#set document(title: "1.2 Similar Triangles", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 1.2#h(0.6em)Similar Triangles === Congruent Triangles Two triangles are #strong[congruent] if they have exactly the same size and shape. This means that their corresponding angles are equal, and their corresponding sides have the same lengths, as shown below. #figure(figph[congruent triangles], alt: "congruent triangles", caption: none) #examplebox("Example 1")[][ The two triangles below are congruent. List the corresponding parts, and find the angles #math.equation(block: false, alt: "θ")[$θ$], #math.equation(block: false, alt: "ϕ")[$ϕ$], and #math.equation(block: false, alt: "χ")[$χ$], and side #math.equation(block: false, alt: "z")[$z$]. #figure(figph[congruent triangles with vertical angles], alt: "congruent triangles with vertical angles", caption: none) left #solutionbox[ In these triangles, #math.equation(block: false, alt: "∠ B equals ∠ D")[$upright(∠) B = upright(∠) D$] because they are both right angles, and #math.equation(block: false, alt: "∠ B C A equals ∠ D C E")[$upright(∠) B C A = upright(∠) D C E$] because they are vertical angles, so #math.equation(block: false, alt: "θ equals 25 degrees")[$θ = 25^(∘)$]. The third angles, #math.equation(block: false, alt: "∠ A")[$upright(∠) A$] and #math.equation(block: false, alt: "∠ E")[$upright(∠) E$], must also be equal, so #math.equation(block: false, alt: "ϕ equals χ equals 65 degrees")[$ϕ = χ = 65^(∘)$]. (Do you see why?) The sides opposite each pair of corresponding angles are equal, so #math.equation(block: false, alt: "A B equals D E")[$A B = D E$], #math.equation(block: false, alt: "B C equals C D")[$B C = C D$] and #math.equation(block: false, alt: "A C equals C E")[$A C = C E$]. In particular, we find that #math.equation(block: false, alt: "z equals 9")[$z = 9$]. ] ] The two triangles at right are congruent. Find the values of #math.equation(block: false, alt: "α")[$α$], #math.equation(block: false, alt: "β")[$β$], and #math.equation(block: false, alt: "γ")[$γ$]. #figure(figph[two congruent triangles], alt: "two congruent triangles", caption: none) #math.equation(block: true, alt: "α equals 180 minus open parenthesis 75 plus 20 close parenthesis equals 85 degrees , β equals 75 degrees , γ equals 85 degrees")[$α = 180 − ( 75 + 20 ) = 85^(∘) , " " β = 75^(∘) , " " γ = 85^(∘)$] Recall that the #strong[altitude] of a triangle is the segment from one vertex of the triangle perpendicular to the opposite side. #examplebox("Example 2")[][ Show that the altitude of an equilateral triangle divides it into two congruent right triangles. #figure(figph[equilateral triangle with altitude], alt: "equilateral triangle with altitude", caption: none) #solutionbox[ Consider for example an equilateral triangle of side 8 inches, as shown above. The altitude is perpendicular to the base, so each half of the original triangle is a right triangle. Because each right triangle contains a #math.equation(block: false, alt: "60 degrees")[$60^(∘)$] angle, the remaining angle in each triangle must be #math.equation(block: false, alt: "90 degrees minus 60 degrees equals 30 degrees")[$90^(∘) − 60^(∘) = 30^(∘)$]. Both triangles have a side of length 8 between the angles of 30° and 60°, so they are congruent. (Consequently, the short sides of the congruent triangles are equal, so each is half the original base.) ] ] The triangles in the previous example are a special type of right triangle called #math.equation(block: false, alt: "30 degrees")[$30^(∘)$]–#math.equation(block: false, alt: "60 degrees")[$60^(∘)$]–#math.equation(block: false, alt: "90 degrees")[$90^(∘)$] triangles. Notice that in these triangles, the #strong[leg] opposite the #math.equation(block: false, alt: "30 degrees")[$30^(∘)$] angle is half the length of the #strong[hypotenuse]. The diagonal of a #strong[parallelogram] divides it into two congruent triangles, as shown at right. List the corresponding parts of the two triangles, and explain why each pair is equal. #figure(figph[parallelogram with diagonal], alt: "parallelogram with diagonal", caption: none) #math.equation(block: false, alt: "∠ B C A equals ∠ C A D")[$upright(∠) B C A = upright(∠) C A D$] and #math.equation(block: false, alt: "∠ B A C equals ∠ A C D")[$upright(∠) B A C = upright(∠) A C D$] because they