28Tylerr
Suppose the side lengths and height of the triangle were divided by three. What effect would this have on the perimeter? & The Area? Justify your answer @Geometry
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28Tylerr
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The Orginial Perimeter was 33m and the Area was 27.
28Tylerr
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This is the last part of the question.
amistre64
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the perimeter is just like before; just divide it by 3; the area tho is nit a linear measurement and needs some work to it
28Tylerr
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P would equal 11?
28Tylerr
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Our base was 12? I think
amistre64
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12 sounds right, youd have to dbl chk tho :)
28Tylerr
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Our base was 12? I think
28Tylerr
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Here let me draw it out real quick (:
amistre64
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\[Area = \frac{1}{2}*base*height\]
\[Area = \frac{1}{2}*\frac{base}{3}*\frac{height}{3}=\frac{1}{18}bh\]
28Tylerr
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Our base was 12? I think
28Tylerr
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|dw:1320881287444:dw|
amistre64
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base = 6 and height = 9 it looks like in that
28Tylerr
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Our base was 12? I think
28Tylerr
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P = 1/2 x 54
28Tylerr
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A =*
amistre64
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|dw:1320881588115:dw|
amistre64
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if you just divided the measures in half, then your not matching this problem.
It wants the measures divided by 3
amistre64
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12/3 = 4
18/3 = 6
28Tylerr
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My bad lol
amistre64
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i just checked out your other post; it was 6 for the bottom and 9 for the height ....
amistre64
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the normal area, is then: 1/2 x 54 as you stated
amistre64
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if we divide all the sides by 3; that will go to:
1/(2*3*3) x 54
or simply
1/18 x 54
amistre64
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which is also:
1/2 * 6/3 * 9/3
1/2 * 2 * 3
1/2 x 6
28Tylerr
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A = 3
amistre64
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yep
28Tylerr
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Man thanks your real good at this math.
amistre64
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we can say that if the sides are divided by 3, then the area is divided by 3^2