#set document(title: "2.1 Side and Angle Relationships", 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")) == 2.1#h(0.6em)Side and Angle Relationships === Introduction From geometry we know that the sum of the angles in a triangle is 180°. Are there any relationships between the angles of a triangle and its sides? First of all, you have probably observed that the longest side in a triangle is always opposite the largest angle, and the shortest side is opposite the smallest angle, as illustrated below. #figure(figph[triangle sides], alt: "triangle sides", caption: none) #figure(figph[triangle angles], alt: "triangle angles", caption: none) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ It is usual to label the angles of a triangle with capital letters, and the side opposite each angle with the corresponding lower-case letter, as shown at right. We will follow this practice unless indicated otherwise. #figure(figph[triangle notation], alt: "triangle notation", caption: none) ] #examplebox("Example 1")[][ In #math.equation(block: false, alt: "△ F G H , ∠ F equals 48 degrees ,")[$upright(△) F G H , upright(∠) F = 48^(∘) ,$] and #math.equation(block: false, alt: "∠ G")[$upright(∠) G$] is obtuse. Side #math.equation(block: false, alt: "f")[$f$] is 6 feet long. What can you conclude about the other sides? #solutionbox[ Because #math.equation(block: false, alt: "∠ G")[$upright(∠) G$] is greater than #math.equation(block: false, alt: "90 degrees")[$90^(∘)$], we know that #math.equation(block: false, alt: "∠ F plus ∠ G")[$upright(∠) F + upright(∠) G$] is greater than #math.equation(block: false, alt: "90 degrees plus 48 degrees equals 138 degrees")[$90^(∘) + 48^(∘) = 138^(∘)$], so #math.equation(block: false, alt: "∠ H")[$upright(∠) H$] is less than #math.equation(block: false, alt: "180 degrees minus 138 degrees equals 42 degrees .")[$180^(∘) − 138^(∘) = 42^(∘) .$] Thus, #math.equation(block: false, alt: "∠ H less than ∠ F less than ∠ G ,")[$upright(∠) H < upright(∠) F < upright(∠) G ,$] and consequently #math.equation(block: false, alt: "h less than f less than g")[$h < f < g$]. We can conclude that #math.equation(block: false, alt: "h less than 6")[$h < 6$] feet long, and #math.equation(block: false, alt: "g greater than 6")[$g > 6$] feet long. ] ] In isosceles triangle #math.equation(block: false, alt: "△ R S T")[$upright(△) R S T$], the vertex angle #math.equation(block: false, alt: "∠ S equals 72 degrees")[$upright(∠) S = 72^(∘)$]. Which side is longer, #math.equation(block: false, alt: "s")[$s$] or #math.equation(block: false, alt: "t")[$t$]? Because the two base angles are each #math.equation(block: false, alt: "54 degrees")[$54^(∘)$], #math.equation(block: false, alt: "∠ S")[$upright(∠) S$] is the largest angle, so #math.equation(block: false, alt: "s")[$s$] is the longest side and #math.equation(block: false, alt: "s")[$s$] is longer than #math.equation(block: false, alt: "t")[$t$]. === The Triangle Inequality It is also true that the sum of the lengths of any two sides of a triangle must be greater than the third side, or else the two sides will not meet to form a triangle. This fact is called the triangle inequality. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Triangle Inequality] In any triangle, we must have that #math.equation(block: true, alt: "p plus q greater than r")[$p + q > r$] where #math.equation(block: false, alt: "p , q , and r")[$p , q , "and" " " r$] are the lengths of the sides of the triangle. #figure(figph[triangle], alt: "triangle", caption: none) ] We cannot use the triangle inequality to find the #emph[exact] lengths of the sides of a triangle, but when two sides are known, the triangle inequality allows us to find upper and lower bounds for the length of the third side. #examplebox("Example 2")[][ Two sides of a triangle have lengths 7 inches and 10 inches, as shown at right. What can you say about the length of the third side? #figure(figph[triangle], alt: "triangle", caption: none) #solutionbox[ We let #math.equation(block: false, alt: "x")[$x$] represent the length of the third side of the triangle. By looking at each side in turn, we can apply the triangle inequality three different ways, to get #math.equation(block: true, alt: "7 less than x plus 10 , 10 less than x plus 7 , and x less than 10 plus 7")[$7 < x + 10 , " " " " " " 10 < x + 7 , " " " " " " "and" " " " " " " x < 10 + 7$] We solve each of these inequalities to find #math.equation(block: true, alt: "minus 3 less than x , 3 less than x , and x less than 17")[$− 3 < x , " " " " " " 3 < x , " " " " " " "and" " " " " " " x < 17$] We already know that #math.equation(block: false, alt: "x greater than minus 3")[$x > − 3$] because #math.equation(block: false, alt: "x")[$x$] must be positive, but the other two inequalities do give us new information. The third side must be greater than 3 inches but less than 17 inches long. ] ] Can you make a triangle with three wooden sticks of lengths 14 feet, 26 feet, and 10 feet? Sketch a picture, and explain why or why not. No, #math.equation(block: false, alt: "10 plus 14")[$10 + 14$] is not greater than 26. === Right Triangles: The Pythagorean Theorem In Chapter 1 we used the Pythagorean theorem to derive the distance formula. We can also use the Pythagorean theorem to find one side of a right triangle if we know the other two sides. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pythagorean Theorem] In a right triangle, if #math.equation(block: false, alt: "c")[$c$] stands for the length of the hypotenuse, and the lengths of the two legs are denoted by #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$], then #math.equation(block: true, alt: "a squared plus b squared equals c squared")[$bold(italic(a))^(2) + bold(italic(b))^(2) = bold(italic(c))^(2)$] #figure(figph[right triangle], alt: "right triangle", caption: none) ] #examplebox("Example 3")[][ A 25-foot ladder is placed against a wall so that its foot is 7 feet from the base of the wall. How far up the wall does the ladder reach? #solutionbox[ We make a sketch of the situation, as shown below, and label any known dimensions. We'll call the unknown height #math.equation(block: false, alt: "h")[$h$]. The ladder forms the hypotenuse of a right triangle, so we can apply the Pythagorean theorem, substituting 25 for #math.equation(block: false, alt: "c")[$c$], 7 for #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "h")[$h$] for #math.equation(block: false, alt: "a")[$a$]. #math.equation(block: true, alt: "a squared plus b squared, equals c squared; h squared plus 7 squared, equals 25 squared")[$a^(2) + b^(2) & = c^(2) \ h^(2) + 7^(2) & = 25^(2)$] Now solve by extraction of roots: #figure(figph[ladder], alt: "ladder", caption: none) #math.equation(block: true, alt: "h squared plus 49, equals 625, Subtract 49 from both sides.; h squared, equals 576, Extract roots.; h, equals plus or minus the square root of 576, Simplify the radical.; h, equals plus or minus 24")[$h^(2) + 49 & = 625 & & "Subtract 49 from both sides." \ h^(2) & = 576 & & "Extract roots." \ h & = ± sqrt(576) & & "Simplify the radical." \ h & = ± 24$] The height must be a positive number, so the solution #math.equation(block: false, alt: "minus 24")[$− 24$] does not make sense for this problem. The ladder reaches 24 feet up the wall. ] ] A baseball diamond is a square whose sides are 90 feet long. The catcher at home plate sees a runner on first trying to steal second base, and throws the ball to the second-baseman. Find the straight-line distance from home plate to second base. Use the Pytthagorean theorem to find the diagonal of a square whose sides are 90 feet long. #math.equation(block: true, alt: "c squared equals 90 squared plus 90 squared equals 2 open parenthesis 90 squared close parenthesis")[$c^(2) = 90^(2) + 90^(2) = 2 ( 90^(2) )$] , so #math.equation(block: false, alt: "c equals 90 the square root of 2 