Precalculus Review Booklet Topics 14-19

Topics • (14) Areas of various two-dimensional objects • (15) Volumes of various three-dimensional objects • (16) Triangle congruence and similarity • (17) The Pythagorean Theorem • (18) 45º-45º-90º- and 30º-60º-90º-triangles • (19) Pythagorean triples XIV. Areas of various two-dimensional objects We now move from algebra to geometry. Let’s review the areas of some important twodimensional figures and the surface areas of some important three-dimensional figures. square: If one side of a square has length x, every side has length x. The square’s area is the product of two adjacent sides. Its area is x*x = x2. rectangle: The area of a rectangle, too, is the product of the lengths of two adjacent sides. The square is a special case of the rectangle. (Rectangle means right angle: every angle in a rectangle is right.) A rectangle’s area is length*width. triangle: Let’s consider right triangles before we consider other triangles. The right angle in a right triangle allows us to regard the triangle as one-half of a rectangle. To find the area of the triangle, we find the product of the lengths of the two sides about the right angle and multiply by ½. In any triangle, one can drop at least one altitude (a line which extends from a vertex to a side and meets the side in a right angle) that does not extend outside the triangle. (Call such an altitude a non-escaping altitude.) The length of an altitude gives us a height of the triangle. The indefinite article a is important because other heights might result from the dropping of altitudes from other vertices. In a right triangle, either of the sides about the right angle may be regarded as an altitude since these two sides meet each other in a right angle. The symbolic representation of the area of a right triangle is ½h1h2. In a non-right triangle, we can drop at least one internal altitude, one that stays within the triangle. (A right triangle has two altitudes which are in the boundary of the triangle but not within the triangle. Both of these altitudes are non-escaping but also non-internal.) Assume we have a non-right triangle in which we have dropped an internal altitude. Note that such an altitude divides this triangle into two right triangles. We now can determine the areas of both of these right triangles by employing what we established above. Assume that the length of the side which receives the altitude is b and that the altitude divides b into b1 and b2. The area of one of the right triangles is ½b1h and the area of the other is ½b2h. The sum of these areas will give us the area of the whole triangle. ½b1h + ½b2h = ½h(b1 + b2) = ½hb = ½bh. In a right triangle, either of h1 or h2 may be regarded as a base b because either may be regarded as an altitude-receiving side. We now are able to say that the area of any planar straight-sided triangle is ½bh. trapezoid: A trapezoid is a four-sided two-dimensional figure which has two parallel sides. The other sides may be parallel or unparallel. (If the other two sides are parallel, the figure is called a parallelogram.) We can calculate the area of a trapezoid by dividing it into figures whose areas we know. Again we will find useful the idea of an altitude. We are 2 guaranteed two parallel sides in a trapezoid. Let’s call these sides b1 and b2. Drop two nonescaping altitudes (ones which do not extend outside the trapezoid). The other two sides will make our altitude-dropping such that (1) both altitudes go to one base or (2) one goes to each base. In case (1), one of b1 and b2 will be divided into three pieces, and in case (2), each of b1 and b2 will be divided into two pieces. Since b1 and b2 are parallel, the two altitudes will have equal length. (Case 1) Assume b1 receives both altitudes. b1 will be divided into three parts: p1, p2, and p3. p2, the central part of b1, and b2 have the same length since the altitudes are parallel. That is, p2 = b2. The altitudes divide the trapezoid into two triangles and one rectangle. The triangles have areas ½p1h and ½p3h and the rectangle has area p2h = b2h. The sum of the areas of the triangles and the rectangle is the area of the trapezoid. That area is ½p1h + ½p3h + p2h = ½p1h + ½p3h + ½p2h + ½p2h [I divided p2h into two halves.] = ½h(p1 + p2 + p3) + ½p2h = ½hb1 + ½p2h [Recall that p1 + p2 + p3 = b1.] = ½h(b1 + p2) = ½h(b1 + b2) = ½(b1 + b2)h = (average of the lengths of the bases)(height). (Case 2) In this case, each of b1 and b2 receives an altitude – each is divided into two pieces. Let’s name b1’s pieces p11 and p12 and b2’s pieces p21 and p22. Again, the trapezoid is divided into two triangles and one rectangle. The triangles have areas ½p12h and ½p21h and the rectangle has area p11h = p22h. (p11 = p22 since the altitudes are parallel.) The trapezoid’s area is the sum of the areas of the two triangles and the rectangle. That area is ½p12h + ½p21h + p11h = ½p12h + ½p21h + ½p11h + ½p11h [I divided p11h into two halves.] = ½p12h + ½p21h + ½p11h + ½p22h [since p11 = p22] = ½h(p11 + p12) + ½h(p21 + p22) = ½hb1 + ½hb2 = ½h(b1 + b2) = ½(b1 + b2)h = (average of the lengths of the bases)(height). parallelogram: A parallelogram is a figure with two pairs of parallel, opposite sides. A parallelogram is a special case of a trapezoid. (A trapezoid has at least one pair of parallel, opposite sides. A parallelogram surpasses this requirement because it has two pairs of parallel, opposite sides.) Therefore, we may use the formula for the area of a trapezoid to calculate the area of a parallelogram. Denote one pair of parallel, opposite sides b1 and b2 and drop the altitude from b1 to b2. A parallelogram’s area = ½(b1 + b2)h = ½(b1 + b1)h [that the other two sides are parallel gives us that b1 = b2] = ½(2b1)h = b1h. A parallelogram’s area is the product of the length of one side and the height measured perpendicularly to that side. circle: A circle is the set of all points in a plane that are equidistant from one point in that plane. (A circular disk is the circle and all points interior to the circle.) All circles are similar. The ratio of the circumference (i.e., distance around the circle) to the diameter is the number . ( is approximately equal to 3.14.) Therefore, the circumference of a circle is d = (2r) [recall that the diameter of a circle is two times the length of the radius of a circle] = 2r. Let’s estimate the area of a circle. Imagine dividing a circle with radius r into onethousand triangular slivers. The base of each triangle is a chord of the circle (i.e., a straight line connecting two points of a circle) and there are 1000 chords of equal length. We may approximate the length of a chord by the length of the arc of the circle bound off by the chord. Each such arc is one one-thousandth of the circle’s circumference = 2r/1000 = r/500. We may approximate the height of each triangle by the radius of the circle = r. Therefore, our approximation of the circle’s area is 1000*(area of one triangle), which equals 1000*½bh = 1000*½(r/500)r = 1000*(r2/1000) = r2. The actual area of the 1000-gon is less than r2. However, as the number of sides of an approximating n-gon increases toward infinity, the n-gon approaches the shape of the circle. That is, the limit of the n-gon’s shape as n approaches infinity is the shape of a circle. The limit of the n-gon’s area as n approaches infinity is r2. That is, the area of a circle is r2. 3 surface area of a (right circular) cylinder: A (right circular) cylinder is a circle given height in a third dimension. (right in right circular cylinder gives that the walls of the cylinder are perpendicular to its ends; circular gives that the ends are circles – rather than [non-circular] ellipses, for example.) Imagine opening a cylinder by cutting along the height in a straight line perpendicularly to the circular base. Then lay the opened cylinder flatly on a table. The result is a rectangle with base equal to the circle’s circumference and height equal to the cylinder’s height. Therefore, a cylinder’s surface area is 2rh. If one is asked to compute the surface area of a cylindrical container (i.e., a cylinder plus its two circular ends), simply add the area of the two circular ends to the surface area of the cylinder: 2rh + 2r2. surface area of a right circular cone: If one were to draw a line from the vertex of a right circular cone (e.g., an ice-cream cone) to the center of its circular base, the line would intersect the circle perpendicularly (i.e., all angles between radii and the line would be right). Non-right (slant) circular cones do not have this property. I wish to consider here only right circular cones. (A non-circular cone is a cone whose base is not circular. Imagine an elliptical cone.) Imagine placing a right circular cone which has been doused in ink on a piece of paper and rolling it – vertex fixed – through exactly one revolution of the circle’s circumference. After one revolution, exactly all of the surface area will have left ink on the paper. The shape we would see is a wedge of a circle. If we knew the full area of the circle and we knew how much of the circle our wedge represented, we could calculate the area of the wedge. The wedge has a radius equal to the slant height of the cone; i.e., the length from the cone’s vertex to the edge of the cone’s circular base. Let’s name the slant height s. The full area of the circle that is partially represented by the wedge is s2. The circumference of the full circle is 2s. In printing the wedge, the cone rolled through an arc-length equal to the circumference of its circular base, 2r. Now we shall compare this arc-length to the circumference of the full circle of which the wedge represents a part. We have that 2r/2s = r/s is the fraction of the full circle represented by the wedge. Then (r/s)(s2) = rs = (the fraction r/s)(the