Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 2 - Solving Equations - 2-8 Proportions and Similar Figures - Lesson Check - Page 133: 1

Answer

Part a: 32.5 cm Part b: 1 cm : 2.5 cm

Work Step by Step

Part a: Using proportions will make this problem easier to visualize: base of the original triangle/base of the enlarged triangle=height of the original triangle/height of the enlarged triangle Substitute the given values into the proportion: 13/x=7/17.5 Cross Products Property (AKA cross multiply and divide): 13(17.5)=7x Simplify by multiplying the left side of the equation and then dividing 7 on each side: 227.57=32.5 The base of the enlarged triangle is 32.5 cm Part b: The scale in this problem would be found by using a ratio of two corresponding sides of the triangles. For example, to get from the height of the original triangle to the height of the enlarged triangle, write out the equation 7x=17.5 (7 times what number will give us 17.5) Divide 7 on both sides of the equation so that x=2.5 Also, you can use the bases to determine the scale. To get from the base of the original triangle to the base of the enlarged triangle, write out the equation 13x=32.5 (13 times what number will give us 32.5) Divide 13 on both sides of the equation so that x=2.5 The scale of the model is 1 cm : 2.5 cm. For every 1 cm on the original triangle, it is 2.5 cm on the enlarged triangle.
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