A right-angled triangle is triangle with one of the inner angles 90°.
The sides that make a right angle are called catheti. The third side
opposite to the right angle is called hypotenuse.
Right-angled triangles have specific properties that set them apart from other
triangles. One of the most famous theorems is the Pythagoras Theorem. It
refers to right-angled triangles and states that if a and b are
catheti and c is a hypotenuse, then:
a² + b² = c².
The geometric interpretation of the theorem is that if we construct squares on
the sides of a right-angled triangle, then the total area of the squares next to
the catheti is equal to the area of the square next to the hypotenuse.
The Pythagoras theorem is a case of the more general formula for finding an
angle in a triangle using the length of its sides. If we look for angle γ,
opposite to the side c and concluded between a and b,
then:
cos(γ) = (a² + b² - c²)/(2ab).
For a right-ngled triangle, γ = 90°, therefore, cos(γ) = 0
and a²+b²-c² = 0, which corresponds to the Pythagoras Theorem.
For the next problems use the formulae in II and the interactive
canvas in IV. Instructions for using the interactive canvas are in
V, and proposed solutions are in VI.
When verifying an assertion through an experiment in the interactive canvas, it
should be noted that the data shown is rounded and the results are not formal evidence,
but just on observations.
Problem №1
Construct the following triangles and check which ones are right-angled:
Triangle with sides AB=5, BC=4 and CA=3.
Triangle with sides CA=6 and AB=4, and concluded angle ∢A=30°.
Triangle with side AB=10 and adjacent angles 45°.
Problem №2
Suggest a way to create an approximately right-angled triangle
in the interactive canvas without using the calculated lengths of
sides and angles.
Problem №3
Using only the interactive canvas, calculate with precision of
two digits after the decimal point √13(square root
of 13). Use the Pythagoras Theorem and a suitable triangle.
Problem №4
Suggest an algorithm that can easily determine with the Pythagoras
theorem, whether a triangle is acute or obtuse. Demonstrate the
algorithm with examples.
The interactive canvas contains 3 points – A, B and
C, which are vertices of a triangle. They can only be dragged
within the canvas.
Below the canvas there is a panel of calculated values – the lengths of the
sides of the triangle, the area of the square to side AB, the sum of
the areas of the squares to the other two sides, and the measures of the angles
in the three vertices.
Above the canvas there is a toolbar for canvas configuration.
Lines – if this option is enabled, lines are drawn as
extensions to the sides of the triangle.
Squares – if this option is enabled, squares are drawn
to the sides of the triangle.
Grid – if this option is enabled, a grid is drown behind
the triangle. Each square is 1x1 in size.
Alignment – if this option is enabled, draged vertex will
automatically align to the nearest position with integer or half-integer
coordinates. This allows for the quick positioning of vertices and side
parallel to the axes.
Precision – if this option is enabled, the dragging of
the vertices is slowed down tenfold. This way you can fine-tune the
positions of the vertices.
When attempting to drag a vertex outside the interactive canvas, it is limited to
half a unit from the end of the canvas.
We construct the required triangle for each subproblem.
Click on the illustration to zoom it.
Sides BC and CA are long integer number of unite. The mode of automatic alignment can be used to set the side along the grid axes. The the constructed triangle is right-angled. We check the length of the third side in the panel below the canvas. It is as requested in the condition, thus the contructed triangle is the solution and it is right-angled.
We estimate the approximate locations of the triangle vertices, so that the lengths and the angles are close to the given values. In rprecise mode we align the vertices until we get the "exact" values. After constructing the triangle we can observe that no one of its angles is 90°, thus the triangle is no right-angled.
Both angles are 45°, thus the third angle is 90°. by definition this triangle is right-angled, independent on the length of the side. The illustrations shows a triangle with AB = 10 and ∢A = ∢B = 30°
Problem №2
The triangle is created in mode with shown lines and squares. When a line coincides with the side of a square, then the corresponding angle is approximately 90°. The illustration shows lines for sides BC and CA coincide with squares' sides. The Pythagoras theorem is also applicable and both areas are 16.5 square units.
Problem №3
A possible approach to calculate square root of 13 is to generate a right-angled triangle with hypotenuse square exactly 13. Then the length of the hypotenuse is √13.
In this case we need triangle with overall area of catheti squares 13. One such triangle has catheti with length 2 and 3; and ares of the squares 4 and 9. Following the Pythagoras theorem we get that the hypotenuse square has area 13.
After constructing the right-angled triangle with catheti 2 and 3 we find the length of the hypothenuse to be 3.61. Thus we have that √13 ≈ 3.61. Higher precision calculation yeilds √13 ≈ 3.6055512….
Problem №4
We will present an algorithm that determines whether an angle in a triangle is acute, right or obtuse. By applying this algorithm at most three times ist is possible to determine the type of a triangle.
The value of function cos(γ) is:
positive, if 0≤γ<90°;
zero, if γ=90°;
negative, if 90°<γ<180°.
Because the sign of 2ab is always positive, thne the sign of cos(γ) will be the same as the sign of a² + b² - c². Thus, for the angle opposing side c:
if a² + b² > c², the angle is acute;
if a² + b² = c², the angle is right;
if a² + b² < c², the angle is obtuse.
To determine the type of the triangle it is neede to know the type of the largest angle. This angle is opposite to the longest side.
The three illustrations below present these three cases. In each case side AB is the longest. In the left illustration the sum of areas of catheti squares is bigger, thus the triangle is acute. In the middle illustration the areas are equal, so the triangle is right-angled. The right illustration features smaller sum of areas and an obtuse triangle.
Sides BC and CA are long integer number of unite. The mode of automatic alignment can be used to set the side along the grid axes. The the constructed triangle is right-angled. We check the length of the third side in the panel below the canvas. It is as requested in the condition, thus the contructed triangle is the solution and it is right-angled.
We estimate the approximate locations of the triangle vertices, so that the lengths and the angles are close to the given values. In rprecise mode we align the vertices until we get the "exact" values. After constructing the triangle we can observe that no one of its angles is 90°, thus the triangle is no right-angled.
Both angles are 45°, thus the third angle is 90°. by definition this triangle is right-angled, independent on the length of the side. The illustrations shows a triangle with AB = 10 and ∢A = ∢B = 30°
