How to determine the value of x in similar triangles?

Triangles are called similar if their corresponding angles are equal and their corresponding sides are in proportion. This property of similar triangles can be used to determine the value of unknown variables, such as x.

Understanding Similar Triangles

Corresponding Angles and Sides

Similar triangles have the same shape but may differ in size. For example, if triangle ABC is similar to triangle DEF, we write this as $triangle ABC sim triangle DEF$. This notation means:

  • $angle A = angle D$
  • $angle B = angle E$
  • $angle C = angle F$
  • $frac{AB}{DE} = frac{BC}{EF} = frac{AC}{DF}$

Proportionality of Sides

The key property of similar triangles is that their corresponding sides are proportional. This means that the ratio of the lengths of corresponding sides in one triangle is equal to the ratio of the lengths of the corresponding sides in the other triangle.

Solving for x in Similar Triangles

Step-by-Step Example

Let’s solve an example to make this clearer. Suppose we have two similar triangles, $triangle ABC$ and $triangle DEF$, and we know the following measurements:

  • $AB = 6$
  • $BC = 8$
  • $DE = 3$
  • $EF = 4$
  • $AC = x$
  • $DF = y$

Since the triangles are similar, the sides are proportional:

$frac{AB}{DE} = frac{BC}{EF} = frac{AC}{DF}$

We can set up the proportion using the known values to find $x$:

$frac{AB}{DE} = frac{AC}{DF}$

Substituting the known values:

$frac{6}{3} = frac{x}{y}$

Simplifying, we get:

$2 = frac{x}{y}$

If we also know the value of $y$, we can solve for $x$. Suppose $y = 5$, then:

$2 = frac{x}{5}$

Multiplying both sides by 5:

$x = 10$

Another Example

Consider another example where $triangle PQR sim triangle XYZ$ and we know:

  • $PQ = 10$
  • $QR = 15$
  • $XY = 5$
  • $XZ = x$
  • $YZ = 8$

Using the proportionality of sides:

$frac{PQ}{XY} = frac{QR}{YZ}$

Substituting the values:

$frac{10}{5} = frac{15}{8}$

Simplifying the left-hand side, we get:

$2 = frac{15}{8}$

This shows that the triangles are indeed similar. Now, to find $x$:

$frac{PQ}{XY} = frac{XZ}{YZ}$

Substituting the values:

$frac{10}{5} = frac{x}{8}$

Simplifying, we get:

$2 = frac{x}{8}$

Multiplying both sides by 8:

$x = 16$

Conclusion

Determining the value of x in similar triangles involves understanding the concept of proportionality. By setting up and solving proportions based on the corresponding sides, you can find the unknown lengths. Similar triangles are a powerful tool in geometry, helping solve many practical problems.

3. CK-12 Foundation – Similar Triangles

Citations

  1. 1. Khan Academy – Similar Triangles
  2. 2. Math is Fun – Similar Triangles

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