anonymous
  • anonymous
Two rectangles are similar. If the height of the first rectangle is 3 inches, and the height of the second rectangle is 9 inches, how much larger is the second rectangle's area? They have the same area. 3 times as large 6 times as large 9 times as large
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathstudent55
  • mathstudent55
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mathstudent55
  • mathstudent55
Notice the figure above. In it a segment is 1 inch long, and a segment is 4 inches long. The right segment is 4 times the length of the left segment. Below you have squares. The left square is 1 inch by 1 inch. Its area is 1 square inch. On the right, the square is 4 inches by 4 inches. Its are is 16 square inches. Notice that the side is 4 times longer, but the area id 16 times larger. 16 is the same as 4^2. When a side of a polygon increases by 4 times, the area increases by 4^2 times. In general. if the scale factor between the sides of similar polygons is k, then the scale factor of the areas is k^2. If you do it for the volumes, you'll see the scale factor of the volumes is k^3.
anonymous
  • anonymous
a

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anonymous
  • anonymous
i think
mathstudent55
  • mathstudent55
Did you read anything I wrote? At least read this part again: "When a side of a polygon increases by 4 times, the area increases by 4^2 times. In general. if the scale factor between the sides of similar polygons is k, then the scale factor of the areas is k^2."
anonymous
  • anonymous
c
mathstudent55
  • mathstudent55
No. If the scale factor of the sides is k, the scale factor of the areas is k^2. If the scale factor of the sides is 3, the scale factor of the areas is 3^2. What is 3^2? 3^2 = 3 * 3 = ?
anonymous
  • anonymous
d
mathstudent55
  • mathstudent55
Notice that the height was 3 and became 9, Since 9/3 = 3, the scale factor of the heights is 3. If the scale factor of the sides is 3, the scale factor of the areas is 3^2.
mathstudent55
  • mathstudent55
Correct. Answer is D.
anonymous
  • anonymous
thanks
mathstudent55
  • mathstudent55
You're welcome.

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