are alternate interior angles. If two pairs of angles in a triangle are equal, so is the third pair, so #math.equation(block: false, alt: "∠ B equals ∠ D")[$upright(∠) B = upright(∠) D$]. #math.equation(block: false, alt: "B C equals A D")[$B C = A D$] and #math.equation(block: false, alt: "A B equals C D")[$A B = C D$] because they are opposite sides of a parallelogram, and #math.equation(block: false, alt: "A C equals A C")[$A C = A C$]. === Similar Triangles Two triangles are #strong[similar] if they have the same shape but not necessarily the same size. The corresponding angles are equal, and the corresponding sides are #strong[proportional]. We can think of one similar triangle as an enlargement or a reduction of the other. (See the figures below.) The enlargement this sentence describes, on a slider. The gray right triangle is fixed, with legs 4 and 3; the blue triangle beside it is its copy scaled by the factor k, with legs 4k and 3k. The two marked corner angles are identical at every k — that is 'corresponding angles are equal' — because scaling changes lengths, never directions. Meanwhile every blue side is exactly k times its gray partner, so the ratios of corresponding sides all equal k, and the ratio within each triangle — height to base, 3k to 4k — stays 3/4 at every k. That constancy of ratios within a shape is the fact the whole of trigonometry is built on. At k = 1 the copy has exactly the original's size as well as its shape: congruent triangles, the special case of similar.{"functions":\[{"color":"\#6a6a6a","expression":"t\*4","expression2":"0","id":"base-fixed","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"4","expression2":"t\*3","id":"upright-fixed","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"t\*4","expression2":"t\*3","id":"hyp-fixed","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#6a6a6a","expression":"0.7\*cos(0.6435\*t)","expression2":"0.7\*sin(0.6435\*t)","id":"angle-fixed","inequality":"lt","lineDash":\[\],"lineWidth":2,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"6 + t\*4\*k","expression2":"0","id":"base-scaled","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"6 + 4\*k","expression2":"t\*3\*k","id":"upright-scaled","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"6 + t\*4\*k","expression2":"t\*3\*k","id":"hyp-scaled","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true},{"color":"\#1f4e79","expression":"6 + 0.7\*cos(0.6435\*t)","expression2":"0.7\*sin(0.6435\*t)","id":"angle-scaled","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"parametric","tMax":1,"tMin":0,"variable":"t","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-k","label":"Scale factor k","max":1.75,"min":0.5,"name":"k","step":0.25,"value":1.5}\],"title":"A right triangle and its scaled copy, side by side","version":1,"viewport":{"centerX":6.2,"centerY":2.2,"scale":44}}#figure(figph[similar triangles], alt: "similar triangles", caption: none) #math.equation(block: true, alt: "∠ A, equals ∠ A prime; ∠ B, equals ∠ B prime; ∠ C, equals ∠ C prime")[$upright(∠) A & = upright(∠) A^(′) \ upright(∠) B & = upright(∠) B^(′) \ upright(∠) C & = upright(∠) C^(′)$] #math.equation(block: true, alt: "the fraction a over a prime equals the fraction b over b prime equals the fraction c over c prime")[$frac(a, a^(′)) = frac(b, b^(′)) = frac(c, c^(′))$] #figure(figph[similar triangles projected], alt: "similar triangles projected", caption: none) To decide whether two triangles are similar, it turns out that we need to verify only one of the two conditions for similarity, and the other condition will be true automatically. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Similar Triangles] Two triangles are similar if either + their corresponding angles are equal, or + their corresponding sides are proportional. ] #examplebox("Example 3")[][ Which of the pairs of triangles shown below are similar? + #figure(figph[similar triangles], alt: "similar triangles", caption: none) + #figure(figph[not similar triangles], alt: "not similar triangles", caption: none) + #figure(figph[similar triangles 42 degrees], alt: "similar triangles 42 degrees", caption: none) #solutionbox[ + We will check whether the corresponding sides are proportional. We compute the ratios of the corresponding sides, making sure to write each ratio in the same