Notes Changes 2023-08-30T17:59:58Z tag:notes.mathforge.org,2021-01-30:Notes/changes An Instiki Wiki Instiki two triangles overlapping a rectangle solution (Revision 2) 2023-08-30T17:59:58Z 2023-08-30T17:59:59Z tag:notes.mathforge.org,2023-08-30:Notes/revision/2006/two+triangles+overlapping+a+rectangle+solution/2 Andrew Stacey

Revised on August 30, 2023 17:59:58 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Two Triangles Overlapping a Rectangle Puzzle

Two Triangles Overlapping a Rectangle

What fraction of the rectangle is not covered by these two equilateral triangles?

Solution by Lengths in an Equilateral Triangle

Two triangles overlapping a rectangle labelled

Let xx be the side length of the right-hand triangle, so the length of BEB E. From lengths in an equilateral triangle, the length of GDG D is 32x\frac{\sqrt{3}}{2} x. As this is the length of IAI A, the length of JKJ K is 32\frac{\sqrt{3}}{2} times this, so is 34x\frac{3}{4} x.

Since KK is horizontally aligned with JJ which is the midpoint of IAI A, it is also the midpoint of BGB G and so CC is the midpoint of BDB D. Therefore, CEC E is 34x\frac{3}{4} x and so the width of the rectangle is 32x\frac{3}{2} x.

The blue regions can be matched with the white (inside the rectangle) except for the strip between HCH C and GDG D: triangle IHKI H K matches IJKI J K, triangle GFEG F E matches GDEG D E, the combination of triangle KHGK H G with ABKA B K matches AKJA K J. Therefore the blue regions comprise half of the rectangle without the strip between HCH C and GDG D. This strip has width 14x\frac{1}{4} x , so meaning that the rest of the rectangle (which is split between the white and blue regions regions) are is equivalent to a block of the rectangle of with width32x14x=54x \frac{5}{4} \frac{3}{2} x - \frac{1}{4} x = \frac{5}{4}x . Therefore (and the same height as the main rectangle). Half of this is blue, so the blue regions form are equivalent to a rectangle of width5 4 8x \frac{5}{4} \frac{5}{8}x. Therefore the blue regions form 58x÷32x=512\frac{5}{8}x \div \frac{3}{2} x = \frac{5}{12}ths of the rectangle.

a square and a rectangle overlapping solution (Revision 2) 2023-08-16T20:21:34Z 2023-08-16T20:21:34Z tag:notes.mathforge.org,2023-08-16:Notes/revision/2005/a+square+and+a+rectangle+overlapping+solution/2 Andrew Stacey

Revised on August 16, 2023 20:21:34 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

A Square and a Rectangle Overlapping

A Square and a Rectangle Overlapping

75%75\% of the purple square is shaded. What percentage of the red rectangle is shaded?

Solution by Pythagoras' Theorem and Area of a Triangle and Rectangles

A square and a rectangle overlapping labelled

With the points labelled as above, the fact that 75%75\% of the purple square is shaded means that the length of CDC D is one quarter of the length of ADA D. Since the purple quadrilateral is a square, EDE D is the same length as ADA D. Let xx be the length of CDC D, so EDE D has length 4x4 x. Applying Pythagoras' theorem to triangle EDCE D C shows that the length of ECE C is 17x\sqrt{17} x. The ratios of the sides of triangle EDCE D C is therefore 1:4:171 : 4 : \sqrt{17}.

Triangles CBAC B A and CDEC D E are both right-angled triangles and angles AC^BA \hat{C} B and DC^ED \hat{C} E are equal as they are vertically opposite so triangles CBAC B A and CDEC D E are similar. As the length of CAC A is 3x3 x, the lengths of triangle CBAC B A are 317x\frac{3}{\sqrt{17}} x, 1217x\frac{12}{\sqrt{17}}x, and 3x3x. The length of EBE B is therefore 17x+317x=2017x\sqrt{17} x + \frac{3}{\sqrt{17}}x = \frac{20}{\sqrt{17}} x.

The ratio of the lengths of CBC B and EBE B is therefore 3:203 : 20 and so the percentage of the red rectangle that is shaded is 15 85% 15\% 85\%.

three circles solution (Revision 2) 2023-08-16T20:19:20Z 2023-08-16T20:19:21Z tag:notes.mathforge.org,2023-08-16:Notes/revision/2004/three+circles+solution/2 Andrew Stacey

Revised on August 16, 2023 20:19:20 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Three Circles

Three Circles

The small circles each have area 11. What’s the area of the large circle?