area of the full circle) = the area of the wedge = the surface area of the cone. For the surface area of a right circular conic container, add the area of the cone’s circular base to the surface area of the cone: rs + r2. frustum: Imagine dividing a cone into two pieces by making a cut parallel to the base. Of the two resulting pieces, the one containing the initial base is a frustum. (The other is a new cone.) We can calculate the surface area of a frustum by determining the surface area of the initial cone (before bisection) and subtracting from it the surface area of the new cone. Let’s work here with only right circular frusta. Assume we are given a right circular frustum with the following qualities: it slants inward 30º (i.e., from an imaginary vertical wall) and its slant height is s; its larger circular base has radius r. We must determine the slant height of the full cone of which this frustum represents a post-bisectional part. We may construct part of a right triangle with our given information: one leg of the triangle is r and a bit of the hypotenuse is represented by s; the angle between r and s is 90º – 30º = 60º. Recall from the trigonometry of a 30º-60º-90º-triangle that the side opposite the 30º-angle has one-half the length of the hypotenuse. In our case, the hypotenuse is the slant height of the initial cone and the side opposite the 30º-angle is r. Therefore, we know that ½(hypotenuse) = ½(slant height of initial cone) = r. Then 2r = slant height of initial cone = si. Now we may calculate the surface area of the initial cone: rsi = r(2r) = 2r2. 4 The initial cone is the resection of the frustum and a smaller cone. We can calculate the surface area of the smaller cone once we know its slant height and the radius of its base. Its slant height is the difference between the frustum’s slant height and that of the full cone: si – s. The radius of the smaller cone’s base can be gotten from another 30º-60º-90º-triangle. Within the previous 30º-60º-90º-triangle, draw a vertical line from the tip of s (the bit of the full triangle’s hypotenuse represented by the frustum’s slant height) to r. This vertical line creates a smaller 30º-60º-90º-triangle within the larger triangle. The smaller triangle’s hypotenuse is s. Therefore, we know that the side created from r has length ½s because it is opposite the 30º-angle. Since we dropped a vertical line to r from the tip of the smaller cone’s radius, we now know that the smaller cone’s radius is ½s less than r. That is, the smaller cone’s radius = r – ½s. We now have all the necessary ingredients to calculate the surface area of the smaller cone. The smaller cone’s surface area = (r – ½s)(si – s) = (r – ½s)(2r – s) = ½(2r – s)(2r – s) = ½(2r – s)2. The last expression is not a general formula for the surface area of frustum. Recall that we fixed the angle of the frustum’s inward slant (from an imaginary vertical wall) at 30º in our example. The example offers a general method for working with given information (specifically, larger base’s radius, slant height, and angle of inward slant) to find the surface area of a frustum. To write a general formula, though, one needs a fuller employment of trigonometry than reference to 30º-60º-90º-triangles. sphere: A sphere is the set of all points in three-space equidistant from one point in threespace. A spherical ball is the sphere and all points interior to the sphere. I leave the approximation of the surface area of a sphere as an exercise. One method is to use frusta and two cones to build an approximation to a sphere. To optimize one’s utilization of this method, one needs to employ frusta of varying slant angles. To enable ourselves to do this smoothly, we need to review trigonometry. (Our review begins in section XVI.) XV. Volumes of various three-dimensional objects rectangular box: A rectangular box is one whose edges meet perpendicularly. Its volume is the product of its length, width, and height. A cube is a rectangular box whose length, width, and height are equal. Therefore, its volume is l*w*h = l3 = w3 = h3. cylinder: We described a (right circular) cylinder above as a circle given height in a third dimension. To calculate a cylinder’s volume, we find the product of the height and the circular base’s area. That is, the volume of a cylinder = r2h. right circular cone: I leave this as an exercise. spherical ball: I leave this as an exercise. XVI. Triangle congruence and similarity Assume we have a triangle T with sides of lengths a, b, c and angles of degrees , , and . Another triangle is congruent to T if it has sides a, b, and c. Another triangle is similar to T if it has angles , , and . If T is a triangle with sides 3, 4, and 5, any triangle congruent to T has sides 3, 4, and 5. If T has sides 3, 4, and 5, any triangle similar to T has sides 3r, 4r, and 5r for some positive real 5 number r; the similar triangle will be an enlarged or shrunken