Maths and Physics Tuition/Tests/Notes

Proof of Thin Lens Equation

of the object from the lens is the distance of the image from the lens can be easily derived using properties of similar triangles. The triangles CAB and CDE are similar, so (1)where O is the size of the object and I is the size of the image. The...

Pythagoras Theorem

With Pythagoras theorem we can find the lengths of a side in a right angled triangle given the other two sides. With the sides of the triangle as labelled above, we use the formula For example It is important to label the sides of the triangle a, b and...

Three Forces in Equilibrium

force of -12, equal and opposite to the first force of -5 horizontally and +12 vertically. 3. We can draw a vector triangle with the sides representing the forces. The triangle will be closed. All the forces will follow each other around the triangle in...

Constructing a Right Angle

To construct a right angle, we can use the circle theorem which says that a triangle inscribed in a circle, one edge of which is a diameter, is always a right angled triangle. Therefore, start by drawing a circle with a compass. The centre of the...

The Length, Midpoint and Gradient of a Line

A and B in the diagram below. We draw a straight line between A and B, and make this rhe hypotenuse of a right angled triangle by drawing a horizontal line from A to a point directly below B and another line down from B to a point right of A. This...

Constructing a Right Angle

To construct a right angle, we can use the circle theorem which says that a triangle inscribed in a circle, one edge of which is a diameter, is always a right angled triangle. Therefore, start by drawing a circle with a compass. The centre of the...

Problem Solving With Areas

at the centre of a circle, radius and this circle also passes through B, the distance By symmetry, is then an equolateral triangle with side of length and area Cut the shaded area up into an equilateral triangle and three equal segments 1, 2, 3 as shown...

Trigonometry in Three Dimensions

the distance AG, use Pythagoras Theorem in three dimensions. AB, BC and CG are mutually at right angles, so Consider the triangle ABG. AB is horizontal and BC and BF are both at right angles to AB, so every line in the plane BFGC is also at right angles...

Pythagorean Triples With Area Equals Perimeter

which Pythagorean Triples is the area of the associated right angled triangle equal to it's perimeter? All Pythagorean triples are generated by \[(m^2-n^2,2mn,m^2+n^2\] . The perimeter of the triangle is \[m^2-n^2+2mn+m^2+n^2=2mn+2m^2=2m(n+m)\] . The...

Sample Data-Weblinks

Proof of Pythagoras Theorem

The above diagrams represent rearrangements of sets of shapes. The blue triangles are all right angled, with the right angles at the corners of the yellow square, and congruent so all have the same area and the yellow squares are congruent so have the...

Proof That a Line Drawn From the Centre of a Circle to a Chord Bisects the Chord

The theorem is illustrated below. is the midpoint of The proof is quite simple. Draw radii to complete the triangles and These triangles have a side in common ( ) and Then from Pythagoras Theorem for triangle and for triangle

Proof That a Line Drawn From a Point to the Centre of a Circle Bisects the Angle Between the Two Tangents Drawn From That Point

The theorem is illustrated below. Proof: Construct the triangles and by drawing radii as below. since both are radii of the circle and is common to both. Further, angle since these are between a tangent and a radius. From Pythagoras theorem, so and the...

Proof of Pythagoras Theorem

above both have equal areas since both are of side To produce the large square on the right arrange the red and blue triangles from the square above left around the outside as shown. The area of the large square above left is given by The area of the...

Proof of Formula Between Extended Chords

a circle with the extended chords that intersect outside a circle are equal. Proof Draw the chords and Angle is common to triangles and Angles and are the same, since both are subtended by the points and on the circumference. Hence angles and are the...

Problem Solving

The ladder is 10m long and rests against a 1 m 3 box. The ladder reaches x m up the wall. Find We can extract similar triangles from the above diagram and use the ratios of their lengths to derive a polynomial equation in From the diagram above, using...

Regular Subdivision of a Sphere

3 4 12 6 8 Cube 3 5 30 12 20 Dedecahedron 4 3 12 8 6 Octahedron 5 3 30 20 12 Icosahedron In fact, if f=3 each face is a triangle. Triangles can be subdivided into triangles to form more regular subdivisions. Similarly, ff f=4 each face is a rectangle....

Area of Kite Formed By Tangents to Circle and Radii

is \[A= \frac{1}{2}( AO \times BC)\] \[BC= \sqrt{(3-1)^2+(4-1)^2}= \sqrt{13}\] . To find the other diagonal we use that triangles ABC and BQO are similar triangles, so \[\frac{BQ}{BO}=\frac{AB}{AO}\] . \[AB=\sqrt{AO^2-BO^2}= \sqrt{13-2^2}=3\] , then...

The Ambiguous Case

The ambiguous case arises when using the Sine Rule to find an angle in a triangle. It occurs because the Sin function is symmetric about 90°, so that When we solve for there is an acute solution, and an obtuse solution, Example: Find the angle A in the...

The Distrance Between Two Points

A and B in the diagram below. We draw a straight line between A and B, and make this rhe hypotenuse of a right angled triangle by drawing a horizontal line from A to a point directly below B and another line down from B to a point right of A. This...