Solution by Similar Triangles and Angle Between a Radius and Tangent

Three circles labelled

As the two small circles have the same area, they are the same size. Let rr be the radius of the small circles and RR of the larger one. Then πr 2=1\pi r^2 = 1.

With the points labelled as above, angles OD^AO \hat{D} A and OC^BO \hat{C} B are right-angles as they are the angle between a radius and tangent. So triangles ODAO D A and OCBO C B are similar. In particular, the ratios of the lengths of ADA D to OAO A and BCB C to OBO B are the same. In terms of the radii of the circles, this means that R:R+ 2 3r=1:2 R : R + 2 3 r = 1 : 2. Equivalently, R+ 2 3r=2R R + 2 3 r = 2 R and so R= 2 3r R = 2 3 r. The area of the larger circle is then πR 2=π( 2 3r) 2= 4 9πr 2= 4 9 \pi R^2 = \pi (2 (3 r)^2 = 4 9 \pi r^2 = 4 9.

three circles in a rectangle ii solution (Revision 2) 2023-08-08T20:27:41Z 2023-08-08T20:27:41Z tag:notes.mathforge.org,2023-08-08:Notes/revision/2003/three+circles+in+a+rectangle+ii+solution/2 Andrew Stacey

Revised on August 8, 2023 20:27:41 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Three Circles in a Rectangle II Puzzle

Three Circles in a Rectangle II

I’ve stacked three right-angled triangles to make this rectangle. What’s its area?

Solution by Similar Triangles

Three circles in a rectangle II labelled

All three triangles are right-angled triangles, angles ED^CE \hat{D} C and AD^BA \hat{D} B add up to 90 90^\circ, and angles CB^EC \hat{B} E and DB^AD \hat{B} A also add up to 90 90^\circ, so all three triangles are similar. The length scale factors from the largest to the other two are 45\frac{4}{5} and 35\frac{3}{5}. Let xx be the length of the diagonal, BDB D, so then ADA D has length 4x5\frac{4 x}{5} and ABA B has length 3x5\frac{3 x}{5}.

To compute xx, consider the area of the largest triangle. Using the formula for the area of a triangle, it is 12×4x5×3x5=6x 225\frac{1}{2} \times \frac{4 x}{5} \times \frac{3 x}{5} = \frac{6 x^2}{25}. Using the centre of the circle, the triangle can be decomposed into three triangles each with height the radius of the in-circle? and bases the sides of the original triangle. The area calculated this way is 12×552×(3x5+4x5+x)= 6 3x \frac{1}{2} \times 5 \frac{5}{2} \times \left(\frac{3 x}{5} + \frac{4 x }{5} + x\right) = 6 3 x. Equating these gives x=25252 x = 25 \frac{25}{2}.

The sides of the rectangle are then 20 10 20 10 and 15 7.5 15 7.5 so its area is 300 75 300 75.

two overlapping triangles and a circle solution (Revision 2) 2023-08-08T19:18:16Z 2023-08-08T19:18:16Z tag:notes.mathforge.org,2023-08-08:Notes/revision/2002/two+overlapping+triangles+and+a+circle+solution/2 Andrew Stacey

Revised on August 8, 2023 19:18:16 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Two Overlapping Triangles and a Circle Puzzle

Two Overlapping Triangles and a Circle

What’s the sum of these three angles?

Solution by Vertically Opposite Angles, Alternate Segment Theorem, and Angles in a Triangle

Two overlapping triangles and a circle labelled

With the points labelled as above, angles FH^CF \hat{H} C and EH^BE \hat{H} B are equal as they are vertically opposite, then angles FH^CF \hat{H} C and HG^CH \hat{G} C are equal by the alternate segment theorem. So angle HG^CH \hat{G} C is equal to angle EH^BE \hat{H} B. By a similar argument, angle CH^GC \hat{H} G is equal to angle AG^DA \hat{G} D. Angles CH^GC \hat{H} G, HG^CH \hat{G} C, and GC^HG \hat{C} H are the three angles interior to triangle CGHC G H and so sum to 180 180^\circ.

