Triangle Inequality

• The sum of the lengths of any two sides of a tri­angle is greater than the length of the third side.
• The difference of the lengths of any two sides of a triangle is less than the length of the third side.

EXAMPLE 7: A teacher asked her class to draw triangles in which the lengths of two of the sides were 6 inches and 7 inches and the length of the third side was also a whole number of inches. To determine how many different triangles the class could draw, note that if x represents the length of the third side, then by KEY FACT H9, 6 + 7 > x, so x < 13. Also, 7 – 6 < x, so x > 1. So, x could have 11 different values: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. The perimeters of the triangles are the integers from 15 to 25. The following diagram illustrates four of the triangles the class could have drawn.

Frequently, questions on the Math 1 test require you to calculate the area of a triangle.

Key Fact H10

The area of a triangle is given by $A=\frac{1}{2}bh$, where b and h are the lengths of the base and height, respectively.

1. Any side of the triangle can be taken as the base.
2. The height (which is also called an altitude) is a line segment drawn perpendicular to the base from the opposite vertex.
3. In a right triangle, either leg can be the base and the other the height.
4. If one endpoint of the base is the vertex of an obtuse angle, then the height will be outside the triangle.
5. In each figure at the top of page 133:

• If $\overline{AC}$ is the base, $\overline{BD}$ is the height.
• If $\overline{AB}$ is the base, $\overline{CE}$ is the height.
• If $\overline{BC}$ is the base, $\overline{AF}$ is the height.

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The height can be outside the triangle. See $\overline{AF}$ in the diagram on page 133.

EXAMPLE 8: To find the area of equilateral triangle PQR, below, whose sides are 12, draw in altitude $\overline{PS}\text{.}$ ΔPQS is a 30-60-90 right triangle, and so by KEY FACT H8, QS = 6 and $PS=6\sqrt{3}$. Finally, the area of $\Delta PQS=\frac{1}{2}\left(12\right)\left(6\sqrt{3}\right)=36\sqrt{3}$.

Replacing 12 by s in Example 8 yields a useful formula.

Key Fact H11

If A represents the area of an equilateral triangle with side s, then $\mathit{A}\mathbf{=}\frac{{\mathbf{s}}^{\mathbf{2}}\sqrt{\mathbf{3}}}{\mathbf{4}}.$

There is another formula for the area of a triangle that can be used whenever you know the lengths of all three sides. The formula given in KEY FACT H12 is known as Heron’s formula.

TIP

Learning this formula for the area of an equilateral triangle can save you time.

Key Fact H12

If a, b, and c are the lengths of the three sides of a triangle, and if s represents the semiperimeter, ${\displaystyle \frac{a+b+c}{2}}$, then the area of the triangle is given by $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}.$

So, to find the area of an equilateral triangle whose sides are 12, you have three choices.

• You can use the formula from KEY FACT H10, $A=\frac{1}{2}bh$, as we did in Example 8.
• You can use the formula from KEY FACT H11, $A=\frac{s^2\sqrt{3}}{4}$. Then $A=\frac{{12}^2\sqrt{3}}{4}= \frac{{}^{36}\cancel{144}\sqrt{3}}{4}=36\sqrt{3}$.
• You can use the formula from KEY FACT H12, $A=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$. Since the perimeter of the triangle is 36, its semiperimeter, s, is 18 and
  $A=\sqrt{18\left(18-12\right)\left(18-12\right)\left(18-12\right)}=\sqrt{18\left(6\right)\left(6\right)\left(6\right)}=\stackrel{ 6 6}{\sqrt{\left(3\right)\cancel{\left(6\right)\left(6\right)}\cancel{\left(6\right)\left(6\right)}}}=36\sqrt{3}$
  .