approximately equals 127.3")[$c = 90 sqrt(2) ≈ 127.3$] feet #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Keep in mind that the Pythagorean theorem is true only for right triangles, so the converse of the theorem is also true. In other words, if the sides of a triangle satisfy the relationship #math.equation(block: false, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$], then the triangle must be a right triangle. We can use this fact to test whether or not a given triangle has a right angle. ] #examplebox("Example 4")[][ Delbert is paving a patio in his back yard, and would like to know if the corner at #math.equation(block: false, alt: "C")[$C$] is a right angle. He measures 20 cm along one side from the corner, and 48 cm along the other side, placing pegs #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$] at each position, as shown at right. The line joining those two pegs is 52 cm long. Is the corner a right angle? #figure(figph[triangle for patio corner], alt: "triangle for patio corner", caption: none) #solutionbox[ If is a right triangle, then its sides must satisfy #math.equation(block: false, alt: "p squared plus q squared equals c squared")[$p^(2) + q^(2) = c^(2)$]. We find #math.equation(block: true, alt: "p squared plus q squared, equals 20 squared plus 48 squared equals 400 plus 2304 equals 2704; c squared, equals 52 squared equals 2704")[$p^(2) + q^(2) & = 20^(2) + 48^(2) = 400 + 2304 = 2704 \ c^(2) & = 52^(2) = 2704$] Yes, because #math.equation(block: false, alt: "p squared plus q squared equals c squared")[$p^(2) + q^(2) = c^(2)$], the corner at #math.equation(block: false, alt: "C")[$C$] is a right angle. ] ] The sides of a triangle measure 15 inches, 25 inches, and 30 inches long. Is the triangle a right triangle? No, #math.equation(block: false, alt: "15 squared plus 25 squared not equal to 30 squared")[$" " 15^(2) + 25^(2) ≠ 30^(2)$]. The Pythagorean theorem relates the sides of #emph[right] triangles. However, for information about the sides of other triangles, the best we can do (without trigonometry!) is the triangle inequality. Nor does the Pythagorean theorem help us find the #emph[angles] in a triangle. In the next section we discover relationships between the angles and the sides of a right triangle. Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Solve the inequality. 1. #math.equation(block: false, alt: "6 minus x greater than 3")[$#h(0.222em) 6 − x > 3$] 2. #math.equation(block: false, alt: "the fraction minus 3 x over 4 greater than or equal to minus 6")[$#h(0.222em) display(frac(− 3 x, 4)) ≥ − 6$] 3. #math.equation(block: false, alt: "3 x minus 7 less than or equal to minus 10")[$#h(0.222em) 3 x − 7 ≤ − 10$] 4. #math.equation(block: false, alt: "4 minus 3 x less than 2 x plus 9")[$#h(0.222em) 4 − 3 x < 2 x + 9$] If #math.equation(block: false, alt: "x less than 0")[$x < 0$], which of the following expressions are positive, and which are negative? 5. #math.equation(block: false, alt: "minus x")[$#h(0.222em) − x$] 6. #math.equation(block: false, alt: "minus open parenthesis minus x close parenthesis")[$#h(0.222em) − ( − x )$] 7. #math.equation(block: false, alt: "vertical bar x vertical bar")[$#h(0.222em) | x |$] 8. #math.equation(block: false, alt: "minus vertical bar x vertical bar")[$#h(0.222em) − | x |$] 9. #math.equation(block: false, alt: "minus vertical bar minus x vertical bar")[$#h(0.222em) − | − x |$] 10. #math.equation(block: false, alt: "x to the power minus 1")[$#h(0.222em) x^(− 1)$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "x less than 3")[$#h(0.222em) x < 3$] + #math.equation(block: false, alt: "x less than or equal to 8")[$#h(0.222em) x ≤ 8$] + #math.equation(block: false, alt: "x less than or equal to minus 1")[$#h(0.222em) x ≤ − 1$] + #math.equation(block: false, alt: "x greater than minus 1")[$#h(0.222em) x > − 1$] + Positive + Negative + Positive + Negative + Negative + Negative ] === Section 2.1 Summary ==== Vocabulary - Converse - Extraction of roots - Inequality ==== Concepts + The longest side in a triangle is opposite the largest angle, and the shortest side is opposite the smallest angle. + Triangle Inequality: In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. + Pythagorean Theorem: In a right triangle with hypotenuse #math.equation(block: false, alt: "c , a squared plus b squared equals c squared")[$c , " " " " a^(2) + b^(2) = c^(2)$]. + If the sides of a triangle satisfy the relationship #math.equation(block: false, alt: "a squared plus b squared equals c squared")[$" " a^(2) + b^(2) = c^(2) " "$], then the triangle is a right triangle. ==== Study Questions + Is it always true that the hypotenuse is the longest side in a right triangle? Why or why not? + In #math.equation(block: false, alt: "△ D E F")[$upright(△) D E F$], is it possible that #math.equation(block: false, alt: "d plus e greater than f")[$" " d + e > f " "$] and #math.equation(block: false, alt: "e plus f greater than d")[$" " e + f > d " "$] are both true? Explain your answer. + In a right triangle with hypotenuse #math.equation(block: false, alt: "c")[$c$], we know that #math.equation(block: false, alt: "a squared plus b squared equals c squared")[$" " a^(2) + b^(2) = c^(2) " "$]. Is it also true that #math.equation(block: false, alt: "a plus b equals c")[$" " a + b = c " "$]? Why or why not? + The two shorter sides of an obtuse triangle are 3 in and 4 in. What are the possible lengths for the third side? ==== Skills + Identify inconsistencies in figures \#1-12 + Use the triangle inequality to put bounds on the lengths of sides \#13-16 + Use the Pythagorean theorem to find the sides of a right triangle \#17-26 + Use the Pythagorean theorem to identify right triangles \#27-32 + Solve problems using the Pythagorean theorem \#33-42 === Homework 2.1 For Problems 1–12, explain why the measurements shown cannot be accurate. #figure(figph[triangle], alt: "triangle", caption: none) The sum of the angles is not #math.equation(block: false, alt: "180 degrees")[$180^(∘)$]. #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) The exterior angle is not equal to the sum of the opposite interior angles. #figure(figph[rectangle], alt: "rectangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) The sum of the acute angles is not #math.equation(block: false, alt: "90 degrees")[$90^(∘)$]. #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) The largest side is not opposite the largest angle. #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) The Pythagorean theorem is not satisfied. #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangles], alt: "triangles", caption: none) #math.equation(block: false, alt: "5 squared plus 12 squared equals 13 squared")[$5^(2) + 12^(2) = 13^(2)$], but the angle opposite the side of length 13 is #math.equation(block: false, alt: "85 degrees")[$85^(∘)$]. #figure(figph[triangle], alt: "triangle", caption: none) If two sides of a triangle are 6 feet and 10 feet long, what are the largest and smallest possible values for the length of the third side? #math.equation(block: true, alt: "4 less than x less than 16")[$4 < x < 16$] Two adjacent sides of a parallelogram are 3 cm and 4 cm long. What are the largest and smallest possible values for the length of the diagonal? If one of the equal sides of an isosceles triangle is 8 millimeters long, what are the largest and smallest possible values for the length of the base? #math.equation(block: true, alt: "0 less than x less than 16")[$0 < x < 16$] The town of Madison is 15 miles from Newton, and 20 miles from Lewis. What are the possible values for the distance from Lewis to Newton? For Problems 17–22, + Make a sketch of the situation described, and label a right triangle. + Use the Pythagorean Theorem to solve each problem. The size of a TV screen is the length of its diagonal. If the width of a 35-inch TV screen is 28 inches, what is its height? 