Solution by Angles in a Quadrilateral, Angle at the Centre is Twice the Angle at the Circumference, Angle Between a Radius and Tangent

With the points labelled as above, let us write aa for angle AG^DA \hat{G} D, bb for angle EH^BE \hat{H} B, and cc for angle GC^HG \hat{C} H.

The anti-clockwise angle GO^ F H G \hat{O} F H is twice angle GC^ F H G \hat{C} F H as the angle at the centre is twice the angle at the circumference, so is 2c2 c. So the anti-clockwise angle HO^ F G H \hat{O} F G is 360 2c360^\circ - 2 c.

Angles FH^OF \hat{H} O and OG^FO \hat{G} F are 90 90^\circ as they are both the angles between a radius and tangent. So in quadrilateral OHFGO H F G, angle GF^HG \hat{F} H is 180 (360 2c)=2c180 180^\circ - (360^\circ - 2 c) = 2 c - 180^\circ.

Angle FH^CF \hat{H} C is equal to angle EH^BE \hat{H} B, so is bb, as they are vertically opposite angles. Similarly, angle CG^FC \hat{G} F is equal to aa.

The angle sum of quadrilateral GCHFG C H F is then a+(360 c)+b+2c180 =a+b+c+180 a + (360^\circ - c) + b + 2 c - 180^\circ = a + b + c + 180^\circ. Since the sum of the angles in a quadrilateral is 360 360^\circ, this shows that a+b+c=180 a + b + c = 180^\circ.

Solution by Invariance Principle

Two overlapping triangles and a circle special

In the above configuration, triangle DEFD E F is equilateral. Angle HO^GH \hat{O} G is 120 120^\circ, so triangles HOCH O C and COGC O G are also equilateral making angle GC^HG \hat{C} H equal to 120 120^\circ. Triangle ABCA B C is isosceles meaning that angles HB^EH \hat{B} E and DA^GD \hat{A} G are 30 30^\circ. Angles BE^HB \hat{E} H and GD^AG \hat{D} A are 120 120^\circ, meaning that angles EH^BE \hat{H} B and AG^DA \hat{G} D are 30 30^\circ. So the marked angles sum to 120 +30 +30 =180 120^\circ + 30^\circ + 30^\circ = 180^\circ.

four quarter circles solution (Revision 2) 2023-08-08T19:14:13Z 2023-08-08T19:14:13Z tag:notes.mathforge.org,2023-08-08:Notes/revision/2001/four+quarter+circles+solution/2 Andrew Stacey

Revised on August 8, 2023 19:14:13 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Four Quarter Circles Puzzle

Four Quarter Circles

Four quarter circles. What fraction is shaded?

Solution by Area of a Circle and Pythagoras' Theorem

Four quarter circles labelled

Let aa, bb, cc, dd be the radii of the quarter circles in increasing length. By considering the width of the rectangle OACEO A C E then a+b=ca + b = c. The diagonals AEA E and OCO C have the same length, so d=b+c=a+2bd = b + c = a + 2 b.

The equations are slightly simpler with everything expressed in terms of bb and cc, so a=cba = c - b and d=b+cd = b + c. Applying Pythagoras' theorem to triangle OACO A C yields:

d 2=c 2+(c+a) 2=c 2+(2cb) 2=5c 24cb+b 2 d^2 = c^2 + (c + a)^2 = c^2 + (2 c - b)^2 = 5 c^2 - 4 c b + b^2

But also d 2=(b+c) 2=b 2+2bc+c 2d^2 = (b + c)^2 = b^2 + 2 b c + c^2 so 4c 2=6cb4c^2 = 6 c b and so 2c=3b2 c = 3 b. Then a=cb=12ba = c - b = \frac{1}{2} b so b=2ab = 2 a, c=3ac = 3 a, and d=5ad = 5 a.

If using aa and bb the resulting equation is 4a 2+2ab2b 2=04 a^2 + 2 a b - 2 b^2 = 0 which leads to (2ab)(a+b)=0(2 a - b)(a + b) = 0 and so, as aa and bb are both lengths, b=2ab = 2 a then c=3ac = 3 a, and d=5ad = 5 a.

