Revised on August 30, 2023 17:59:58 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |What fraction of the rectangle is not covered by these two equilateral triangles?
Let be the side length of the right-hand triangle, so the length of . From lengths in an equilateral triangle, the length of is . As this is the length of , the length of is times this, so is .
Since is horizontally aligned with which is the midpoint of , it is also the midpoint of and so is the midpoint of . Therefore, is and so the width of the rectangle is .
The blue regions can be matched with the white (inside the rectangle) except for the strip between and : triangle matches , triangle matches , the combination of triangle with matches . Therefore the blue regions comprise half of the rectangle without the strip between and . This strip has width , 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 width . 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 width. Therefore the blue regions form ths of the rectangle.
Revised on August 16, 2023 20:21:34 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |of the purple square is shaded. What percentage of the red rectangle is shaded?
With the points labelled as above, the fact that of the purple square is shaded means that the length of is one quarter of the length of . Since the purple quadrilateral is a square, is the same length as . Let be the length of , so has length . Applying Pythagoras' theorem to triangle shows that the length of is . The ratios of the sides of triangle is therefore .
Triangles and are both right-angled triangles and angles and are equal as they are vertically opposite so triangles and are similar. As the length of is , the lengths of triangle are , , and . The length of is therefore .
The ratio of the lengths of and is therefore and so the percentage of the red rectangle that is shaded is .
Revised on August 16, 2023 20:19:20 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |The small circles each have area . What’s the area of the large circle?
As the two small circles have the same area, they are the same size. Let be the radius of the small circles and of the larger one. Then .
With the points labelled as above, angles and are right-angles as they are the angle between a radius and tangent. So triangles and are similar. In particular, the ratios of the lengths of to and to are the same. In terms of the radii of the circles, this means that . Equivalently, and so . The area of the larger circle is then .
Revised on August 8, 2023 20:27:41 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |I’ve stacked three right-angled triangles to make this rectangle. What’s its area?
All three triangles are right-angled triangles, angles and add up to , and angles and also add up to , so all three triangles are similar. The length scale factors from the largest to the other two are and . Let be the length of the diagonal, , so then has length and has length .
To compute , consider the area of the largest triangle. Using the formula for the area of a triangle, it is . 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 . Equating these gives .
The sides of the rectangle are then and so its area is .
Revised on August 8, 2023 19:18:16 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |What’s the sum of these three angles?
With the points labelled as above, angles and are equal as they are vertically opposite, then angles and are equal by the alternate segment theorem. So angle is equal to angle . By a similar argument, angle is equal to angle . Angles , , and are the three angles interior to triangle and so sum to .
With the points labelled as above, let us write for angle , for angle , and for angle .
The anti-clockwise angle is twice angle as the angle at the centre is twice the angle at the circumference, so is . So the anti-clockwise angle is .
Angles and are as they are both the angles between a radius and tangent. So in quadrilateral , angle is .
Angle is equal to angle , so is , as they are vertically opposite angles. Similarly, angle is equal to .
The angle sum of quadrilateral is then . Since the sum of the angles in a quadrilateral is , this shows that .
In the above configuration, triangle is equilateral. Angle is , so triangles and are also equilateral making angle equal to . Triangle is isosceles meaning that angles and are . Angles and are , meaning that angles and are . So the marked angles sum to .
Revised on August 8, 2023 19:14:13 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |Four quarter circles. What fraction is shaded?
Let , , , be the radii of the quarter circles in increasing length. By considering the width of the rectangle then . The diagonals and have the same length, so .
The equations are slightly simpler with everything expressed in terms of and , so and . Applying Pythagoras' theorem to triangle yields:
But also so and so . Then so , , and .
If using and the resulting equation is which leads to and so, as and are both lengths, then , and .
The area of the outer quarter circle is and the sum of the areas of the shaded quarter circles is . Hence the fraction that is shaded is .
Revised on August 8, 2023 19:09:04 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |The hexagon is regular. What’s the area of the quarter circle?
In the above diagram, the line segment 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 to , and to . Since is a straight line, so also must be. Triangle is therefore formed by extending two edges of a regular hexagon and so is an equilateral triangle. In particular, has length and so has length .