order, #math.equation(block: false, alt: "the fraction larger triangle over smaller triangle")[$display(frac("larger triangle", "smaller triangle"))$]. (The other order, #math.equation(block: false, alt: "the fraction smaller triangle over larger triangle")[$display(frac("smaller triangle", "larger triangle"))$], would also work, as long as we use the same order for all the ratios.) #linebreak() #math.equation(block: false, alt: "shorter legs: the fraction 9 over 6")[$" " " " " " " " " " "shorter legs:" " " " " display(frac(9, 6)) " " " " " " " " " "$] longer legs: #math.equation(block: false, alt: "the fraction 12 over 8")[$" " " " display(frac(12, 8)) " " " " " " " " " "$] hypotenuses: #math.equation(block: false, alt: "the fraction 15 over 10")[$" " " " display(frac(15, 10))$] #linebreak() Because all of these ratios are equal to #math.equation(block: false, alt: "1.5")[$1.5$], the triangles are similar. + The ratios of corresponding sides are not equal: the ratio of the longest sides is #math.equation(block: false, alt: "the fraction 6 over 5")[$display(frac(6, 5))$], but the ratio of the smallest sides is #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$]. The triangles are not similar. + The missing angle of the first right triangle is #math.equation(block: false, alt: "48 degrees")[$48^(∘)$], and the missing angle in the second right triangle is #math.equation(block: false, alt: "42 degrees")[$42^(∘)$], so three pairs of angles match. The triangles are similar. ] ] Are the triangles below similar? Explain why or why not in each case. + #figure(figph[similar triangles by sides], alt: "similar triangles by sides", caption: none) + #figure(figph[similar triangles by angles], alt: "similar triangles by angles", caption: none) a. The triangles are similar because #math.equation(block: false, alt: "the fraction 4 over 6 equals the fraction 6 over 9 equals the fraction 8 over 12")[$" " display(frac(4, 6) = frac(6, 9) = frac(8, 12))$], so the sides are proportional. b. The third angle in both triangles is #math.equation(block: false, alt: "80 degrees")[$80^(∘)$], so the triangles are similar because their corresponding angles are equal. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In part (b) of the previous exercise, note that the third angle in each triangle must be #math.equation(block: false, alt: "80 degrees")[$80^(∘)$], because the sum of the angles is #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. Thus, we need only show that #emph[two] pairs of angles are equal to show that two triangles are similar. ] === Using Proportions with Similar Triangles The figure in the next example shows a parallelogram #math.equation(block: false, alt: "A B C D")[$A B C D$] and two triangles, #math.equation(block: false, alt: "△ A B E")[$upright(△) A B E$] and #math.equation(block: false, alt: "△ F C E")[$upright(△) F C E$]. Can we find the unknown lengths #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$] in the larger triangle? First note that two pairs of corresponding angles in the triangles are equal: #math.equation(block: false, alt: "∠ B E A")[$upright(∠) B E A$] and #math.equation(block: false, alt: "∠ F E C")[$upright(∠) F E C$] are vertical angles, and #math.equation(block: false, alt: "∠ E F C")[$upright(∠) E F C$] and #math.equation(block: false, alt: "∠ B A E")[$upright(∠) B A E$] are alternate interior angles. But if two pairs of corresponding angles are equal, then the third pair must be equal also. This means that the two triangles are similar, and we can use the fact that their corresponding sides are proportional to find #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. #examplebox("Example 4")[][ Find the value of #math.equation(block: false, alt: "x")[$x$] in the figure at right. #figure(figph[parallelogram with triangles], alt: "parallelogram with triangles", caption: none) #solutionbox[ We see that #math.equation(block: false, alt: "x")[$x$] is the length of the shortest side in #math.equation(block: false, alt: "△ A B E")[$upright(△) A B E$]. We know the short side in #math.equation(block: false, alt: "△ F C E")[$upright(△) F C E$] and the lengths of the medium sides in each triangle. If we form the ratios of the short sides and the medium sides, we obtain the following proportion. #math.equation(block: true, alt: "the fraction