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Rectangle ABCD is dilated to create rectangle A'B'C'D' . The width of rectangle ABCD is 20 feet. The width of rectangle A'B'C'D' is 16 feet. The area of rectangle ABCD is 100 square feet. What is the area of rectangle A'B'C'D' ?

Respuesta :

calculista

Answer:

The area of rectangle A'B'C'D' is 64 square feet

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar, then the ratio of its corresponding sides is equal to the scale factor

Let

z ----> the scale factor

x ----> the width of rectangle A'B'C'D' in feet

y ----> the width of rectangle ABCD in feet

so

[tex]z=\frac{x}{y}[/tex]

we have

[tex]x=16\ ft\\y=20\ ft[/tex]

substitute the values

[tex]z=\frac{16}{20}[/tex]

Simplify

[tex]z=\frac{4}{5}[/tex]

step 2

Find the area of rectangle A'B'C'D'

we know that

If two figures are similar, then the ratio of its areas is equal to the scale factor squared

Let

z ----> the scale factor

x ----> the area of rectangle A'B'C'D' in square feet

y ----> the area of rectangle ABCD in square feet

so

[tex]z^2=\frac{x}{y}[/tex]

we have

[tex]z=\frac{4}{5}[/tex]

[tex]y=100\ ft^2[/tex]

substitute  the values

[tex](\frac{4}{5})^2=\frac{x}{100}[/tex]

solve for x

[tex]\frac{16}{25}=\frac{x}{100}[/tex]

[tex]x=\frac{16}{25}(100)\\x=64\ ft^2[/tex]

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