The line segment is the height of a regular hexagon which has length times its side length, and so the length of is . Applying Pythagoras' theorem to triangle shows that the square of the length of is . The area of the quarter circle is therefore .
Created on February 17, 2023 17:09:36 by Andrew Stacey (81.109.65.126).
These are the techniques used in solving the problem shaded spikes in a hexagon.
Created on January 6, 2023 18:20:23 by Andrew Stacey (167.98.65.106).
What fraction is shaded? The hexagon is regular, with equally spaced dots around its perimeter.
Consider the diagram labelled as below.
From the properties of a regular hexagon, the triangles with apex at and side one of those of the hexagon are all equilateral. Each has area th of that of the hexagon.
Let be the height of one equilateral triangle, so the length of , and let be the base, so the length of . Then the area of the hexagon is .
As the points and are the midpoints of their sides, the length of is half way between that of , namely , and of , namely . So has length . Similarly, the height of above the line is , so the area of triangle is .
The length of is and of is , so the area of triangle is .
The height of above is half of the height of (and of ) so is . Since has length this means that triangle has area .
The total shaded area is therefore:
Since the area of the hexagon is this means that one third of the hexagon is shaded.
Consider the diagram labelled as below.
The (orange) shaded areas are cut and reassembled as follows.
This leaves the shaded area being parallelogram which consists of two of the six equilateral triangles that make up the hexagon. The shaded area is therefore one third of the total.
Created on August 24, 2022 18:01:23 by Andrew Stacey (81.109.65.126).
These are the techniques used in solving the problem semi-circle in a square ii.
Created on August 24, 2022 18:00:24 by Andrew Stacey (81.109.65.126).
These are the techniques used in solving the problem semi-circle in a square.
Revised on August 24, 2022 17:59:35 by Andrew Stacey (81.109.65.126).
Changes are formatted as follows: Added |Is more of this square pink or blue? How would you prove it?
SemiCircleinaSquareLabelled.png?
In the above diagram, the point labelled is the centre of the semi-circle. The point labelled is where the semi-circle touches the top edge of the square, so as the angle between a radius and tangent is , is perpendicular to . Similarly, is perpendicular to .
The line segments and are radii of the circle, so are of the same length. Therefore, and are of the same length. Triangles and are congruent since is parallel to and so angles and are corresponding angles, and and have the same length. So and have the same length so triangle is isosceles.
Let be the radius of the semi-circle. Then as is an isosceles right-angled triangle with hypotenuse of length , its shorter sides both have length . The side length of the square is therefore . Its area is thus
The area of the semi-circle is .
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:
Therefore the blue area is larger.
Created on August 24, 2022 17:56:16 by Andrew Stacey (81.109.65.126).
A semicircle in a square. What’s the missing length?
SemiCircleinaSquareIIlabelled.png?
With the points labelled as above, angle is the angle in a semi-circle and so is . Therefore, rotating the square by (either clockwise or anti-clockwise) rotates line segment so that it is parallel to line segment . It also still runs between two sides of the square, since rotating the square leaves the square looking the same. Therefore, line segments and have the same length, and so the missing length is .
Created on May 6, 2022 16:11:30 by Andrew Stacey (167.98.65.106).
Is more of this square pink or blue? How would you prove it?
SemiCircleinaSquareLabelled.png?
In the above diagram, the point labelled is the centre of the semi-circle. The point labelled is where the semi-circle touches the top edge of the square, so as the angle between a radius and tangent is , is perpendicular to . Similarly, is perpendicular to .
The line segments and are radii of the circle, so are of the same length. Therefore, and are of the same length. Triangles and are congruent since is parallel to and so angles and are corresponding angles, and and have the same length. So and have the same length so triangle is isosceles.
Let be the radius of the semi-circle. Then as is an isosceles right-angled triangle with hypotenuse of length , its shorter sides both have length . The side length of the square is therefore . Its area is thus
The area of the semi-circle is .
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:
Therefore the blue area is larger.
Revised on May 6, 2022 15:59:17 by Andrew Stacey (167.98.65.106).
Changes are formatted as follows: Added |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.
The perimeter of a circle is also called its circumference. A circle with radius has circumference .
The area of a circle of radius is .
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.