Lesson 7 · from §76 of the 1910 text
A Series of Equal Ratios
A series of equal ratios
If \dfrac ab = \dfrac cd = \dfrac ef, then \frac{a + c + e}{b + d + f} = \frac ab = \frac cd = \frac ef.
Proof. Call the common value r: \frac ab = \frac cd = \frac ef = r. Then a = br, c = dr, and e = fr. Adding, a + c + e = (b + d + f)r, so \dfrac{a + c + e}{b + d + f} = r.
In words: in a series of equal ratios, the sum of the antecedents is to the sum of the consequents as any antecedent is to its consequent.
Example. \frac23 = \frac46 = \frac{10}{15}, and \frac{2 + 4 + 10}{3 + 6 + 15} = \frac{16}{24} = \frac23. ✓
A useful consequence: if two triangles have their sides in the same ratio, their perimeters are in that ratio too.
Try one:
Quick check
\tfrac23 = \tfrac46 = \tfrac{10}{15}. Find \dfrac{2 + 4 + 10}{3 + 6 + 15} as a reduced fraction.
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Test the truth of the rule in Exercises 1–4.
\tfrac12 = \tfrac36 = \tfrac5{10}. Find (sum of the antecedents) : (sum of the consequents), reduced. Is it equal to each ratio?
\dfrac1a = \dfrac2{2a} = \dfrac{3b}{3ab}. Find (sum of the antecedents) : (sum of the consequents), reduced. Is it equal to each ratio?
3 : 4 = 6 : 8 = 12 : 16. Find (sum of the antecedents) : (sum of the consequents), reduced. Is it equal to each ratio?
\dfrac1{x - y} = \dfrac{a}{ax - ay} = \dfrac{b}{bx - by}. Find (sum of the antecedents) : (sum of the consequents), reduced. Is it equal to each ratio?
Taken in the same order, the sides of two triangles are 3, 4, 5, and 9, 12, 15 respectively. What is the ratio of the sides of the first triangle to the corresponding sides of the second? Compare this ratio with the ratio of the perimeter of the first triangle to the perimeter of the second.
If \dfrac ab = \dfrac cd = \dfrac ef, prove \dfrac{2a + 3c + 4e}{2b + 3d + 4f} = \dfrac ab. Then test it: when a = 2, b = 3, c = 4, d = 6, e = 6, f = 9, what is the value of \dfrac{2a + 3c + 4e}{2b + 3d + 4f}?