or unaltered (in the case that r = 1) version of T. If r = 1, the two triangles will be congruent. Two congruent triangles have the same angles. Congruence implies similarity, but similarity does not imply congruence. There are a few tests to determine whether two triangles are congruent or not. SSS (sideside-side): If two triangles’ sides have the same lengths, then the two triangles are congruent by definition. SAS (side-angle-side): If two triangles share two sides and the angle between the two sides, they are congruent. The angle between the two sides constrains the third side to have a particular length. ASA (angle-side-angle): If two triangles share two angles and the side between the two angles, they are congruent. The two angles constrain the meeting point of the other two sides and, therefore, the lengths of the other two sides. The tests for similarity include all of the tests for congruence (since congruence implies similarity) and one other. Recall that the sum of the degrees of the angles of any planar triangle is 180º. Therefore, if one is given the measures of two angles, one can calculate the third angle’s measure by subtracting the sum of the given angles’ degree-measures from 180º. The other test for similarity is AA (angle-angle). If two triangles share two angles, they share their third angles, too. XVII. The Pythagorean Theorem Right triangles are triangles which have a right angle, which is a 90º-angle. The hypotenuse is the side opposite the 90º-angle in a right triangle. The Pythagorean Theorem presents a relationship among the sides of all right planar triangles. It states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. If the hypotenuse has length c and the other two sides have lengths a and b, we have: a2 + b2 = c2. The following is a quick proof of the Pythagorean Theorem. (I was led in my construction of the following proof by my vague memory of a proof I read in William Dunham’s The Mathematical Universe. Likely my proof and the one in his book are the same.) The sum of the degree-measures of the two non-right angles in a right triangle is 90º. Call the vertices of any right triangle r (for right angle) and o1 and o2 (for other1 and other2). If two copies of this arbitrary right triangle are situated so that o1 and o2 are touching and side a and side b are resting on a line (a line is a 180º-angle), the angle between the hypotenuses will be a 90º-angle. Therefore, situating four copies of an arbitrary right triangle as those in the drawing below are situated gives a square with sides of length a + b inside of which rests a smaller square with side c. 6 What is the larger square’s area? (a+b)2 = [F.O.I.L.] a2 + 2ab + b2. To determine the area of the inner square in terms of a and b, we need to subtract from a2 + 2ab + b2 the area of the four right triangles surrounding the inner square. Each triangle has area ½ab. Therefore, the four triangles have area 4(½ab) = 2ab. Thus the area of the larger square minus the area of the four triangles within it = a2 + 2ab +b2 – 2ab = a2 + b2. We know the area of the inner square is c2. Now we have that a2 + b2 = c2. Our having worked with an arbitrary triangle provides us with a general result: for any right planar triangle, a2 + b2 = c2. XVIII. 45º-45º-90º- and 30º-60º-90º-triangles Two types of right triangles occur frequently in trigonometry. The first is the isosceles right triangle. An isosceles triangle is one with two sides of equal length - and therefore two angles of equal magnitude. The isosceles right triangle has angles with degree-measures 90º, 45º, and 45º (because the non-right angles’ degree-measures must sum to 90º). The Pythagorean Theorem gives that a2 + b2 = c2. In the case of an isosceles right triangle, a = b. Therefore, we have a2 + a2 = 2a2 = c2. If we take the square root of each side, we have 2½a = c. If we standardize the isosceles right triangle by making c = 1, we know a = b = 1/2½ = 2½/2 [I rationalized 1/2½ by multiplying both numerator and denominator by 2½.] The second frequently occurring triangle is the 30º-60º-90º-triangle. Start with an equilateral triangle, which is a triangle whose three sides have equal length – and whose three angles therefore have equal magnitude (60º = 180/3). Assume that each side of the equilateral triangle has length c. Drop an altitude from any vertex. The altitude divides the equilateral triangle into two triangles. Since an equilateral triangle is an isosceles triangle, our altitude is a line of symmetry – dividing the triangle into two congruent parts. Each part is a right triangle since the altitude meets the opposite side perpendicularly. Each right triangle has a right angle, a 60º-angle (original angle of the equilateral triangle) and a (180º – 90º – 60º)-angle = 30º-angle. We have a side of length c, and a side of length ½c. What is the length of the third side; i.e., the altitude? The Pythagorean Theorem