QUESTION IMAGE
Question
kuta software - infinite geometry
using similar polygons
the polygons in each pair are similar. find the missing sid
1)
image of two rectangles: first with top 10, left side x ?; second with top 15, right side 24, bottom 15
handwritten work: x/24 = 10/15, 15x=240, x=16
2)
partially visible
3)
image of two triangles: first with sides 10, 8, 14; second with sides x ?, 4, 7
4)
partially visible
5)
image of two parallelograms: first with sides 12, 10, 12; second with sides 6, ?, 6
6)
partially visible
Problem 1:
Step1: Define the scale factor
Since the rectangles are similar, the ratios of corresponding sides are equal. Let the missing side be \( x \). The ratio of the lengths of the first rectangle to the second is \( \frac{10}{15} \), and the ratio of the heights should be \( \frac{x}{24} \).
\[
\frac{10}{15} = \frac{x}{24}
\]
Step2: Solve for \( x \)
Cross - multiply: \( 15x = 10\times24 \)
\[
15x = 240
\]
Divide both sides by 15: \( x=\frac{240}{15}=16 \)
Problem 3:
Step1: Find the scale factor
The triangles are similar. The ratio of the corresponding sides with known lengths: for the side of length 8 in the first triangle and 4 in the second, the scale factor \( k=\frac{4}{8}=\frac{1}{2} \)
Step2: Find the missing side \( x \)
The side of length 10 in the first triangle corresponds to \( x \) in the second. Using the scale factor \( \frac{x}{10}=\frac{1}{2} \) (or we can also use the ratio of the bases \( \frac{7}{14}=\frac{1}{2} \)). Solving for \( x \): \( x = 10\times\frac{1}{2}=5 \)
Problem 5:
Step1: Determine the scale factor
The parallelograms are similar. The ratio of the corresponding sides with known lengths: for the side of length 12 in the first parallelogram and 6 in the second, the scale factor \( k = \frac{6}{12}=\frac{1}{2} \)
Step2: Calculate the missing side
The side of length 10 in the first parallelogram corresponds to the missing side \( x \) in the second. Using the scale factor \( \frac{x}{10}=\frac{1}{2} \)
\[
x=10\times\frac{1}{2} = 5
\]
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