Tenth Grade Mathematics: Ratio of Areas of Similar Triangles Theorem
When two figures are scaled by a linear factor k, how does their area scale? While lengths such as perimeter, altitudes, and medians scale linearly by k, two-dimensional area is an areal measure formed by multiplying two linear dimensions. Consequently, area scales quadratically by k^2. The Ratio of Areas of Similar Triangles Theorem proves that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Statement and Formal Proof of the Theorem
Theorem (Ratio of Areas of Similar Triangles):
The ratio of the areas of two similar triangles is equal to the square of the
ratio of their corresponding sides:
Area(triangle ABC) ( AB )^2 ( BC )^2 ( AC )^2
--------------------- = ( ---- ) = ( ---- ) = ( ---- )
Area(triangle PQR) ( PQ ) ( QR ) ( PR )
Geometric Construction for the Proof:
A P
/|\ /| / | \ / | / | \ / | / | \ / | B +----+----+ C Q +----+----+ R
M N
Given: Two triangles ABC and PQR such that triangle ABC ~ triangle PQR. To Prove: Area(ABC) / Area(PQR) = (AB / PQ)^2 = (BC / QR)^2 = (AC / PR)^2.
Construction: Draw altitude AM perpendicular to BC (AM | BC) and altitude PN perpendicular to QR (PN | QR).
Mathematical Proof
By the standard area formula:
Area(triangle ABC) = (1/2) * BC * AM
Area(triangle PQR) = (1/2) * QR * PN
Divide the two area equations:
Area(ABC) / Area(PQR) = [(1/2) * BC * AM] / [(1/2) * QR * PN]
= (BC / QR) * (AM / PN) --- (Equation 1)
Now consider the right triangles ABM and PQN: Angle B = Angle Q (since triangle ABC ~ triangle PQR). Angle AMB = Angle PNQ = 90 degrees (by construction). Therefore, by the AA similarity criterion:
triangle ABM ~ triangle PQN
Because corresponding sides of similar triangles are proportional:
AM / PN = AB / PQ --- (Equation 2)
However, because the original triangles ABC and PQR are given to be similar, all corresponding sides are in the same ratio:
AB / PQ = BC / QR = AC / PR --- (Equation 3)
Substituting Equation 3 into Equation 2 reveals that the ratio of the altitudes is equal to the ratio of the base sides:
AM / PN = BC / QR --- (Equation 4)
Now substitute Equation 4 into Equation 1:
Area(ABC) / Area(PQR) = (BC / QR) * (BC / QR) = (BC / QR)^2
Using the side equality from Equation 3, we obtain the full theorem:
Area(ABC) / Area(PQR) = (AB / PQ)^2 = (BC / QR)^2 = (AC / PR)^2
This completes the proof.
Corollaries: Ratios of Altitudes, Medians, and Perimeters
Because all linear dimensions in similar triangles scale proportionally:
1. Ratio of Altitudes:
Area(ABC) / Area(PQR) = (Altitude_1 / Altitude_2)^2
2. Ratio of Medians:
Area(ABC) / Area(PQR) = (Median_1 / Median_2)^2
3. Ratio of Angle Bisectors:
Area(ABC) / Area(PQR) = (Bisector_1 / Bisector_2)^2
4. Ratio of Perimeters:
Perimeter(ABC) / Perimeter(PQR) = AB / PQ (Linear!)
Area(ABC) / Area(PQR) = (Perimeter(ABC) / Perimeter(PQR))^2
Detailed Application Example
Let triangle ABC ~ triangle DEF and their areas be respectively 64 cm^2 and 121 cm^2. If EF = 15.4 cm, find BC.
By the theorem:
Area(ABC) / Area(DEF) = (BC / EF)^2
64 / 121 = (BC / 15.4)^2
Take square root of both sides:
sqrt(64 / 121) = BC / 15.4
8 / 11 = BC / 15.4
BC = (8 * 15.4) / 11 = 8 * 1.4 = 11.2 cm.
Chapter Exercises
Exercise 1: The areas of two similar triangles are 81 cm^2 and 49 cm^2. If the altitude of the first triangle is 6.3 cm, find the corresponding altitude of the second triangle.
Exercise 2: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2*CD, find the ratio of the areas of triangles AOB and COD.
Exercise 3: Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solutions and Explanations
Solution 1: We know that the ratio of the areas of two similar triangles equals the square of the ratio of their corresponding altitudes: Area1 / Area2 = (h1 / h2)^2. Given Area1 = 81 cm^2, Area2 = 49 cm^2, and h1 = 6.3 cm. Substituting: 81 / 49 = (6.3 / h2)^2. Taking square roots of both sides: 9 / 7 = 6.3 / h2. Solving for h2: h2 = (7 * 6.3) / 9 = 7 * 0.7 = 4.9 cm. Thus, the corresponding altitude of the second triangle is 4.9 cm.
Solution 2: In trapezium ABCD with AB || DC, consider triangle AOB and triangle COD. As shown in earlier chapters, because AB || DC, alternate interior angles are equal: Angle OAB = Angle OCD, and Angle OBA = Angle ODC. By AA similarity, triangle AOB ~ triangle COD. By the area ratio theorem: Area(triangle AOB) / Area(triangle COD) = (AB / CD)^2. We are given AB = 2CD. Substituting AB: Area(AOB) / Area(COD) = (2CD / CD)^2 = (2 / 1)^2 = 4 / 1. Therefore, the ratio of the areas of triangles AOB and COD is 4 : 1.
Solution 3: Let the side length of the square be s. The diagonal d of the square, by the Pythagorean theorem, has length d = sqrt(s^2 + s^2) = sqrt(2s^2) = ssqrt(2). Let T1 be the equilateral triangle constructed on side s of the square, and T2 be the equilateral triangle constructed on diagonal d. Because all equilateral triangles are equiangular (all angles 60 degrees), T1 is similar to T2 (T1 ~ T2). By the area ratio theorem, the ratio of their areas is equal to the square of the ratio of their side lengths: Area(T1) / Area(T2) = (Side of T1 / Side of T2)^2 = (s / d)^2 = (s / (s*sqrt(2)))^2 = (1 / sqrt(2))^2 = 1 / 2. Multiplying by Area(T2) gives: Area(T1) = (1/2) * Area(T2). This proves that the area of the equilateral triangle described on a side of the square is exactly half the area of the equilateral triangle described on its diagonal.