Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
1. Section 4.1
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Exercise 41 Page 225

Practice makes perfect
a In similar shapes, the ratio of corresponding sides is always the same. Therefore, to determine x we have to identify the corresponding sides of the two shapes.
With this information we can write an equation that involves x. x/10=30/12 Let's solve this equation for x.
x/10=30/12
â–Ľ
Solve for x
x/10=5/2
x=5/2* 10
x=50/2
x=25
We found that x = 25''.
b Let's find the area of the shapes one at the time.

Small Shape

By dividing the shape into a triangle and a rectangle, we can calculate these areas and then add the results.

Large Shape

We can find the area of the larger shape by using the same method as for the smaller shape. However, the common ratio of the areas between similar figures will always be the square of the ratio of two corresponding sides. A_(large)/A_(small)=(25/10)^2 By substituting the area of the small shape into this equation, we can solve for A_(large).
A_(large)/A_(small)=(25/10)^2
A_(large)/56=(25/10)^2
â–Ľ
Solve for A_(large)
A_(large)=56(25/10)^2
A_(large)=56(625/100)
A_(large)=35 000/100
A_(large)=350
The area of the larger shape is 350 in^2.