How to Calculate Distance in a Triangle?

Calculating distances in a triangle is a fundamental skill in geometry. Depending on the information available, there are various methods to determine these distances. Let’s explore some of the most common techniques.

Using the Distance Formula

If you know the coordinates of two points, you can use the distance formula to find the distance between them. The distance formula is derived from the Pythagorean theorem and is given by:

$d = sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$

For example, if you have points A(1, 2) and B(4, 6), the distance between them is:

$d = sqrt{(4 – 1)^2 + (6 – 2)^2} = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$

Using the Pythagorean Theorem

The Pythagorean theorem is useful for right-angled triangles. It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. The formula is:

$c^2 = a^2 + b^2$

Where c is the hypotenuse, and a and b are the other two sides. For example, if you have a triangle with sides of lengths 3 and 4, the hypotenuse is:

$c = sqrt{3^2 + 4^2} = sqrt{9 + 16} = sqrt{25} = 5$

Using the Law of Cosines

The Law of Cosines is useful when you know two sides and the included angle or all three sides of the triangle. The formula is:

$c^2 = a^2 + b^2 – 2ab cdot cos(C)$

Where a and b are the sides, C is the included angle, and c is the side opposite the angle. For example, if you have a triangle with sides 7 and 10 and an included angle of 45 degrees, the distance of the third side is:

$c = sqrt{7^2 + 10^2 – 2 cdot 7 cdot 10 cdot cos(45^circ)}$

Using a calculator for the cosine value:

$c = sqrt{49 + 100 – 140 cdot 0.707} = sqrt{149 – 98.98} = sqrt{50.02} approx 7.07$

Using Heron’s Formula

Heron’s formula is useful when you know all three sides of the triangle. First, calculate the semi-perimeter (s):

$s = frac{a + b + c}{2}$

Then, use Heron’s formula to find the area (A):

$A = sqrt{s(s – a)(s – b)(s – c)}$

For example, for a triangle with sides 6, 8, and 10, the semi-perimeter is:

$s = frac{6 + 8 + 10}{2} = 12$

The area is:

$A = sqrt{12(12 – 6)(12 – 8)(12 – 10)} = sqrt{12 cdot 6 cdot 4 cdot 2} = sqrt{576} = 24$

Conclusion

Understanding these methods allows you to calculate distances in various types of triangles, enhancing your problem-solving skills in geometry. Whether using the distance formula, Pythagorean theorem, Law of Cosines, or Heron’s formula, each method provides a reliable way to find distances in triangles.

3. Wikipedia – Law of Cosines

Citations

  1. 1. Khan Academy – Distance Formula
  2. 2. Math is Fun – Pythagorean Theorem
  3. 4. Khan Academy – Heron’s Formula

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