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The relation between the sides of a right-angled triangle
Example 1
What can we state about areas \latex{ A_1 }, \latex{ A_2 } and \latex{ A_3 } of the green squares shown in the figure?
\latex{ A_3 }
\latex{ A_1 }
\latex{ A_2 }
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\alpha}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\beta}
\latex{\gamma}
\latex{\gamma}
\latex{\gamma}
\latex{\gamma}
\latex{\alpha+\beta= 90º}
Figure 34
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ a }
\latex{ b }
\latex{ c }
\latex{ c }
\latex{ c }
\latex{ c }
Solution
A1 + A2 = A3, since the omitted parts of both squares with sides a + b consist of four right-angled triangles with legs a and b, so their areas are mutually equal.
If the hypotenuse of these right-angled triangles is \latex{ c }, then A1 = a2, A2 = b2, A3 = c2, and thus a2 + b2 = c2.
This way we proved the following theorem of PYTHAGORAS, a Greek mathematician and philosopher from the \latex{ 6 }th century BC:
PYTHAGOREAN THEOREM (OR PYTHAGORAS’ THEOREM):
In a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the other \latex{ 2 } sides.
It can be proven that the converse of the Pythagorean theorem is also true.
THEOREM: If the sum of squares of the length of two sides in a triangle is equal to the square of the length of the third side, then the triangle is a right-angled triangle.
The combination the Pythagorean theorem and its converse:
THEOREM: A triangle is a right-angled triangle if and only if the sum of squares of the length of two sides is equal to the square of the length of the third side.
Example 2
We fix the two ends of a \latex{ 1,001\, m } long rope at the end-points of a \latex{ 1\, km } long straight path. A machine lifts up the mid-point of the rope as much as possible. Is it possible for a \latex{ 180\, cm } tall person to walk across underneath the rope without bending down?
Solution
When applying the Pythagorean theorem we can calculate height \latex{ h } of the mid-point of the rope. (Figure 35)
rope
\latex{ 500.5\, m }
\latex{ 500\, m }
path
Figure 35
\latex{ C }
\latex{ F }
\latex{ h }
\latex{ B }
\latex{ A }
AC2 = h2 + AF2, which implies
\latex{h=\sqrt{AC^{2}-AF^{2} } =\sqrt{500.25}\;m\approx 22.37\;m}
This surprising result shows that someone can freely walk across underneath the rope and a crane is needed to lift up the mid-point of the rope.
Example 3
Let us express the altitude and the area of a regular triangle with sides \latex{ a } with the help of \latex{ a }.
Solution
We know that the altitude of a regular triangle perpendicularly bisects the corresponding side, and it also bisects the angle opposite this side. (Figure 36)
Using the Pythagorean theorem in triangle \latex{ AFC }:
Figure 36
\latex{\frac{a}{2}}
\latex{ h }
\latex{ F }
\latex{ B }
\latex{ A }
\latex{ a }
\latex{ a }
\latex{ C }
\latex{a^{2} = h^{2} + \left(\frac{a}{2} \right)^{2}},
which implies
\latex{h = \sqrt{a^{2}-\left(\frac{a}{2} \right)^{2}} = \sqrt{a^{2}-\frac{a}{4}^{2}}= \sqrt{\frac{3a^2}{4}}=\frac{a\sqrt{3}}{2}}.
The ALTITUDE
of the regular triangle:
\latex{h=\frac{a\sqrt{3}}{2}}
of the regular triangle:
\latex{h=\frac{a\sqrt{3}}{2}}
The altitude of the regular triangle with sides \latex{a}: \latex{h=\frac{a\sqrt{3}}{2}}.
The area of the triangle can also be expressed with the help of the length of the side:
\latex{A=\dfrac{a\times h}{2}=\dfrac{a\times\dfrac{a\sqrt{3}}{2}}{2}}
which implies
\latex{A=\frac{a^2\sqrt{3} }{4}}
The AREA
of the regular triangle:
\latex{A=\frac{a^2\sqrt{3} }{4}}
of the regular triangle:
\latex{A=\frac{a^2\sqrt{3} }{4}}
Exercises
{{exercise_number}}. Two interior angles of a triangle are
- \latex{ 30º } and \latex{ 60º ;}
- \latex{ 45º } and \latex{ 75º ;}
- \latex{ 82º } and \latex{ 26º ;}
- \latex{ 90º } and \latex{ 10º ;}
- \latex{ 23.5º } and \latex{ 114.6º ;}
- \latex{ 88º } and \latex{ 91º }.
