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activity 3 math connection topic 1 lesson 4 similar figures figures tha…

Question

activity 3 math connection
topic 1 lesson 4
similar figures
figures that are proportional in size, or that have proportional dimensions, are called similar figures.
when timmons photo company prints photo packages, they include several sizes of photos that are all mathematically similar. the largest size is 12 in.×16 in.
16 in.
12 in.

analyze frans and joes calculations of the width of a mathematically similar photo that has a height of 8 inches.

fran
width of larger photo / height of larger photo = width of smaller photo / height of smaller photo
12 inches / 16 inches = w inches / 8 inches
÷2
w = 6
i calculated that the width of the smaller photo is 6 inches.
joe
width of larger photo / width of smaller photo = height of larger photo / height of smaller photo
12 inches / w inches = 16 inches / 8 inches
×6
w = 6
i calculated that the width of the smaller photo is 6 inches.

Explanation:

Step1: Analyze Fran's method

Fran set up the proportion $\frac{\text{width of larger photo}}{\text{height of larger photo}}=\frac{\text{width of smaller photo}}{\text{height of smaller photo}}$. Substituting the values: $\frac{12}{16}=\frac{w}{8}$. Cross - multiplying gives $16w = 12\times8$, then $16w=96$, and $w = 6$. Her proportion is set up correctly based on the property of similar figures where the ratios of corresponding sides are equal.

Step2: Analyze Joe's method

Joe set up the proportion $\frac{\text{width of larger photo}}{\text{width of smaller photo}}=\frac{\text{height of larger photo}}{\text{height of smaller photo}}$. Substituting values: $\frac{12}{w}=\frac{16}{8}$. Cross - multiplying gives $16w=12\times8$, then $16w = 96$, and $w = 6$. His proportion is also set up correctly as it is another valid way to express the ratio of corresponding sides of similar figures.

Answer:

Both Fran and Joe's calculations are correct. They both correctly used the property of similar figures (ratios of corresponding sides are equal) to find that the width $w$ of the smaller photo with height 8 inches is 6 inches, just setting up the proportions in different but valid ways.