larger triangle over smaller triangle : the fraction x over 4 equals the fraction 15 over 6")[$display(frac("larger triangle", "smaller triangle")) : " " " " " " " " " " display(frac(x, 4)) " " " " " " = " " " " " " display(frac(15, 6))$] To solve the proportion, we cross-multiply to get #math.equation(block: true, alt: "6 x, equals 4 open parenthesis 15 close parenthesis equals 60, Divide both sides by 6.; x, equals the fraction 60 over 6 equals 10")[$6 x & = 4 ( 15 ) = 60 & & "Divide both sides by 6." \ x & = frac(60, 6) = 10 & &$] ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Remember that “cross-multiplying” is a short-cut technique for solving proportions. It does not work for other operations involving fractions. ] Find the value of #math.equation(block: false, alt: "y")[$y$] in the previous example. Setting the ratios of corresponding sides equal, we find #math.equation(block: false, alt: "the fraction y over 8 equals the fraction 15 over 6")[$display(frac(y, 8)) = display(frac(15, 6))$], and solving for #math.equation(block: false, alt: "y")[$y$] yields #math.equation(block: false, alt: "y equals 20")[$y = 20$]. === Similar Right Triangles If two #emph[right] triangles have one pair of corresponding acute angles with the same measure, then the triangles are similar. We can use this fact about right triangles to make indirect measurements. #examplebox("Example 5")[][ Delbert would like to know the height of a certain building. He gets Francine to hold up a 5-foot pole near the building and measures the length of its shadow. The shadow of the pole is 3 feet long, and the shadow of the building is 12 feet long. #figure(figph[building], alt: "building", caption: none) + Use similar triangles to write a proportion involving the height of the building. + Solve the proportion to find the height of the building. #solutionbox[ + In the figure above we see two right triangles: One triangle is formed by the building and its shadow, and the other by the pole and its shadow. Because the light rays from the sun are parallel, the two angles at the tips of the shadows are equal. Thus, the two right triangles are similar, and their corresponding sides are proportional. The ratios of heights and bases in the two triangles yield the proportion #math.equation(block: true, alt: "the fraction larger triangle over smaller triangle : the fraction h over 5 equals the fraction 12 over 3")[$display(frac("larger triangle", "smaller triangle")) : " " " " " " " " " " display(frac(h, 5)) " " " " " " = " " " " " " display(frac(12, 3))$] + To solve the proportion, we cross-multiply to get #math.equation(block: true, alt: "3 h, equals 5 open parenthesis 12 close parenthesis equals 60, Divide both sides by 3.; h, equals the fraction 60 over 3 equals 20")[$3 h & = 5 ( 12 ) = 60 & & "Divide both sides by 3." \ h & = frac(60, 3) = 20 & &$] The building is 20 feet tall. ] ] In Example 1.19 we created a 30°-60°-90° triangle in which the shorter leg was 4 inches and the hypotenuse was 8 inches. The hypotenuse of another 30°-60°-90° triangle is 5 feet. What is the length of the side opposite the 30° angle? The side opposite the 30° angle is the shorter leg, which is half the length of the hypotenuse. So the its length is 2.5 feet. === Overlapping Triangles In some applications, similar triangles may share a side or an angle. #examplebox("Example 6")[][ Identify two similar triangles in the figure at right, and write a proportion to find #math.equation(block: false, alt: "H")[$H$]. #figure(figph[overlapping triangles], alt: "overlapping triangles", caption: none) #solutionbox[ The two triangles overlap, sharing the marked angle, as shown below. Because each triangle also has a right angle, they are similar. #figure(figph[separated triangles], alt: "separated triangles", caption: none) Note that the base of the larger triangle is #math.equation(block: false, alt: "24 plus 12 equals 36")[$24 + 12 = 36$]. The ratio of the heights and the ratio of the bases must be equal, so we write the following proportion. #math.equation(block: true, alt: "the fraction H over 10, equals the fraction 36 over 24, Cross-multiply.; 24 H, equals 360, Divide both sides by 24.; H, equals the fraction 360 over 24 equals 15")[$frac(H, 10) & = frac(36, 24) & & "Cross-multiply." \ 24 H & = 360 & & "Divide both sides by 24." \ H & = frac(360, 24) = 15 & &$] ] ] Heather wants to know the height of a street lamp. She discovers that when she is 12 feet from the lamp, her shadow is 6 feet long. Find the height of the street lamp. #figure(figph[streetlamp], alt: "streetlamp", caption: none) Because the two triangles are similar, #math.equation(block: false, alt: "the fraction 5 over 6 equals the fraction h over 18")[$display(frac(5, 6)) = display(frac(h, 18))$]. Solving for #math.equation(block: false, alt: "h")[$h$] gives a height of 15 feet. Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Which of the following expressions and equations are proportions? 1. #math.equation(block: false, alt: "the fraction 7 over x equals the fraction 3 over 5")[$display(frac(7, x)) = display(frac(3, 5))$] 2. #math.equation(block: false, alt: "the fraction x over 2 equals the fraction 8 over x plus 2")[$display(frac(x, 2)) = display(frac(8, x + 2))$] 3. #math.equation(block: false, alt: "1 plus the fraction x over 4 equals the fraction 2 x over 3")[$1 + display(frac(x, 4)) = display(frac(2 x, 3))$] 4. #math.equation(block: false, alt: "the fraction 6 over x plus the fraction x over 5")[$display(frac(6, x)) + display(frac(x, 5))$] 5. #math.equation(block: false, alt: "the fraction 3 over x plus 1 times the fraction 2 x over 5")[$display(frac(3, x + 1)) ⋅ display(frac(2 x, 5))$] 6. #math.equation(block: false, alt: "the fraction 1 over x plus the fraction 2 over 3 x equals the fraction x minus 2 over 2")[$display(frac(1, x)) + display(frac(2, 3 x)) = display(frac(x − 2, 2))$] Solve each equation. Begin by writing an equivalent equation without fractions: multiply both sides by the LCD. 7. #math.equation(block: false, alt: "the fraction x over 12 equals the fraction 3 over x")[$display(frac(x, 12)) = display(frac(3, x))$] 8. #math.equation(block: false, alt: "1 plus the fraction x over 2 equals the fraction 2 x over 5")[$1 + display(frac(x, 2)) = display(frac(2 x, 5))$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers Only 1 and 2 are proportions. 7. #math.equation(block: false, alt: "plus or minus 6")[$± 6$] 8. #math.equation(block: false, alt: "minus 10")[$− 10$] ] === Section 1.2 Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Congruent - Altitude - Leg - Hypotenuse - Parallelogram - Similar - Proportional ==== Concepts + Two triangles are #strong[congruent] if they have exactly the same size and shape. + The #strong[altitude] of an equilateral triangle divides it into two congruent right triangles. + In a 30°-60°-90° right triangle, the #strong[leg] opposite the 30° angle is half the length of the #strong[hypotenuse]. + Two triangles are #strong[similar] if they have the same shape but not necessarily the same size. The corresponding angles are equal, and the corresponding sides are #strong[proportional]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Similar Triangles] Two triangles are similar if either + their corresponding angles are equal, or + their corresponding sides are proportional. ] + If two #emph[right] triangles have #emph[one] pair of corresponding acute angles with the same measure, then the triangles are similar. ==== Study Questions + What is the difference between congruent triangles and similar triangles? + What is the name of the short-cut method for solving proportions? Why does the method work? + In two triangles, if two corresponding pairs of angles are equal, are the triangles similar? How do you know? + For the triangles shown, which of the following equations is true? Explain why. #figure(figph[simtri], alt: "simtri", caption: none) + #math.equation(block: false, alt: "the fraction 4 over x equals the fraction 6 over 8")[$display(frac(4, x)) = display(frac(6, 8))$] + #math.equation(block: false, alt: "the fraction x over 4 equals the fraction 6 over 8")[$display(frac(x, 4)) = display(frac(6, 8))$] + #math.equation(block: false, alt: "the fraction x over x plus 4 equals the fraction 6 over 8")[$display(frac(x, x + 4)) = display(frac(6, 8))$] + #math.equation(block: false, alt: "the fraction x over x plus 4 equals the fraction 6 over 14")[$display(frac(x, x + 4)) = display(frac(6, 14))$] ==== Skills Practice each skill in the Homework Problems listed. + Identify congruent triangles and find unknown parts \#1-6 + Identify similar triangles \#7-10 + Find