Similarity #2 - Questions and Answers

Question 1 Identify a similarity statement for the two triangles. [Image of two triangles, one labeled DEF and the other labeled WXY with angle labels and side lengths.] The triangles are not similar △DEF ~ △WXY △FED ~ △WXY △DEF ~ △XYW △FED ~ △XYW Answer Step 1 Two triangles DEF and WXY is given in the problem. The ratio of side DE and WX is, DE/WX = 8/6 = 1.33 The ratio of side FD and YW is, FD/YW = 12/9 = 1.33 Step 2 The angle between the corresponding sides, ∠D and ∠W are the same right angles. So, The ratio of the two corresponding sides of the triangle is the same for two pair sides. The angle between the two corresponding sides of the triangles are congruent. The above statements satisfy the condition of the SAS similarity rule. So, △DEF ~ △WXY. Thus, the second option "△DEF ~ △WXY" is the correct option. Question 2 Using the figure below, solve for x and y. There are two triangles shown in the figure below. [Image of two triangles labeled ABC and GFE with side lengths and angle measures labeled.] Answer There are two corresponding angles that are equal in both triangles which represent that both triangles follow the AA similarity criteria. So, △ABC ~ △GFE The formula for similarity for AB and AC is, AB/GF = AC/GE Substitute the values in the above equation. 14/x = 8/10 x = 14(10)/8 x = 70/4 x= 17.5 m Step 2 The formula for similarity for AB and BC is, AB/GF = BC/FE Substitute the values in the above equation. 14/21 = 8/y y = 8(21)/14 y = 8(3)/2 y= 4(3) y=12 m Thus, the value of x is 17.5 m and the value of y is 12 m. Question 3 In the diagram of △GKJ below, LH || KJ, GL=4, LK=24, and HJ=30. What is the lenght of GJ? [Image of a triangle labeled GKJ with a line segment LH contained in it and side lengths labeled] Answer Step 1 Here, the objective is to find the length of the side using the similarity of the triangle. Step 2 In the figure, check the similarity of the triangles △GLH and △GKJ. Here, LH || KJ, therefore, ∠GLH = ∠GKJ ∠GHL = ∠GJK ∠LGH = ∠KGJ Therefore, both the triangles △GLH ~ △GKJ So, the ratio of sides of the triangles will be equal. Step 3 Now, use the ratio of the sides as, GL/LK = GH/HJ 4/24 = GH/30 GH = 4 * 30 / 24 GH=5 The length of the side GJ is, GJ=GH+HJ =5+30 =35 Therefore, the length of side GJ=35 Question 4 Write a similarity statement relating the three triangles in the diagram. [Image of a triangle with vertices labeled N, O, and P. A line segment from O to the side NP creates points Q] Answer Step 1 In △NOP and △NQO ∠ONP = ∠QNO (same angle) ∠PON = ∠OQN (right angles) So, △NOP ~ △NQO (by AA property) Step 2 In △NOP and △OQP ∠OPN = ∠OPQ (same angle) ∠PON = ∠PQO (right angle) So, △NOP ~ △OQP (by AA property) So, △NOP ~ △OQP ~ △NQO Result: △NOP ~ △OQP ~ △NQO. Question 5 Solve the point on the line y = 2x + 3 that is closest to the origin. Answer 1 Point on line y=2x+3 closest to (0,0) A point on the line is (x,2x+3). Using the distance formula we get D= √(2x+3-0)²+(x-0)² = √(2x+3)²+x² = √5x²+12x+9 We can simply look at the expression inside the Radical which will make finding the derivative simpler. The minimum x value of that gives will also give the minimum for √(5x²+12x+9) y = 5x²+12x+9 Now find where y'=0 y'=10x+12 0=10x+12 10x=-12 x=-12/10 = -6/5 Since y=5x²+12x+9 is an upward opening parabola this is the minimum. Now plug it into the equation to find the y-coordinate. y=2(-6/5)+3 = 0.6 = 3/5 [Image of a coordinate plane showing the line y=2x+3] Result (-6/5, 3/5) Answer 2 Explanation: To solve the problem of finding the point on the line y = 2x + 3 that is closest to the origin, we can use the concept of distance between two points. Let's assume that the point on the line closest to the origin is (x₀, y₀). The distance between the origin (0,0) and the point (x₀,y₀) is given by the formula: distance = √(x₀-0)² + (y₀ - 0)² Since the point (x₀,y₀) lies on the line y = 2x + 3, we can substitute y in terms of x: distance = √(x₀-0)² + ((2x₀ + 3) - 0)² Simplifying the equation further: distance = √(x₀)² + (2x₀ + 3)² To minimize the distance, we can differentiate the equation with respect to x₀ and set the derivative equal to zero: d(distance)/dx₀ = 0 Differentiating and solving for x₀: d/dx₀ (√(x₀)² + (2x₀ + 3)²) = 0 2x₀ + (2(2x₀ + 3)2)/2√(x₀)²+(2x₀+3)² = 0 x₀ + 4x₀ + 6 = 0 5x₀ + 6 = 0 x₀ = -6/5 Now, substituting this value back into the equation for the line y = 2x + 3 to find y₀: y₀ = 2(-6/5) + 3 y₀ = 3/5 Therefore, the point on the line y = 2x + 3 that is closest to the origin is (-6/5, 3/5).