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WCCUSD Geometry 1 Benchmark 2 Study Guide G.SRT.1-Some Things To Know Dilations affect the size of the pre-image. The pre-image will enlarge or reduce by the ratio given by the scale factor. A dilation with a scale factor of −1 > x > 1 enlarges it. A dilation of −1 < x < 1 reduces it. 1´ You Try: A. Dilate the following figure using a scale factor of 2 with center of dilation at the origin. Ex: Dilate the following figure using a scale factor of 3 with center of dilation at (5,-6). B. Dilate the following figure using a scale 1 factor of with center at (4,-2). 2 One Solution: Plot (5, −6) . Draw lines from center of dilation through vertices of the preimage. Since the scale factor is 3, each distance from the center of dilation to the image will triple. Plot image’s vertices and connect them to complete the image. C. Dilate the following figure using a scale factor of -2 with center of dilation at the origin. For all dilations centered at (a, b) with a scale factor of k, the image’s coordinates can be found using (a + k(x − a), b + k(y − b)) . If the center of dilation is at the origin, then a and b are zero, resulting in the new image location coordinates as (k ⋅ a, k ⋅ b) . G.SRT.1 G.SRT.1 Page 1 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry 2 Benchmark 2 Study Guide G.SRT.2-Some Things To Know 2´ When a figure is dilated to make an image, corresponding angles are equal and corresponding sides are proportional relative to the scale factor used to dilate. You try: A. Prove the following figures are similar by describing a series of transformations that will map the smaller triangle to the larger triangle. Two different-sized figures can be shown to be similar by using transformations if one of the figures can be mapped onto the other using a series of transformations, one of which is a dilation and the other(s) a reflection, rotation and/or translation. Ex #1: Prove the following figures are similar by describing a series of transformations that will map Figure 1 onto Figure 2. B. Are these triangles similar? Justify your reasoning. One possible solution: Dilate Figure 1 by a scale factor of 2 with center of dilation at . Then translate Figure 1 four units right and four units down. Ex #2: If 𝑨𝑩𝑪𝑫~𝑬𝑭𝑮𝑯, find . Solution: Since the figures are similar, then a dilation has occurred using a scale factor. This creates corresponding sides that are proportional: BD CD = FG HG 30 CD = 6 16 ∴ 𝐶𝐷 = 80 𝑐𝑚 G.SRT.2 G.SRT.2 Page 2 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry 3 Benchmark 2 Study Guide G.SRT.3-Some Things To Know 3´ This standard asks the student to establish that AA is a similarity criterion for two triangles using a series of transformations. You try: Omar thinks that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. To show this, he drew the figure below. Ex: Phillip draws two triangles. Two pairs of corresponding angles are congruent. Select each statement that is true for all such pairs of triangles. A. A sequence of rigid motions carries one triangle onto the other. B. A sequence of rigid motions and dilations carries one triangle onto the other. C. The two triangles are similar because the triangles satisfy Angle-Angle criterion. D. The two triangles are congruent because the triangles satisfy Angle-Angle criterion. E. All pairs of corresponding angles are congruent because triangles must have an angle sum of 180° . F. All pairs of corresponding sides are congruent because of the proportionality of corresponding side lengths. Solutions: True—B, C, and E. A is only true for congruent triangles and not for similar triangles. D is not true because Angle-Angle does not prove triangle congruency. F is not true because corresponding sides are not congruent for similar triangles. G.SRT.3 Page 3 of 11 Which set of transformations maps ΔABC to ΔDEC and supports Omar’s thinking? A. A rotation of 180° clockwise about point C followed by a dilation with a center of point C and a scale factor of 2. B. A rotation of 180° clockwise point C followed by a dilation with a center of 1 point C and a scale factor of . 