21 in If a 30-meter pine tree casts a shadow of 30 meters, how far is the tip of the shadow from the top of the tree? The diagonal of a square is 12 inches long. How long is the side of the square? #math.equation(block: false, alt: "6 the square root of 2")[$6 sqrt(2) " "$]in The length of a rectangle is twice its width, and its diagonal is #math.equation(block: false, alt: "4 the square root of 5")[$4 sqrt(5)$] meters long. Find the dimensions of the rectangle. What size rectangle can be inscribed in a circle of radius 30 feet if the length of the rectangle must be three times its width? #figure(figph[circle], alt: "circle", caption: none) The rectangle is #math.equation(block: false, alt: "6 the square root of 10")[$6 sqrt(10)$] inches by #math.equation(block: false, alt: "18 the square root of 10")[$18 sqrt(10)$] inches. What size square can be inscribed inside a circle of radius 8 inches, so that its vertices just touch the circle? #figure(figph[circle], alt: "circle", caption: none) For Problems 23–26, find the unknown side of the triangle. #figure(figph[triangle], alt: "triangle", caption: none) 29 #figure(figph[triangle], alt: "triangle", caption: none) #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "the square root of 3")[$sqrt(3)$] #figure(figph[triangle], alt: "triangle", caption: none) For Problems 27–32, decide whether a triangle with the given sides is a right triangle. 9 in, 16 in, 25 in No 12 m, 16 m, 20 m 5 m, 12 m, 13 m Yes 5 ft, 8 ft, 13 ft #math.equation(block: false, alt: "5 squared")[$5^(2)$] ft, #math.equation(block: false, alt: "8 squared")[$8^(2)$] ft, #math.equation(block: false, alt: "13 squared")[$13^(2)$] ft No #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$] ft, #math.equation(block: false, alt: "the square root of 8")[$sqrt(8)$] ft, #math.equation(block: false, alt: "the square root of 13")[$sqrt(13)$] ft Show that the triangle with vertices #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], #math.equation(block: false, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 3 , 3 close parenthesis")[$( 3 , 3 )$] is an isosceles right triangle, that is, a right triangle with two sides of the same length. The distance from #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 3 , 3 close parenthesis")[$( 3 , 3 )$] is #math.equation(block: false, alt: "3 the square root of 2")[$3 sqrt(2)$], and the distance from #math.equation(block: false, alt: "open parenthesis 3 , 3 close parenthesis")[$( 3 , 3 )$] to #math.equation(block: false, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$] is also #math.equation(block: false, alt: "3 the square root of 2")[$3 sqrt(2)$], so the triangle is isosceles. The distance from #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] to #math.equation(block: false, alt: "open parenthesis 6 , 0 close parenthesis")[$( 6 , 0 )$] is 6, and #math.equation(block: false, alt: "open parenthesis 3 the square root of 2 close parenthesis squared plus open parenthesis 3 the square root of 2 close parenthesis squared equals 6 squared")[$( 3 sqrt(2) )^(2) + ( 3 sqrt(2) )^(2) = 6^(2)$] so the triangle is a right triangle. Two opposite vertices of a square are #math.equation(block: false, alt: "A open parenthesis minus 9 , minus 5 close parenthesis")[$A ( − 9 , − 5 )$] and #math.equation(block: false, alt: "C open parenthesis 3 , 3 close parenthesis")[$C ( 3 , 3 )$]. + Find the length of a diagonal of the square. + Find the length of the side of the square. A 24-foot flagpole is being raised by a rope and pulley, as shown in the figure. The loose end of the rope can be secured to a ring on the ground 7 feet from the base of the pole. From the ring to the top of the pulley, how long should the rope be when the flagpole is vertical? #figure(figph[flagpole], alt: "flagpole", caption: none) 25 ft To check whether the corners of a frame are square, carpenters sometimes measure the sides of a triangle, with two sides meeting at the join of the