Unit 7, lesson 60

Similar Triangles and Changing Dimensions

How matching angles prove two triangles similar, how to pair up corresponding sides, and how one fixed ratio links the radius and depth of water in a cone or the height of a tree and its shadow.

  • Theory part 119
  • Practice part 120
  • Geometry and Models

On a sunny afternoon a short pole and a tall tree both cast shadows. You can measure the pole and both shadows with a tape, but not the tree. Because the sun’s rays make the same angle with the ground everywhere nearby, the two shadow triangles have the same shape, and one ratio gives you the tree’s height. The same idea ties together the radius and depth of water in a cone, which calculus uses again and again when quantities change together.

Key terms

Equal angles

Definition Two angles are equal when they have the same measure. If two angles of one triangle are equal to two angles of another triangle, then the third angles are equal too, because the angles of every triangle add up to 180∘180^\circ, or π\pi radians. For nondegenerate triangles, two pairs of equal corresponding angles are enough to prove that the triangles are similar. This is the angle angle criterion.

In plain words If two triangles share two angles, they have the same shape. One may be a scaled copy of the other.

Example A triangle has angles 50∘50^\circ and 60∘60^\circ, and another has angles 60∘60^\circ and 70∘70^\circ. The first one’s third angle is 180∘−50∘−60∘=70∘180^\circ - 50^\circ - 60^\circ = 70^\circ, so both triangles have angles 50∘50^\circ, 60∘60^\circ and 70∘70^\circ and are similar.

Similar triangles

Definition Triangles ABCABC and DEFDEF are similar, written △ABC∼△DEF\triangle ABC \sim \triangle DEF, when the angles at AA, BB and CC equal the angles at DD, EE and FF respectively, and the corresponding sides are proportional,

DEAB=EFBC=FDCA.\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}.

The order of the letters records which vertex matches which.

In plain words Similar triangles are the same shape at possibly different sizes, like a photo and its enlargement.

Example Draw a line parallel to side BCBC of a triangle ABCABC, crossing ABAB at DD and ACAC at EE. The small triangle ADEADE shares the angle at AA, and the angles at DD and BB are corresponding angles at parallel lines, so they are equal. By the angle angle criterion, △ADE∼△ABC\triangle ADE \sim \triangle ABC.

Corresponding sides

Definition In similar triangles, corresponding sides are the sides opposite equal angles. In △ABC∼△DEF\triangle ABC \sim \triangle DEF the pairs are ABAB with DEDE, BCBC with EFEF, and CACA with FDFD. Only ratios of corresponding sides are guaranteed to be equal.

In plain words Match sides by the angles they face, not by where they happen to sit in the drawing.

Example Let △ABC∼△DEF\triangle ABC \sim \triangle DEF with AB=6AB = 6, BC=8BC = 8, CA=10CA = 10 and EF=12EF = 12. The side EFEF corresponds to BCBC, so every length is multiplied by 128=1.5\frac{12}{8} = 1.5. That gives DE=9DE = 9 and FD=15FD = 15.

Scale factor

Definition If △ABC∼△DEF\triangle ABC \sim \triangle DEF, the common ratio

k=DEAB=EFBC=FDCAk = \frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}

is the scale factor from ABCABC to DEFDEF. It is a positive number. Every length in the second triangle is kk times the matching length in the first, and the area is k2k^2 times the area of the first.

In plain words One number tells you how much bigger or smaller the second triangle is. Lengths scale by that number and areas by its square.

Example A triangle with area 66 square centimeters is enlarged with scale factor k=3k = 3. Its sides become three times as long, and its area becomes 9⋅6=549 \cdot 6 = 54 square centimeters, not 1818.

Proportion

Definition A proportion is an equation stating that two ratios are equal,

ab=cd,b≠0, d≠0.\frac{a}{b} = \frac{c}{d}, \qquad b \ne 0, \ d \ne 0.

Multiplying both sides by bdbd shows that it is equivalent to ad=bcad = bc under those conditions. Similar triangles give a proportion for any two pairs of corresponding sides, as long as both ratios are taken in the same direction, for example second triangle over first.

In plain words Put lengths from the same triangle on top in both ratios, then cross multiply to solve.

Example In similar triangles a side of 44 corresponds to xx, and a side of 66 corresponds to 1515. Then

x4=156,\frac{x}{4} = \frac{15}{6},

so 6x=606x = 60 and x=10x = 10.