The area of the outer quarter circle is 14πd 2=254πa 2\frac{1}{4} \pi d^2 = \frac{25}{4} \pi a^2 and the sum of the areas of the shaded quarter circles is 14πc 2+14πb 2+14πa 2=14π 15 14a 2 \frac{1}{4} \pi c^2 + \frac{1}{4} \pi b^2 + \frac{1}{4} \pi a^2 = \frac{1}{4} \pi 15 14 a^2. Hence the fraction that is shaded is 15 1425=35 \frac{15}{25} \frac{14}{25} = \frac{3}{5}.

hexagon in a quarter circle solution (Revision 2) 2023-08-08T19:09:04Z 2023-08-08T19:09:04Z tag:notes.mathforge.org,2023-08-08:Notes/revision/2000/hexagon+in+a+quarter+circle+solution/2 Andrew Stacey

Revised on August 8, 2023 19:09:04 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Hexagon in a Quarter Circle Puzzle

Hexagon in a Quarter Circle

The hexagon is regular. What’s the area of the quarter circle?

Solution by Properties of Chords, Symmetries of Regular Polygons, and Pythagoras' theorem

Hexagon in a quarter circle labelled

In the above diagram, the line segment CDC D is a chord of the quarter circle and so its perpendicular bisector passes through the centre of the circle. It is therefore a line of symmetry of the full circle. It is also a line of symmetry of the hexagon. Reflecting in this line brings ABA B to FEF E, and OAO A to OFO F. Since OABO A B is a straight line, so also must OFEO F E be. Triangle OAFO A F is therefore formed by extending two edges of a regular hexagon and so is an equilateral triangle. In particular, OAO A has length 22 and so OBO B has length 44.

The line segment BDB D is the height of a regular hexagon which has length 3\sqrt{3} times its side length, and so the length of BDB D is 232 \sqrt{3}. Applying Pythagoras' theorem to triangle OBDO B D shows that the square of the length of ODO D is 4 2+(23) 2= 8 16+12= 20 28 4^2 + (2 \sqrt{3})^2 = 8 16 + 12 = 20 28. The area of the quarter circle is therefore 5 7π 5 7 \pi.

shaded spikes in a hexagon techniques 2023-02-17T17:09:36Z 2023-02-17T17:09:37Z tag:notes.mathforge.org,2023-02-17:Notes/revision/1999/shaded+spikes+in+a+hexagon+techniques/1 Andrew Stacey

Created on February 17, 2023 17:09:36 by Andrew Stacey (81.109.65.126).

Techniques Used in a Solution

These are the techniques used in solving the problem shaded spikes in a hexagon.

shaded spikes in a hexagon solution 2023-01-06T18:20:23Z 2023-01-06T18:19:50Z tag:notes.mathforge.org,2023-01-06:Notes/revision/1998/shaded+spikes+in+a+hexagon+solution/1 Andrew Stacey

Created on January 6, 2023 18:20:23 by Andrew Stacey (167.98.65.106).

Solution to the Shaded Spikes in a Hexagon Puzzle

Shaded Spikes in a Hexagon

What fraction is shaded? The hexagon is regular, with equally spaced dots around its perimeter.

Solution by Properties of a Regular Hexagon and the Area of a Triangle

Consider the diagram labelled as below.

Shaded spikes in a hexagon labelled

From the properties of a regular hexagon, the triangles with apex at OO and side one of those of the hexagon are all equilateral. Each has area 16\frac{1}{6}th of that of the hexagon.

Let hh be the height of one equilateral triangle, so the length of OBO B, and let bb be the base, so the length of ACA C. Then the area of the hexagon is 6×12bh=3bh6 \times \frac{1}{2} b h = 3 b h.

As the points LL and DD are the midpoints of their sides, the length of LDL D is half way between that of ACA C, namely bb, and of KEK E, namely 2b2 b. So LDL D has length 32b\frac{3}{2} b. Similarly, the height of AA above the line LDL D is 12h\frac{1}{2} h, so the area of triangle ADLA D L is 12×32b×12h=38bh\frac{1}{2} \times \frac{3}{2} b \times \frac{1}{2} h = \frac{3}{8} b h.

The length of LFL F is 2h2 h and of KLK L is 12b\frac{1}{2} b, so the area of triangle LFKL F K is 12×12b×2h=12bh\frac{1}{2} \times \frac{1}{2} b \times 2 h = \frac{1}{2} b h.

The height of HH above KJK J is half of the height of GG (and of OO) so is 12h\frac{1}{2} h. Since KJK J has length 12b\frac{1}{2} b this means that triangle KHJK H J has area 12×12b×12h=18bh\frac{1}{2} \times \frac{1}{2} b \times \frac{1}{2} h = \frac{1}{8} b h.