boards. Is the corner shown in the figure square? #figure(figph[corner], alt: "corner", caption: none) Find #math.equation(block: false, alt: "α , β")[$α , β$] and #math.equation(block: false, alt: "h")[$h$]. #figure(figph[triangle], alt: "triangle", caption: none) #math.equation(block: true, alt: "α equals 30 degrees , β equals 60 degrees , h equals the square root of 3")[$α = 30^(∘) , β = 60^(∘) , h = sqrt(3)$] Find #math.equation(block: false, alt: "α , β ,")[$α , β ,$] and #math.equation(block: false, alt: "d")[$d$]. #figure(figph[square], alt: "square", caption: none) Find the diagonal of a cube of side 8 inches. Hint: Find the diagonal of the base first. #figure(figph[cube], alt: "cube", caption: none) #math.equation(block: false, alt: "8 the square root of 3")[$8 sqrt(3)$] in Find the diagonal of a rectangular box whose sides are 6 cm by 8 cm by 10 cm. Hint: Find the diagonal of the base first. #figure(figph[box], alt: "box", caption: none) For Problems 41 and 42, make a sketch and solve. + The back of Brian's pickup truck is five feet wide and seven feet long. He wants to bring home a 9-foot length of copper pipe. Will it lie flat on the floor of the truck? + The shell on the pickup is 3 feet tall. Will a 9-foot copper pipe fit diagonally across the back of the truck? + No + Yes What is the longest curtain rod that will fit inside a box 60 inches long by 10 inches wide by 4 inches tall? In this problem, we'll show that any angle inscribed in a semi-circle must be a right angle. The figure shows a triangle inscribed in a unit circle, one side lying on the diameter of the circle and the opposite vertex at point #math.equation(block: false, alt: "open parenthesis p , q close parenthesis")[$( p , q )$] on the circle. #figure(figph[circle], alt: "circle", caption: none) + What are the coordinates of the other two vertices of the triangle? What is the length of the side joining those vertices? + Use the distance formula to compute the lengths of the other two sides of the triangle. + Show that the sides of the triangle satisfy the Pythagorean theorem, #math.equation(block: false, alt: "a squared plus b squared equals c squared")[$a^(2) + b^(2) = c^(2)$]. + #math.equation(block: false, alt: "open parenthesis minus 1 , 0 close parenthesis")[$( − 1 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 1 , 0 close parenthesis")[$( 1 , 0 )$]; 2 + #math.equation(block: false, alt: "the square root of open parenthesis p plus 1 close parenthesis squared plus q squared")[$sqrt(( p + 1 )^(2) + q^(2))$] and #math.equation(block: false, alt: "the square root of open parenthesis p minus 1 close parenthesis squared plus q squared")[$sqrt(( p − 1 )^(2) + q^(2))$] + #math.equation(block: true, alt: "open parenthesis the square root of open parenthesis p plus 1 close parenthesis squared plus q squared close parenthesis squared, plus open parenthesis the square root of open parenthesis p minus 1 close parenthesis squared plus q squared close parenthesis squared; equals p squared plus 2 p plus 1 plus q squared plus p squared minus 2 p plus 1 plus q squared; equals 2 p squared plus 2 plus 2 q squared equals 2 plus 2 open parenthesis p squared plus q squared close parenthesis; equals 2 plus 2 open parenthesis 1 close parenthesis equals 4")[$( sqrt(( p + 1 )^(2) + q^(2)) )^(2) & + ( sqrt(( p − 1 )^(2) + q^(2)) )^(2) \ & = p^(2) + 2 p + 1 + q^(2) + p^(2) − 2 p + 1 + q^(2) \ & = 2 p^(2) + 2 + 2 q^(2) = 2 + 2 ( p^(2) + q^(2) ) \ & = 2 + 2 ( 1 ) = 4$] There are many proofs of the Pythagorean theorem. Here is a simple visual argument. #figure(figph[square], alt: "square", caption: none) + What is the length of the side of the large square in the figure? Write an expression for its area. + Write another expression for the area of the large square by adding the areas of the four right triangles and the smaller central square. + Equate your two expressions for the area of the large square, and deduce the Pythagorean theorem.