2 C. A rotation of 180° clockwise point C followed by a dilation with a center of point C and a scale factor of 3. D. A rotation of 180° clockwise point C followed by a dilation with a center of 1 point C and a scale factor of . 3 G.SRT.3 MCC@WCCUSD 12/11/14 WCCUSD Geometry Benchmark 2 Study Guide G.SRT.5-Some Things To Know 4 When solving problems regarding similarity, remember that corresponding sides of similar figures are proportional. 4´ You try: A. Mark stands next to a tree that casts a 15foot shadow. If Mark is 6 feet tall and casts a 4-foot shadow, how tall is the tree? A. A flagpole 4 meters tall casts a 6-meter shadow. At the same time of day, a nearby building casts a 24meter shadow. How tall is the building? Solution: Draw a picture: h 4m 24 m 6m Write a proportion and solve: h 24 = 4 6 h 24 (6)(4) = (6)(4) 4 6 B. Find the value of x in the figure below. 6h = 24(4) 6h 24(4) = 6 6 h = (4)(4) h = 16 The building is 16 meters tall. B. Find mBE . Let BE = x. Or using SideSplitting Theorem: G.SRT.5 Page 4 of 11 G.SRT.5 MCC@WCCUSD 12/11/14 WCCUSD Geometry 5 Benchmark 2 Study Guide G.SRT.6-Some Things To Know Similar right triangles have side ratios that are equal to each other. For example, every 3060-90 triangle, no matter what size, has a small 1 side to hypotenuse ratio of 1:2 or (or 0.5) . 2 These are the side length ratio definitions of the acute angle, θ : Adjacent Hypotenuse tan θ = Opposite Adjacent You try: Given ΔMAT, match each trigonometric ratio to its equivalent value in the box. 17 M T 8 15 A Opposite sin θ = Hypotenuse cos θ = 5´ Ex: Write each trigonometric ratio using the side lengths of ΔABC below. A. B. C. D. E. Solutions: A. B. C. D. E. 4 5 3 5 3 4 4 3 4 5 G.SRT.6 Page 5 of 11 G.SRT.6 MCC@WCCUSD 12/11/14 WCCUSD Geometry 6 Benchmark 2 Study Guide G.SRT.7-Some Things To Know 6´ The sine of an angle is equal to the cosine of its complement: sin θ = cos(90 − θ ) . You try: The table below shows the approximate values of sine and cosine for selected angles. A. Fill in the rest of the table. Q 12 5 P Angle Value of Sine Value of Cosine 0° 0 0.2588 0.5000 0.7071 1 0.9659 0.8660 15° 13 R 30° 45° According to the figure above: 5 5 sin P = ≈ 0.3846 and cos R = ≈ 0.3846 13 13 60° 75° 90° So, sin P = cos R . B. Explain how you determined the values you used. If sin 32° ≈ 0.5514 then the cosine of its complement is equivalent: sin 32° = cos(90° − 32°) = cos58° ≈ 0.5514 Ex: Determine whether the following statements are true or false: A. sin 43° = cos 47° B. sin 43° = cos 43° C. sin 45° = cos 45° D. sin17° = cos(90 −17)° E. cosθ = sin(90 − θ ) Solutions: A—True, B—False, C—True, D—True, E—True G.SRT.7 G.SRT.7 Page 6 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry Benchmark 2 Study Guide G.SRT.8-Some Things To Know 7 You try: 7´ A. A plane is flying at an elevation of 900 meters. From a point directly underneath the plane, the plane is 1200 meters away from a runway. Drawing and labeling pictures are a great way to solve problems using the trigonometric ratios. Don’t forget the Pythagorean Theorem (𝑎! + 𝑏! = 𝑐 ! )! A. The angle of elevation from a landscaped rock to the top of a 30-foot tall flagpole is . Select all equations that can be used to solve for the angle of depression (θ) from the plane to the runway. Which of the following equations could be used to find the distance between the rock and the base of the flagpole? Select all that apply. A. sin 57° = B. cos57° = C. tan 57° = D. tan 33° = 30 x x 30 30 x x 30 Solution: A. sin θ = 900 1500 B. cosθ = 1200 1500 C. tan θ = 1200 900 D. sin θ = 1200 1500 E. cosθ = 900 1500 F. tan θ = 900 1200 B. Determine the missing side lengths of the following triangles: A and B are not correct because the ratios do not correspond to the definitions of the trig ratios. C and D are correct since the tangent opposite ratios of those angles do show . hypotenuse y x 60° 18 ft 30° 22 60° 18 ft Also, here are some special right triangle ratios to memorize: 45-45-90 18 ft h 30-60-90 1 2 1 1 G.SRT.8 G.SRT.8 End of Study Guide Page 7 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry Benchmark 2 Study Guide You Try Solutions: 1´ A. Dilate the following figure using a scale factor of 2 with center of dilation at the origin. You try: A. Prove the following figures are similar by describing a series of transformations that will map the smaller triangle to the larger triangle. OR multiply each vertex’s coordinate by the scale factor of 2 to find the image’s coordinates: (−1, −1) → (2 ⋅ −1, 2 ⋅ −1) → (−2, −2) (−2, −3) → (2 ⋅ −2, 2 ⋅ −3) → (−4, −6) (0, −3) → (2 ⋅ 0, 2 ⋅ −3) → (0, −6) One solution could be dilated ΔABC by a scale factor of 3 with center of dilation at (2, 1) and then translated 4 units right and 2 units up, then ΔABC maps onto ΔA' B'C ' . B. Dilate the following figure using a scale 1 factor of with center at (4,-2). 2 Another solution could be dilating ΔABC by a scale factor of 3 with center of dilation at the origin. 2´ B. Are these triangles similar? Justify your reasoning. C. Dilate the following figure using a scale factor of -2 with center at the origin. If the triangles are similar, then all corresponding side ratios must be equal since a dilation has occurred. 12 4 3 ÷ = 8 4 2 18 6 3 ÷ = 12 6 2 24 8 3 ÷ = 16 8 2 ∴ ∆𝑇𝑈𝑉~∆𝑊𝑋𝑌. G.SRT.1 Page 8 of 11 G.SRT.2 MCC@WCCUSD 12/11/14 WCCUSD Geometry 3´ Benchmark 2 Study Guide You try: Omar thinks that if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. To show this, he drew the figure below. 4´ You try: A. Mark stands next to a tree that casts a 15-foot shadow. If Mark is 6 feet tall and casts a 4-foot shadow, how tall is the tree? h 6 ft 15 ft 4 ft h 15 = 6 4 Which set of transformations maps ΔABC to ΔDEC and supports Omar’s thinking? h 15 (4)(6) = (4)(6) 6 4 4h = 90 The scale factor is 2 since the corresponding sides have a ratio of 6:3, or 2:1. Therefore, A is the correct answer. G.SRT.3 h = 22.5 The tree is 22.5 ft high. B. Find the value of x in the figure below. (x + 2) + 8 12 + 5 = 8 12 x +10 17 = 8 12 x +10 17 (24) = (24) 8 12 3(x +10) = 34 3x + 30 = 34 3x = 4 4 x= 3 G.SRT.5 Page 9 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry 5´ Benchmark 2 Study Guide You try: 7´ Given ΔMAT, match each trigonometric ratio to its equivalent value in the box. 17 M T A. A plane is flying at an elevation of 900 meters. From a point directly underneath the plane, the plane is 1200 meters away from a runway. Select all equations that can be used to solve for the angle of depression (θ) from the plane to the runway. 8 15 You try: A A. B. C. D. E. F. A B D C A B 900 m 1200 m G.SRT.6 6´ Solution: Use the Pythagorean Theorem to find the length of the hypotenuse: You try: The table below shows the approximate values of sine and cosine for selected angles. A. Fill in the rest of the table. Value of Sine Value of Cosine 0° 0 0.2588 0.5000 1 0.9659 0.8660 0.7071 0.8660 0.9659 1 0.7071 0.5000 0.2588 0 30° 45° 60° 75° 90° 1200 ! + 900 ! = 𝑐! 1440000 + 810000 = 𝑐 ! Angle 15° 𝑎! + 𝑏! = 𝑐 ! 2250000 = 𝑐 ! 1500 = c B. Based on this information, the following are equations that can be used to solve for the angle of depression: B. Explain how you determined the values you used. The sine of an angle is equal to the cosine of its complement. So, sin15° = cos 75° , sin 30° = cos60° , sin 45° = cos 45° and sin 0° = cos90° . C. tan θ = 1200 900 D. sin θ = 1200 1500 E. cosθ = 900 1500 G.SRT.8 G.SRT.7 Page 10 of 11 MCC@WCCUSD 12/11/14 WCCUSD Geometry 7´ Benchmark 2 Study Guide Continued… B. Determine the missing side lengths of the following triangles: y x 60° 30° 22 The shorter side’s length is half the hypotenuse. So x = 11 . The length of the longer leg is 3 longer than the shorter leg, so y = 11 3 . 18 ft h 18 ft 60° 18 ft The altitude of the triangle creates a 30-60-90 triangle with the shorter leg being half of 18 ft, which is 9. Therefore the longer leg is 3 longer than the shorter leg, so h = 9 3 . Since two sides are equal, this is an isosceles triangle, which means that the base angles are equal. Each base angle is 45° , so this is a 45-45-90 triangle. The hypotenuse is 2 longer than the sides, which means the side lengths are 7 units. G.SRT.8 Page 11 of 11 MCC@WCCUSD 12/11/14