Determining Proportions Between Sides of Triangles

In geometry, understanding the relationships between the sides of triangles is crucial. One key concept is proportionality, which describes how the lengths of corresponding sides in similar triangles are related. This concept has wide applications in various fields, including architecture, engineering, and art.

Similar Triangles

Two triangles are considered similar if they have the same shape but not necessarily the same size. This means their corresponding angles are equal, and their corresponding sides are proportional.

Identifying Similar Triangles

You can determine if two triangles are similar using the following postulates:

  1. Side-Angle-Side (SAS) Similarity Postulate: If two sides of one triangle are proportional to two sides of another triangle, and the included angles are congruent, then the triangles are similar.

  2. Side-Side-Side (SSS) Similarity Postulate: If all three sides of one triangle are proportional to the three sides of another triangle, then the triangles are similar.

  3. Angle-Angle (AA) Similarity Postulate: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

Example 1: SAS Similarity

Consider triangles ABC and DEF in the diagram below:

[Diagram of triangles ABC and DEF with corresponding sides labeled and angles marked congruent]

If AB/DE = BC/EF and ∠B = ∠E, then triangles ABC and DEF are similar by the SAS similarity postulate.

Example 2: AA Similarity

Consider triangles GHI and JKL in the diagram below:

[Diagram of triangles GHI and JKL with corresponding angles marked congruent]

If ∠G = ∠J and ∠H = ∠K, then triangles GHI and JKL are similar by the AA similarity postulate.

Proportionality and Solving for Unknown Sides

Once you’ve established that two triangles are similar, you can use the concept of proportionality to solve for unknown side lengths. The corresponding sides of similar triangles are proportional, meaning their ratios are equal.

Example 3: Finding an Unknown Side

Consider triangles MNO and PQR in the diagram below:

[Diagram of triangles MNO and PQR with corresponding sides labeled, some with known lengths, and one side with an unknown length ‘x’]

If we know that MN/PQ = NO/QR, and we know the lengths of MN, PQ, NO, and QR, we can set up a proportion to solve for the unknown side length, ‘x’ (representing QR):

$frac{MN}{PQ} = frac{NO}{x}$

By substituting the known lengths and solving for ‘x’, we can determine the length of the unknown side QR.

Applications of Proportionality

The concept of proportionality in similar triangles has numerous applications in real-world situations, including:

  • Architecture: Architects use similar triangles to scale drawings and blueprints to real-life structures.
  • Engineering: Engineers use similar triangles to calculate distances, heights, and angles in various projects.
  • Photography: The principles of similar triangles are applied in camera lenses to create images that are proportional to the actual objects.
  • Mapmaking: Maps use similar triangles to represent geographic features accurately, maintaining proportions between distances on the map and actual distances on the Earth’s surface.

Conclusion

Understanding the concept of proportionality in similar triangles is essential for solving various geometric problems. By applying the postulates for identifying similar triangles and using proportions, you can determine unknown side lengths and solve practical problems in various fields. This knowledge empowers you to analyze and understand the relationships between shapes and sizes in the world around you.

3. CK-12 – Similar Triangles

Citations

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