Notes on Similar Triangles and Scaling (Transcript-based)

Notes on Similar Triangles and Scaling (from Transcript)

  • Key idea: If two triangles are similar, all corresponding angles are equal and corresponding sides are proportional.

    • In the transcript: moving from "triangle 1" to "triangle 2" is a scaling by a factor of 2 (twice as long).
    • Therefore, all angles remain the same; the two triangles are similar.

  • Setup and notation (as described in the transcript):

    • Let the small triangle have side lengths proportional to the trigonometric values for an angle x:
    • Opposite side length: extOpp=sin⁡xext{Opp} = \sin x
    • Adjacent side length: extAdj=cos⁡xext{Adj} = \cos x
    • Hypotenuse (for the unit triangle): 11
    • The larger (scaled) triangle has corresponding sides scaled by a factor k = 2:
    • Opposite: 2sin⁡x2 \sin x
    • Adjacent: 2cos⁡x2 \cos x
    • Hypotenuse: 22

  • Constant of proportionality (the scale factor):

    • The transcript asks, "What’s the constant of proportionality to go here?" which is the factor used to scale the triangle from 1 to 2.
    • Answer: k=2k = 2
    • This means every linear dimension is multiplied by 2.

  • The expression for the ratio of the corresponding legs in the scaled triangle:

    • The transcript writes: "This would be two sine x over two cosine x." That is:
    • 2sin⁡x2cos⁡x=sin⁡xcos⁡x=tan⁡x\frac{2 \sin x}{2 \cos x} = \frac{\sin x}{\cos x} = \tan x
    • Interpretation: the slope/ratio of the opposite to the adjacent remains the same under scaling, i.e., it equals tan⁡x\tan x in both triangles.

  • Angles and similarity:

    • Since the triangles are similar, the acute angle x is preserved.
    • The third angle also remains the same (sum of angles in a triangle is constant).

  • Area relationship between the two triangles:

    • Area scales with the square of the linear scale factor.
    • For a right triangle with legs as the base and height, area formula is Area=12×(base)×(height)Area = \tfrac{1}{2} \times (\text{base}) \times (\text{height}).
    • If we assume the small triangle has base = (\cos x) and height = (\sin x) (consistent with a unit-hypotenuse right triangle where opposite = sin x and adjacent = cos x):
    • Area1=12sin⁡xcos⁡xArea_1 = \tfrac{1}{2} \sin x \cos x
    • After scaling by k = 2, the larger triangle has base = (2\cos x) and height = (2\sin x):
    • Area2=12(2cos⁡x)(2sin⁡x)=2sin⁡xcos⁡xArea_2 = \tfrac{1}{2} (2\cos x)(2\sin x) = 2 \sin x \cos x
    • Therefore, the area ratio is:
    • Area<em>2Area</em>1=2sin⁡xcos⁡x12sin⁡xcos⁡x=4=k2\frac{Area<em>2}{Area</em>1} = \frac{2 \sin x \cos x}{\tfrac{1}{2} \sin x \cos x} = 4 = k^2
    • In general: Area<em>2=k2⋅Area</em>1  where  k=2Area<em>2 = k^2 \cdot Area</em>1\;\text{where}\;k=2

  • Numerical example (to illustrate the scaling): take x = 30°

    • sin(30°) = 1/2, cos(30°) = (\sqrt{3}/2)
    • Small triangle: Opp = 1/2, Adj = (\sqrt{3}/2), Hyp = 1
    • Large triangle (k = 2): Opp = 1, Adj = (\sqrt{3}), Hyp = 2
    • Area_1 = (\tfrac{1}{2} \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{3}}{8})
    • Area_2 = (\tfrac{1}{2} \cdot 1 \cdot \sqrt{3} = \tfrac{\sqrt{3}}{2})
    • Check: Area2 / Area1 = ((\tfrac{\sqrt{3}}{2})) / ((\tfrac{\sqrt{3}}{8})) = 4 = k^2

  • Connections to foundational principles:

    • Similarity criterion: if an angle and the included side are proportional, triangles are similar; here, equal angles imply similarity.
    • Scale factor and linearity: doubling linear dimensions doubles the lengths but quadruples the area.
    • Trigonometric connections: in a unit-right-triangle representation, opposite = (\sin x) and adjacent = (\cos x); scaling preserves the sine/cosine ratio as tan x.
    • Coordinate interpretation: points on the unit circle scale by k to lie on a circle of radius k; coordinates become ( (k\cos x, k\sin x) ).

  • Practical implications for problem solving:

    • When given a similar triangle enlarged by a factor of 2, you can find new side lengths by multiplying the original sides by 2.
    • If you need the ratio of a corresponding pair of sides after scaling, it remains the same as before scaling (e.g., tan⁡x\tan x).
    • To compute the new area, multiply the original area by the square of the scale factor: Area<em>2=k2⋅Area</em>1Area<em>2 = k^2 \cdot Area</em>1.

  • Summary of key formulas (LaTeX):

    • Scale factor: k=2k = 2
    • Large triangle sides: Opp′=ksin⁡x,Adj′=kcos⁡x,Hyp′=kOpp' = k \sin x, \quad Adj' = k \cos x, \quad Hyp' = k
    • Side ratio (unchanged): Opp′Adj′=ksin⁡xkcos⁡x=tan⁡x\frac{Opp'}{Adj'} = \frac{k \sin x}{k \cos x} = \tan x
    • Areas: Area<em>1=12sin⁡xcos⁡x,Area</em>2=12(ksin⁡x)(kcos⁡x)=k2⋅12sin⁡xcos⁡xArea<em>1 = \tfrac{1}{2} \sin x \cos x, \quad Area</em>2 = \tfrac{1}{2} (k \sin x)(k \cos x) = k^2 \cdot \tfrac{1}{2} \sin x \cos x
    • Area ratio: Area<em>2Area</em>1=k2\frac{Area<em>2}{Area</em>1} = k^2

  • Final takeaway: The transcript demonstrates that when a triangle is scaled by a factor of 2, all linear dimensions double, the area quadruples, and the trigonometric ratio tan x remains the same across the two similar triangles.