unknown parts of similar triangles \#11-20 + Solve problems using proportions and similar triangles \#21-26 + Use proportions to relate sides of similar triangles \#27-38 === Homework 1.2 In Problems 1–4, name two congruent triangles and find the unknown quantities. #math.equation(block: false, alt: "P Q R S")[$P Q R S$] is an isosceles trapezoid. #figure(figph[isosceles trapezoid], alt: "isosceles trapezoid", caption: none) #math.equation(block: true, alt: "△ P Q T ≅ △ S R T")[$upright(△) P Q T ≅ upright(△) S R T$], #math.equation(block: true, alt: "x equals 7")[$x = 7$], #math.equation(block: true, alt: "y equals 3 , α equals 18 degrees")[$y = 3 , α = 18^(∘)$] #figure(figph[parallel lines], alt: "parallel lines", caption: none) #math.equation(block: false, alt: "△ P R U")[$upright(△) P R U$] is isosceles. #figure(figph[isosceles triangle], alt: "isosceles triangle", caption: none) #math.equation(block: true, alt: "△ P R E ≅ △ U R N , z equals 12")[$upright(△) P R E ≅ upright(△) U R N , z = 12$], #math.equation(block: true, alt: "θ equals 10 degrees")[$θ = 10^(∘)$], #math.equation(block: true, alt: "ϕ equals 70 degrees")[$ϕ = 70^(∘)$] #math.equation(block: false, alt: "△ R A N")[$upright(△) R A N$] is isosceles and #math.equation(block: false, alt: "O R equals N G")[$O R = N G$]. Find #math.equation(block: false, alt: "∠ R N G")[$upright(∠) R N G$] and #math.equation(block: false, alt: "∠ R N O")[$upright(∠) R N O$]. #figure(figph[isosceles triangle], alt: "isosceles triangle", caption: none) Delbert and Francine want to measure the distance across a stream. They mark point #math.equation(block: false, alt: "A")[$A$] directly across the stream from a tree at point #math.equation(block: false, alt: "T")[$T$] on the opposite bank. Delbert walks from point #math.equation(block: false, alt: "A")[$A$] down the bank a short distance to point #math.equation(block: false, alt: "B")[$B$] and sights the tree. He measures the angle between his line of sight and the streambank. + Draw a figure showing the stream, the tree, and right triangle #math.equation(block: false, alt: "A B T")[$A B T$]. + Meanwhile Francine, who was still standing at point #math.equation(block: false, alt: "A")[$A$], walks away from the stream at right-angles to Delbert's path. Delbert watches her progress, and tells her to stop at point #math.equation(block: false, alt: "C")[$C$] when the angle between the stream bank and his line of sight to Francine is the same as the angle from the stream bank to the tree. Add triangle #math.equation(block: false, alt: "A B C")[$A B C$] to your figure. + Delbert now measures the distance from point #math.equation(block: false, alt: "A")[$A$] to Francine at point #math.equation(block: false, alt: "C")[$C$]. Explain why this distance is the same as the distance across the stream. #figure(figph[river], alt: "river", caption: none) #math.equation(block: false, alt: "△ A B T ≅ △ A B C ,")[$upright(△) A B T ≅ upright(△) A B C ,$] so #math.equation(block: false, alt: "A T equals A C")[$A T = A C$] If you have a baseball cap, here is another way to measure the distance across a river. Stand at point #math.equation(block: false, alt: "A")[$A$] directly across the river from a convenient landmark, say a large rock, on the other side. Tilt your head down so that the brim of the cap points directly at the base of the rock, #math.equation(block: false, alt: "R")[$R$]. + Draw a figure showing the river, the rock, and right triangle #math.equation(block: false, alt: "A B R")[$A B R$], where #math.equation(block: false, alt: "B")[$B$] is the location of your baseball cap on your head. + Now, without changing the angle of your head, rotate #math.equation(block: false, alt: "90 degrees")[$90^(∘)$] and sight along the bank on your side of the river. Have a friend mark the spot #math.equation(block: false, alt: "C")[$C$] on the ground where the brim of your cap points. Add triangle #math.equation(block: false, alt: "A B C")[$A B C$] to your figure. + Finally, you can measure the distance from point #math.equation(block: false, alt: "A")[$A$] to point #math.equation(block: false, alt: "C")[$C$]. Explain why this distance is the same as the distance across the river. For Problems 7–10, decide whether the triangles are similar, and explain why or why not. #figure(figph[2 triangles], alt: "2 triangles", caption: none) Similar. Corresponding sides are proportional. #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) Similar. Corresponding angles are equal. #figure(figph[2 triangles], alt: "2 triangles", caption: none) Assume the triangles in Problems 11–14 are similar. Solve for the variables. (Figures are not drawn to scale.) