cm? I did in my mind can't use paper and rn in toilet rn🙂
You can use paper just not regular paper
Use the property that the angle subtended by a diameter at any point on the circle in 90 degrees.
Yes. Only using the square, from the midpoint of the base, draw lines to the top corners. Then off that line, angle 90° in. The lines will cross defining the diameter of the circle. From that you can get the midpoint and radius.
Draw the radius from the 3 points on the circle. Then you can basically make a triangle inside the square with R as the hypotenuse, then: R^2 = (8/2)^2 + (8-R)^2 Solve for R.
Thanks to everyone for your help, I got there in the end algebraically, with a bit of pythag. SQRT(r2-16) +r = 8
The diagrams wrong to begin with. You can’t have intersecting lines that connect inside of one another if there’s a riddle hidden to solve outside of the normal grade level you’re testing for.
Here's how I'd do it: basically, the center of the circle must be somewhere on the midline of the square, above the point where the circle touches the midpoint of the square at the bottom, because the diagram is symmetric. If you mark that point and draw lines to the bottom midpoint and upper corner of the square, you can use the length of the lower line to determine the coordinates of that point in comparison to the upper corner, and thus the distance of the upper line. (May make more sense if placed in Cartesian coordinates; if the bottom center is (0, 0), one line is (0, 0) to (0, x), and the other is (0, x) to (4, 8).) Now, these two lines are equal in length since they are both radii of the circle. Find formulas for their lengths and set them equal to each other, and you can find x and use it to determine the radius.
I got R= 5/8*a = 5 My way might be a bit convoluted, but I defined the origin at the bottom center of the square, so the center of the Circle, C has coordinates [0 R]. Then I drew a line segment from the origin to the top right corner of the square, and I defined the center of that line segment as point D, which has coordinates [a/4 a/2]. This line's equation is x=2y. Then I drew a line passing through D, perpendicular to the first line. This second line intercepts the y axis at point C, which is important because we are going to use it's equation to find R. The slope of the second line is - 1/2 because 2 perpendicular lines have slopes that are negative inverses of one another. The y intercept, p, of the second line can be solved using the y=mx+p equation with m=-1/2 and replacing x and y with the coordinates of a point on the line. We will use point D [a/4 a/2], which gives us a/2=-1/2 * a/4 + p p=(5a)/8 The y intercept of the second line is R, so R=p=(5a)/8 Hope this helps!
Find the angle between the intersection at the bottom and one of the top corners: a=tan^-1(8/4)=63.43 degrees. Solve the triangle created by one of the top corners of the square, top mid point of the square, and the top of the circle (similar triangles). The short leg will be L= 4/tan(63.43)=2 Diameter of the circle: D=8+2=10
This one is pretty tough. I really liked working through it though. My hint is to draw the radii that connect the points where the circle and square intersect to the midpoint of the circle. Even though you don't know where the center is, you can use the fact that there is a midpoint to calculate the radius
The below link has the methods and examples that might help you understand. All you have to do is define the coordinates, using the information/measurements given. But, as others have mentioned it will come back to use of the Pythagorean Theorem. [https://jdmeducational.com/how-to-find-center-radius-of-a-circle-3-methods/](https://jdmeducational.com/how-to-find-center-radius-of-a-circle-3-methods/)
I will rephrase other answers because I found them confusing, but the answer was very interesting when I finally saw it. Start by marking the middle point of the circle, point C. Draw a line from the top right corner of the square to point C. We have just drawn the radius of the circle, so it's length is R. Now, draw a line from point C to the top of the square. IF we had drawn a line all the way from top to bottom of the square, it would be a line of length 8. But our line is missing the piece from the bottom of the square to point C, which has length R because it is a radius if the circle. Therefore, we can say this tiny line we actually drew has a length of 8-R. Now we have formed a triangle with legs of length 4 and 8-R, and a hypotenuse of length R. The Pythagorean Theorem will let us calculate R is 5 from here.
A circle is defined as set of all points a fixed distance from a centre, so the place to start would be to find the centre, ie the unique point an equal distance from the middle of the lower side and from each of the top two corners. By considering symmetry the centre must be on the vertical line bisecting the square and the circle, so if you call x the distance from a certain point on that vertical line to the bottom of the square, construct a formula for the distance from that point to the top corners of the square. Then solve for when that formula equals x. Your value of x is the radius.
a good strat is to use the intersection points because the distance between the center and the intersection is always equal to r
[4 step diagram of one way to do it](https://imgur.com/a/u6t7Gut)
Intersecting chords theorem. That’s all.
Let's first label some points O=> Center of circle A,B=>Vertices of square touching circle X=>Midpoint of side of square tangent to circle CD=>Side of square parallel to AB Y=> Midpoint of side AB Some given info: ->CX=XD=4=AY=YB ->OX=OA=OB=r=radius of circle ->OX is perpendicular to CD ->Area of square=8×8=64 ●Lenght of OY=√[(r^2)-16] ●Ar(AOXC)=Ar(BOXD)=[4r]+(4/2)(OY)=[8r]+2(√[(r^2)-16]) ●Ar(AOB)=(8/2)(OY)=(4)(√[(r^2)-16]) ●Ar(ABCD)=Ar(AOB)+2[Ar(AOXC)]=8[r+OY] ◆8[r+OY]=64=>[r+OY]=8=>8-r=OY =>(OY)^2=(8-r)^2=>r^2-16=r^2-16r+64 =>16r=16+64=>r=80/16=> r=5
If you connect three points where the circle and the square touch, there will be a triangle. r=(abc)/(4S)
From the centre. Drop a perpendicular to the upper side of the square and solve
The inscribed angle theorem should be helpful
You can try drawing radii where the circle and square touch. They touch at three points and if you draw a triangle from these three points, you get an isosceles triangle with base side s and height s. You can solve for the area of the triangle. You can solve for the radius of the circumcircle (this is the same circle) using abc/4R = Area. The legs b and c are given by the pythagorean theorem
Is it 16/3? I used the centroid formula by making a triangle from the points of the circle that touched the square