The total shaded area is therefore:

38bh+12bh+18bh=bh \frac{3}{8} b h + \frac{1}{2} b h + \frac{1}{8} b h = b h

Since the area of the hexagon is 3bh3 b h this means that one third of the hexagon is shaded.

Solution by Properties of a Regular Hexagon and Dissection

Consider the diagram labelled as below.

Shaded spikes in a hexagon more labels

The (orange) shaded areas are cut and reassembled as follows.

  1. Triangle KHJK H J is flipped and translated to ABDA B D, where it fits exactly.
  2. Triangle MDNM D N is rotated round to fit in BCMB C M.
  3. Triangle OFPO F P is rotated round to fit in QPKQ P K.
  4. Triangle OQKO Q K is translated to fit in NOLN O L.

This leaves the shaded area being parallelogram KACOK A C O which consists of two of the six equilateral triangles that make up the hexagon. The shaded area is therefore one third of the total.

semi-circle in a square ii techniques 2022-08-24T18:01:23Z 2022-08-24T17:57:08Z tag:notes.mathforge.org,2022-08-24:Notes/revision/1995/semi-circle+in+a+square+ii+techniques/1 Andrew Stacey

Created on August 24, 2022 18:01:23 by Andrew Stacey (81.109.65.126).

Techniques Used in a Solution

These are the techniques used in solving the problem semi-circle in a square ii.

semi-circle in a square techniques 2022-08-24T18:00:24Z 2022-08-24T18:00:24Z tag:notes.mathforge.org,2022-08-24:Notes/revision/1997/semi-circle+in+a+square+techniques/1 Andrew Stacey

Created on August 24, 2022 18:00:24 by Andrew Stacey (81.109.65.126).

Techniques Used in a Solution

These are the techniques used in solving the problem semi-circle in a square.

semi-circle in a square solution (Revision 2) 2022-08-24T17:59:35Z 2022-08-24T17:59:35Z tag:notes.mathforge.org,2022-08-24:Notes/revision/1996/semi-circle+in+a+square+solution/2 Andrew Stacey

Revised on August 24, 2022 17:59:35 by Andrew Stacey (81.109.65.126).

Changes are formatted as follows: Added | Removed | Changed

Solution to the Semi-Circle in a Square Puzzle

Semi-Circle in a Square

Is more of this square pink or blue? How would you prove it?

Solution byAngle Between a Radius and Tangent, Isosceles Right-Angled Triangles, Area of a Circle, and Area of a Square

SemiCircleinaSquareLabelled.png?

In the above diagram, the point labelled OO is the centre of the semi-circle. The point labelled HH is where the semi-circle touches the top edge of the square, so as the angle between a radius and tangent is 90 90^\circ, HCH C is perpendicular to IGI G. Similarly, JEJ E is perpendicular to IAI A.

The line segments OHO H and OJO J are radii of the circle, so are of the same length. Therefore, OEO E and OCO C are of the same length. Triangles OEFO E F and BCOB C O are congruent since JEJ E is parallel to ADA D and so angles CB^OC \hat{B} O and EO^FE \hat{O} F are corresponding angles, and OBO B and OFO F have the same length. So OEO E and EFE F have the same length so triangle OEFO E F is isosceles.

Let rr be the radius of the semi-circle. Then as OEFO E F is an isosceles right-angled triangle with hypotenuse of length rr, its shorter sides both have length r2\frac{r}{\sqrt{2}}. The side length of the square is therefore r(1+12)r\left(1 + \frac{1}{\sqrt{2}}\right). Its area is thus

r 2(1+12) 2 r^2 \left(1 + \frac{1}{\sqrt{2}} \right)^2

The area of the semi-circle is 12πr 2\frac{1}{2} \pi r^2.

To compare these two, consider the decimal representations of the circle area and half the area of the square, since if the blue area is larger than the pink then it will be over half the square and vice-versa:

π2 1.5708 12(1+12) 2 1.4571 \begin{aligned} \frac{\pi}{2} &\simeq 1.5708 \\ \frac{1}{2}\left(1 + \frac{1}{\sqrt{2}}\right)^2 &\simeq 1.4571 \end{aligned}

Therefore the blue area is larger.

semi-circle in a square ii solution 2022-08-24T17:56:16Z 2022-08-24T17:56:16Z tag:notes.mathforge.org,2022-08-24:Notes/revision/1994/semi-circle+in+a+square+ii+solution/1 Andrew Stacey

Created on August 24, 2022 17:56:16 by Andrew Stacey (81.109.65.126).