gives us that altitude2 + (½c) 2 = c2. Then altitude2 = c2 – ¼c2 = ¾c2. Taking the square root of both sides, we have altitude = 3½c/2. If we standardize the 30º-60º-90º-triangle by setting the hypotenuse equal to 1, the side opposite the 30º-angle has length ½ and the side opposite the 60º-angle has length 3½/2. XIX. Pythagorean triples Triangles that have sides a, b, and c such that a2 + b2 = c2 are right triangles because the Pythagorean Theorem holds for only right triangles. Note that 32 + 42 = 52; i.e. 9 + 16 = 25. Also, 52 + 122 = 132; i.e., 25 + 144 = 169. {3, 4, 5} and {5, 12, 13} are examples of Pythagorean triples. A Pythagorean triple is an ordered set of three integers such that the first2 + second2 = third2. {3n, 4n, 5n} and {5n, 12n, 13n} are derivative Pythagorean triples for any positive integer n. For example, a triangle with sides 6, 8, and 10 and a triangle with sides 10, 24, and 26 are right triangles. {7, 24, 25} [49 + 576 = 625] and {9, 40, 41} [81 + 1600 = 1681] are other prime (i.e., non-derivative) Pythagorean triples. What are x and y in the Pythagorean triple {11, x, y}? 7 Homework Problem Set 3 Topics • (14) Areas of various two-dimensional objects • (15) Volumes of various three-dimensional objects • (16) Triangle congruence and similarity • (17) The Pythagorean Theorem • (18) 45º-45º-90º- and 30º-60º-90º-triangles • (19) Pythagorean triples Please solve the following problems without using a calculator. 14a. 14b. 14c. 14d. 14e. 14f. 14g. 14h. 14i. 14j. 15a. 15b. 15c. 15d. 16a. 16b. What is the area of a square that has one side of length 22 units? What is the area of a rectangle that has one side of length 18 units and whose perimeter (the measure of the rectangle’s boundary) is 68 units? Find the area of a right triangle whose legs (the sides which meet in a right angle) have lengths 5 and 6. Draw a triangle with sides of lengths 2, 3 and 4 and estimate its area. Find the area of a triangle with sides of lengths 1, 2, and 3. (Draw this special triangle.) Is there a planar triangle with sides of lengths ½, 2, and 3? In an x-y-plane, draw a straight line from A (the origin) to B (12,0); draw another straight line from B to C (9,4); draw another straight line from C to D (4,4); and draw a final straight line from D to A. Find the area of trapezoid ABCD. In an x-y-plane, draw a straight line from E (0,6) to F (10,6); draw another straight line from F to G (16,14); draw another straight line from G to H (3,14); and draw a final straight line from H to E. Find the area of trapezoid EFGH. Find the area of a parallelogram with sides of lengths 36 and 48 and angles of measures 30º and 150º. Find the surface area of a cylindrical can whose height is 4 inches and whose radius has length 2 inches. What is the (external) surface area of a (right circular) ice cream cone whose height is 14.4 centimeters and whose radius has length 2.75 centimeters? What is the volume of a box whose square bottom has a side of length 15 inches and whose height is 18 inches? Find the volume of a cylinder created by circularly rolling a rectangular poster whose sides have lengths 50 and 75 centimeters. Assume we are presented with a right circular cone whose height is 16 inches and whose radius has length 8 inches. Estimate this cone’s volume by constructing an approximation to it from 16 cylinders. Estimate the volume of a sphere whose radius has length 24 units by constructing an approximation to a sphere from 9 cylinders. Is a triangle with sides of lengths 3, 3, and 4 similar to a triangle with sides of lengths 4, 4, and 5? Draw, in an x-y-plane, the triangle whose vertices are the points (0,0), (4,1), and (2,5), and draw two similar triangles within it. 8 17a. 17b. 18a. 18b. Use the Pythagorean theorem to find the length of the hypotenuse of a right triangles whose legs have lengths 17 and 18. What is the length of the second leg of a right triangle whose first leg has length 22 and whose internal altitude has length 11? Assume we are given two copies of one 45º-45º-90º-triangle. If we paste the hypotenuses together (in a plane), what shape results? If one side of an equilateral triangle has length 100, what is the length of its internal altitude? 18c. What is the center of the circle which circumscribes a 45º-45º-90º-triangle with a leg of length 10? i. Draw this triangle in the x-y-plane, draw a circle through the triangle’s vertices, and give the x-value and an estimate of the y-value of the point at which the circle’s center lies. ii. The distance formula states that the distance between two points (x1, y1) and (x2, y2) in the x-y-plane equals [(x1–x2)2 + (y1–y2)2]½. (This formula is a direct application of the Pythagorean Theorem.) Now use the distance formula and algebra to determine the exact y-value of the point at which the center lies. 19a. List eight prime sets of Pythagorean triples. (The notes give four prime sets. Add four more.)