Determine the third interior angle and the exterior angles of the triangle.
{{exercise_number}}. Let us denote the interior angles of a triangle by \latex{\alpha, \beta, \gamma}, and let \latex{\alpha', \beta', \gamma'} be the corresponding exterior angles. Based on the following data determine the missing interior and exterior angles of the triangle.
- \latex{\alpha = 35º, \beta = 80º}
- \latex{\alpha = 56º, \beta' = 113º}
- \latex{\alpha' = 95º, \gamma = 50º}
- \latex{\beta' = 82º, \gamma' = 142º}
- \latex{\alpha' = 190º, \beta = 30º}
- \latex{\alpha' = 92º, \gamma' = 25º}
{{exercise_number}}. The ratio of the interior angles of a triangle is
- \latex{ 1 : 2 : 3 ;}
- \latex{ 4 : 5 : 6 ;}
- \latex{ 3 : 7 : 10 ;}
- \latex{ 2 : 9 : 13 };
- \latex{ 7 : 10 : 19 ;}
- \latex{ 11 : 12 : 13 }.
Determine the interior and exterior angles of the triangle.
{{exercise_number}}. One of the interior angles of a triangle is \latex{ 38º }. One of the other two interior angles is \latex{ 22º } greater than the other one. Give the interior and exterior angles of the triangle.
{{exercise_number}}. Is there a triangle where the length of the sides are
- \latex{ 2\, cm, 4\, cm, 5\, cm ;}
- \latex{ 10\,m, 12\,m, 21\,m };
- \latex{ \frac{2}{3}\;cm,} \latex{\frac{10}{18}\;cm,} \latex{\frac{11}{13}\;cm;}
- \latex{ 2.5 cm,\, 32 mm, 0.58\, dm ?}
{{exercise_number}}. The length of two sides of a triangle are
- \latex{ 2.8\, cm } and \latex{ 1.7\, cm ;}
- \latex{ 4.31\, cm } and \latex{ 3.92\, cm ;}
- \latex{ 32.1\, cm } and \latex{ 52.3\, cm ;}
- \latex{ 81.9\, cm } and \latex{ 82.1\, cm }.
How long can the third side of the triangle be if it is an integer in centimetres?
{{exercise_number}}. Two sides of an isosceles triangle are
- \latex{ 2\, cm } and \latex{ 4\, cm ;}
- \latex{ 0.3\, m } and \latex{ 0.5\, m ;}
- \latex{ 4\, m } and \latex{ 4\, m ;}
- \latex{ 8\, mm } and \latex{ 18\, mm }.
How long can the third side be? Which one is greater in each case: the base angle or the vertex angle (the angle included by the equal sides)?
{{exercise_number}}. \latex{ 6 } line segments are given with length \latex{ 3\, cm ;} \latex{ 3.4\, cm ;} \latex{ 4.4\, cm ;} \latex{ 5.6\, cm ;} \latex{ 6.3\, cm } and \latex{ 9.2\, cm } respectively. How many different triangles can be constructed from these data if a line segment can be used several times?
{{exercise_number}}. It is true for the three sides of a triangle that \latex{a \leq b \leq c}, and two of its angles are:
- \latex{ 84º } and \latex{ 36º ;}
- \latex{ 30º } and \latex{ 90º ;}
- \latex{ 70º } and \latex{ 40º ;}
- \latex{ 120º } and \latex{ 29º ;}
- \latex{ 135º } and \latex{ 45º ;}
- \latex{ 18º } and \latex{ 11º }.
Which side is opposite the third angle?
{{exercise_number}}. Prove that the sum of the length of the base and of the leg of an isosceles triangle is greater than half of the perimeter of the triangle, but it is less than three-quarters of the perimeter.
{{exercise_number}}. Let \latex{ a } and \latex{ b } denote the two legs of a right-angled triangle, and let \latex{ c } denote the length of the hypotenuse. Fill in the missing cells in the table.
\latex{ c }
\latex{ a }
\latex{ b }
\latex{ 5\, cm }
\latex{ 3\, cm }
\latex{ 4\, cm }
\latex{ 7\, m }
\latex{ 1.2\, m }
\latex{ 1.3\, m }