Solution to the Semi-Circle in a Square II Puzzle

Semi-circle in a square ii

A semicircle in a square. What’s the missing length?

Solution by Angle in a Semi-Circle and Symmetry of a Square

SemiCircleinaSquareIIlabelled.png?

With the points labelled as above, angle AE^DA \hat{E} D is the angle in a semi-circle and so is 90 90^\circ. Therefore, rotating the square by 90 90^\circ (either clockwise or anti-clockwise) rotates line segment ABA B so that it is parallel to line segment CDC D. It also still runs between two sides of the square, since rotating the square leaves the square looking the same. Therefore, line segments ABA B and CDC D have the same length, and so the missing length is 2525.

semi-circle in a square solution 2022-05-06T16:11:30Z 2022-05-06T16:11:30Z tag:notes.mathforge.org,2022-05-06:Notes/revision/1993/semi-circle+in+a+square+solution/1 Andrew Stacey

Created on May 6, 2022 16:11:30 by Andrew Stacey (167.98.65.106).

Solution to the Semi-Circle in a Square Puzzle

Semi-Circle in a Square

Is more of this square pink or blue? How would you prove it?

Solution by

SemiCircleinaSquareLabelled.png?

In the above diagram, the point labelled OO is the centre of the semi-circle. The point labelled HH is where the semi-circle touches the top edge of the square, so as the angle between a radius and tangent is 90 90^\circ, HCH C is perpendicular to IGI G. Similarly, JEJ E is perpendicular to IAI A.

The line segments OHO H and OJO J are radii of the circle, so are of the same length. Therefore, OEO E and OCO C are of the same length. Triangles OEFO E F and BCOB C O are congruent since JEJ E is parallel to ADA D and so angles CB^OC \hat{B} O and EO^FE \hat{O} F are corresponding angles, and OBO B and OFO F have the same length. So OEO E and EFE F have the same length so triangle OEFO E F is isosceles.

Let rr be the radius of the semi-circle. Then as OEFO E F is an isosceles right-angled triangle with hypotenuse of length rr, its shorter sides both have length r2\frac{r}{\sqrt{2}}. The side length of the square is therefore r(1+12)r\left(1 + \frac{1}{\sqrt{2}}\right). Its area is thus

r 2(1+12) 2 r^2 \left(1 + \frac{1}{\sqrt{2}} \right)^2

The area of the semi-circle is 12πr 2\frac{1}{2} \pi r^2.

To compare these two, consider the decimal representations of the circle area and half the area of the square, since if the blue area is larger than the pink then it will be over half the square and vice-versa:

π2 1.5708 12(1+12) 2 1.4571 \begin{aligned} \frac{\pi}{2} &\simeq 1.5708 \\ \frac{1}{2}\left(1 + \frac{1}{\sqrt{2}}\right)^2 &\simeq 1.4571 \end{aligned}

Therefore the blue area is larger.

circle (Revision 4) 2022-05-06T15:59:17Z 2022-05-06T15:59:18Z tag:notes.mathforge.org,2022-05-06:Notes/revision/1992/circle/4 Andrew Stacey

Revised on May 6, 2022 15:59:17 by Andrew Stacey (167.98.65.106).

Changes are formatted as follows: Added | Removed | Changed

Circle

A circle is a shape defined by the property that every point on the circle is a fixed distance from a specified centre point.

A line segment from the centre to a point on the circle is called a radius, and its length is called the radius of the circle.

A diameter of the circle is a line segment that joins two points on the circle which also passes through its centre. Its length is called the diameter of the circle.

Circumference

The perimeter of a circle is also called its circumference. A circle with radius rr has circumference 2πr2 \pi r.

Area

The area of a circle of radius rr is πr 2\pi r^2.

Symmetry

A circle has many symmetries: every rotation about its centre and every reflection in a line through its centre.

Many results involving circles begin with the fact that any triangle with one vertex at the centre and the other vertices on the circle must be isosceles as two of its sides are radii of the circle and hence the same length.

category: shape
category: definition
category: circle theorems