Shadow

Definition In a shadow problem we assume level ground, vertical objects and sunlight arriving in parallel rays, all measured at the same moment. Each object and its shadow then form a right triangle with the right angle at the base of the object, and the angle at the tip of the shadow is the same for every object because the rays are parallel. So the triangles are similar, and

height of object 1shadow of object 1=height of object 2shadow of object 2.\begin{gathered} \frac{\text{height of object 1}}{\text{shadow of object 1}} \\[0.6em] = \frac{\text{height of object 2}}{\text{shadow of object 2}}. \end{gathered}

In plain words At one moment, every vertical thing nearby casts a shadow in the same proportion to its height.

Example A pole 22 m tall casts a shadow 1.51.5 m long, and at the same time a tree casts a shadow 1212 m long. With HH the tree’s height,

H12=21.5,\frac{H}{12} = \frac{2}{1.5},

so H=16H = 16 meters.

Cone, radius and height

Definition A right circular cone has a circular base of radius r>0r > 0 and a vertex directly above the center of the base. Its height h>0h > 0 is the perpendicular distance from the vertex to the plane of the base, measured along the axis. Its volume is

V=13πr2h.V = \frac{1}{3}\pi r^2 h.

A cut through the axis shows an isosceles triangle, which the axis splits into two right triangles with legs hh and rr.

In plain words The height is measured straight along the axis, not along the slanted side, and the radius is measured across the base from the center.

Example A cone with radius 33 cm and height 44 cm has volume

13π⋅32⋅4=12π≈37.70cubic centimeters.\begin{gathered} \frac{1}{3}\pi \cdot 3^2 \cdot 4 = 12\pi \approx 37.70 \\[0.4em] \text{cubic centimeters.} \end{gathered}

Its slanted side has length 55 cm, which is not its height.

Changing dimensions in a cone

Definition Consider a right circular cone tank with its vertex at the bottom, height 1212 cm and top radius 55 cm, holding water with a horizontal surface at depth hh, where 0≤h≤120 \le h \le 12. For h>0h > 0 the water forms a smaller cone with the same vertex and axis. Its right triangle, with legs hh and rr, shares the angle at the vertex with the tank’s triangle, and both have a right angle where the axis meets a horizontal surface. So the two are similar and

rh=512,0<h≤12.\frac{r}{h} = \frac{5}{12}, \qquad 0 < h \le 12.

The ratio needs h>0h > 0, but its consequence r=5h12r = \frac{5h}{12} also holds at h=0h = 0, where the tank is empty and r=0r = 0.

In plain words The water always fills a small cone that is a scaled copy of the whole tank. That fixes the radius as soon as you know the depth.

Example Substituting r=5h12r = \frac{5h}{12} into the cone volume gives the water volume as a function of depth alone,

V(h)=13π(5h12)2h=25πh3432,0≤h≤12.\begin{aligned} V(h) &= \frac{1}{3}\pi\left(\frac{5h}{12}\right)^2 h \\[0.6em] &= \frac{25\pi h^3}{432}, \qquad 0 \le h \le 12. \end{aligned}

At h=12h = 12 this is 100π100\pi cubic centimeters, the full tank, which matches 13π⋅52⋅12\frac{1}{3}\pi \cdot 5^2 \cdot 12. At h=6h = 6, half the depth, it is 12.5π12.5\pi, only one eighth of the full volume.

Check yourself

  1. A person 1.81.8 m tall stands xx meters from the base of a lamppost whose light is 66 m above the ground. Find the length ss of the person’s shadow in terms of xx.
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The person and the lamppost give similar right triangles that share the angle at the tip of the shadow, so

1.8s=6x+s.\frac{1.8}{s} = \frac{6}{x + s}.

Cross multiplying gives 6s=1.8x+1.8s6s = 1.8x + 1.8s, so 4.2s=1.8x4.2s = 1.8x and s=3x7s = \frac{3x}{7}.

  1. A conical tank with its vertex down is 1010 m deep with top radius 44 m. Write the volume of water as a function of the depth hh.
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Similar triangles give r=2h5r = \frac{2h}{5} for 0≤h≤100 \le h \le 10, so

V(h)=13π(2h5)2h=4πh375,0≤h≤10.\begin{aligned} V(h) &= \frac{1}{3}\pi\left(\frac{2h}{5}\right)^2 h \\[0.6em] &= \frac{4\pi h^3}{75}, \qquad 0 \le h \le 10. \end{aligned}
  1. Two similar triangles have scale factor 33, and the smaller one has area 1010. What is the area of the larger one?
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Areas scale by the square of the scale factor, so the larger area is 9⋅10=909 \cdot 10 = 90.

How this lesson connects