Properties of Triangles
There are three types of triangles based on how many sides or angles that are equal. Those are, the Equilateral Triangle with all three sides equal, and all angles equal (always 60°), the Isosceles Triangle with all two sides equal and two angles equal, and lastly, the Scalene Triangle with none of the sides equal and none of the angles equal.
Equilateral Triangle
Isosceles Triangle
Scalene Triangle
Once we classify the triangle, we can use this information to solve for missing sides/angles.
Sum of Interior Angles
In a triangle, the three interior angles always add to \(180°\):
\(\angle \text{A} + \angle \text{B} + \angle \text{C} = 180°\)
Refer to the diagram below. A triangle with angles \(\angle \text{A},\ \angle \text{B}\) and \(\angle \text{C}\) is extended with a parallel line \(\vec{\text{DE}}\), such that \(\vec{\text{AC}}\) is parallel to \(\vec{\text{DE}}\). From this we can use a "Z" pattern rule to see that the three angles make a straight line around point \(\text{B}\). We know that a straight line is \(180°\) so the sum of the interior angles of the triangle must also be \(180°\)!
In an equilateral triangle, all three sides are the same. Therefore, \(\boldsymbol{c = 3}\).
As sum of interior angles in any triangle is \(180^{\circ}\), we can subtract the sum of the known angles to find the missing angle:
\(\angle \text{A} + \angle \text{B} + \angle \text{C} = 180°\)
\(116^{\circ} + 35^{\circ} + \angle \text{C} = 180^{\circ}\)
\(151^{\circ} + \angle \text{C} = 180^{\circ}\)
\(\angle \text{C} = 180^{\circ} - 151^{\circ}\)
\(\angle \text{C} = 29^{\circ}\)
Therefore, we can determine that \(\boldsymbol{\angle \text{C} = 29^{\circ}}\).
To solve this problem we first define the side length of the equilateral triangle as \(x\). Each side is the same and we know the height.
We can focus on half the equilateral triangle with a base of \(x/2\), height of \(2\sqrt{3}\) and hypoteneuse of \(x\). This is a right angle so we can use the Pythagorean theorem!
\(c^2=(a)^2 + (b)^2\)
\(x^2= \left(\cfrac{x}{2}\right)^2+ (2√3)^2\)
\(x^2=\cfrac{x^2}{4}+ 12\)
\(\cfrac{3x^2}{4} = 12\)
\(x^2 = 16\)
\(x = 4\)
Therefore, we can determine the side length of the triangle is \(\boldsymbol{x = 4}\).
In order to verify that this triangle is isosceles, we can determine the respective side lengths of each sides using the following formula:
First, we can determine the side length of side \(MN\):
\(d_{\text{MN}} = \sqrt{(6)^2+(-2)^2}\)
\(d_{\text{MN}} = \sqrt{36+4}\)
\(d_{\text{MN}} = \sqrt{40}\)
Next, we can determine the side length of side \(NO\):
\(d_{\text{NO}} = \sqrt{(0)^2+(10)^2}\)
\(d_{\text{NO}} = \sqrt{10} = 10\)
Finally, we can determine the side length of side \(MO\):
\(d_{\text{MO}} = \sqrt{(6)^2+(8)^2}\)
\(d_{\text{MO}} = \sqrt{36+64}\)
\(d_{\text{MO}} = \sqrt{100} = 10\)
Given that \(\text{MO} = \text{NO}\) but \(\text{MN} \neq \text{MO}\), and \(\text{MN} \neq \text{NO}\), we can determine that this triangle contains 2 equal side lengths. As a result, we can verify that this is an iscoceles triangle.
We can represent the equation of the line in point-intercept form:
First, we need to determine the midpoint of \(\text{XY}\). In this instance, \(\text{X}\) will represent point 1 and \(\text{Y}\) will represent point 2:
\(\text{midpoint} = \left(\cfrac{3+5}{2}, \cfrac{4+(-2)}{2}\right)\)
\(\text{midpoint} = \left(\cfrac{8}{2}, \cfrac{2}{2}\right)\)
\(\text{midpoint} = (4, 1)\)
Next, we can determine the slope between points \(\text{XY}\) and \(\text{Z}\). In this instance, (\text{XY}\) will represent point 1 and \(\text{Z}\) will represent point 2:
\(m = \cfrac{-3-1}{12-4}\)
\(m = \cfrac{-4}{8}\)
\(m = -0.5\)
In order to determine the \(y\)-intercept, we can substitute one of the points into the equation and solve. In this instance, we will choose point \(\text{Z}\):
\(-3 = -0.5(12) + b\)
\(-3 = -6 + b\)
\(b = 6 -3\)
\(b = 3\)
We can write the equation as such:
Therefore, we can determine the equation of the line is \(\boldsymbol{y = -0.5x + 3}\).