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "∠ A equals 37 degrees , ∠ B equals 37 degrees")[$upright(∠) A = 37^(∘) , upright(∠) B = 37^(∘)$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "h equals 12")[$h = 12$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) In Problems 15–20, use properties of similar triangles to solve for the variable. #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "p equals 35")[$p = 35$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "g equals 84")[$g = 84$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "h equals 30")[$h = 30$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) For Problems 21–26, use properties of similar triangles to solve. A rock climber estimates the height of a cliff she plans to scale as follows. She places a mirror on the ground so that she can just see the top of the cliff in the mirror while she stands straight. The angles 1 and 2 formed by the light rays are equal, as shown in the figure. She then measures the distance to the mirror (2 feet) and the distance from the mirror to the base of the cliff (56 feet). If she is 5 feet 6 inches tall, how high is the cliff? #figure(figph[cliff], alt: "cliff", caption: none) 154 feet Edo estimates the height of the Washington Monument as follows. He notices that he can see the reflection of the top of the monument in the reflecting pool. He is 35 feet from the tip of the reflection, and that point is 1080 yards from the base of the monument, as shown below. From his physics class, Edo knows that the angles marked and are equal. If Edo is 6 feet tall, what is his estimate for the height of the Washington Monument? #figure(figph[Wash mon], alt: "Wash mon", caption: none) In the sixth century BC, the Greek philosopher and mathematician Thales used similar triangles to measure the distance to a ship at sea. Two observers on the shore at points #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$] would sight the ship and measure the angles formed, as shown in figure (a). They would then construct a similar triangle as shown in figure (b), with the same angles at #math.equation(block: false, alt: "A prime")[$A^(′)$] and #math.equation(block: false, alt: "B prime")[$B^(′)$], and measure its sides. (This method is called #emph[triangulation].) Use the lengths given in the figures to find the distance from observer to the ship. #figure(figph[triangulation], alt: "triangulation", caption: none) 1 mile The Capilano Suspension Bridge is a footbridge that spans a 230-foot gorge north of Vancouver, British Columbia. Before crossing the bridge, you decide to estimate its length. You walk 100 feet downstream from the bridge and sight its far end, noting the angle formed by your line of sight, as shown in figure (a). You then construct a similar right triangle with a two-centimeter base, as shown in figure (b). You find that the height of your triangle is 8.98 centimeters. How long is the Capilano Suspension Bridge? #figure(figph[bridge], alt: "bridge", caption: none) A conical tank is 12 feet deep and the diameter of the top is 8 feet. If the tank is filled with water to a depth of 7 feet as shown in the figure at right, what is the area of the exposed surface of the water? #figure(figph[cone], alt: "cone", caption: none) 17.1 square feet To measure the distance #math.equation(block: false, alt: "E C")[$E C$] across the lake shown in the figure at right, stand at #math.equation(block: false, alt: "A")[$A$] and sight point #math.equation(block: false, alt: "C")[$C$] across the lake, then mark point #math.equation(block: false, alt: "B")[$B$]. Then sight to point #math.equation(block: false, alt: "E")[$E$] and mark point #math.equation(block: false, alt: "D")[$D$] so that #math.equation(block: false, alt: "D B")[$D B$] is parallel to #math.equation(block: false, alt: "E C")[$E C$]. If #math.equation(block: false, alt: "A D equals 25")[$A D = 25$] yards, #math.equation(block: false, alt: "A E equals 60")[$A E = 60$] yards, and #math.equation(block: false, alt: "B D equals 30")[$B D = 30$] yards, how wide is the lake? #figure(figph[lake], alt: "lake", caption: none) In Problems 27–28, the pairs of triangles are similar. Solve for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. (The figures are not drawn to scale.) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "y equals the fraction 12 over 17 x")[$y = frac(12, 17) x$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) For Problems 29–34, use properties of similar triangles to solve for the variable. #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "h equals 7.5")[$h = 7.5$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "c equals 15")[$c = 15$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "s equals 6")[$s = 6$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) In Problems 35–38,solve for #math.equation(block: false, alt: "y")[$y$] in terms of #math.equation(block: false, alt: "x")[$x$]. #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "y equals the fraction 3 over 5 x")[$y = frac(3, 5) x$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) #figure(figph[2 triangles], alt: "2 triangles", caption: none) #math.equation(block: true, alt: "y equals 5 plus the fraction 3 over 4 x")[$y = 5 + frac(3, 4) x$] #figure(figph[2 triangles], alt: "2 triangles", caption: none) Triangle #math.equation(block: false, alt: "A B C")[$A B C$] is a right triangle, and #math.equation(block: false, alt: "A D")[$A D$] meets the hypotenuse #math.equation(block: false, alt: "B C")[$B C$] at a right angle. + If #math.equation(block: false, alt: "∠ A C B equals 20 degrees")[$upright(∠) A C B = 20^(∘)$], find #math.equation(block: false, alt: "∠ B , ∠ C A D")[$upright(∠) B , upright(∠) C A D$], and #math.equation(block: false, alt: "∠ D A B")[$upright(∠) D A B$]. + Find two triangles similar to #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$]. List the corresponding sides in each of the triangles. #figure(figph[right triangle], alt: "right triangle", caption: none) + #math.equation(block: false, alt: "∠ B equals 70 degrees")[$upright(∠) B = 70^(∘)$], #math.equation(block: false, alt: "∠ C A D equals 70 degrees")[$upright(∠) C A D = 70^(∘)$], #math.equation(block: false, alt: "∠ D A B equals 20 degrees")[$upright(∠) D A B = 20^(∘)$] + #math.equation(block: false, alt: "△ D B A")[$upright(△) D B A$] and #math.equation(block: false, alt: "△ D A C .")[$upright(△) D A C .$] The hypotenuse is #math.equation(block: false, alt: "B C")[$B C$] in #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "B A")[$B A$] in #math.equation(block: false, alt: "△ D B A")[$upright(△) D B A$], and #math.equation(block: false, alt: "A C")[$A C$] in #math.equation(block: false, alt: "△ D A C")[$upright(△) D A C$]. The short leg is #math.equation(block: false, alt: "A B")[$A B$] in #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "D B")[$D B$] in #math.equation(block: false, alt: "△ D B A")[$upright(△) D B A$], and #math.equation(block: false, alt: "D A")[$D A$] in #math.equation(block: false, alt: "△ D A C")[$upright(△) D A C$]. The longer leg is #math.equation(block: false, alt: "A C")[$A C$] in #math.equation(block: false, alt: "△ A B C")[$upright(△) A B C$], #math.equation(block: false, alt: "D A")[$D A$] in #math.equation(block: false, alt: "△ D B A")[$upright(△) D B A$], and #math.equation(block: false, alt: "D C")[$D C$] in #math.equation(block: false, alt: "△ D A C")[$upright(△) D A C$]. Here is a way to find the distance across a gorge using a carpenter's square and a five-foot pole. Plant the pole vertically on one side of the gorge at point #math.equation(block: false, alt: "A")[$A$] and place the angle of the carpenter's square on top of the pole at point #math.equation(block: false, alt: "B")[$B$], as shown in the figure. Sight along one side of the square so that it points to the opposite side of the gorge at point #math.equation(block: false, alt: "P")[$P$]. Without moving the square, sight along the other side and mark point #math.equation(block: false, alt: "Q")[$Q$]. If the distance from #math.equation(block: false, alt: "Q")[$Q$] to #math.equation(block: false, alt: "A")[$A$] is six inches, calculate the width of the gorge. Explain your method. #figure(figph[right triangle], alt: "right triangle", caption: none)