Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 2 - Solving Equations - 2-8 Proportions and Similar Figures - Lesson Check: 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: $\frac{\text{base of the original triangle}}{\text{base of the enlarged triangle}}$=$\frac{\text{height of the original triangle}}{\text{height of the enlarged triangle}}$ Substitute the given values into the proportion: $\frac{\text{13}}{x}$=$\frac{\text{7}}{\text{17.5}}$ Cross Products Property (AKA cross multiply and divide): 13(17.5)=7$x$ Simplify by multiplying the left side of the equation and then dividing 7 on each side: $\frac{\text{227.5}}{7}$=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 7$x$=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 13$x$=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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