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

- anonymous

- jamiebookeater

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- anonymous

The Orginial Perimeter was 33m and the Area was 27.

- anonymous

This is the last part of the question.

- amistre64

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

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## More answers

- anonymous

P would equal 11?

- anonymous

Our base was 12? I think

- amistre64

12 sounds right, youd have to dbl chk tho :)

- anonymous

Our base was 12? I think

- anonymous

Here let me draw it out real quick (:

- amistre64

\[Area = \frac{1}{2}*base*height\]
\[Area = \frac{1}{2}*\frac{base}{3}*\frac{height}{3}=\frac{1}{18}bh\]

- anonymous

Our base was 12? I think

- anonymous

|dw:1320881287444:dw|

- amistre64

base = 6 and height = 9 it looks like in that

- anonymous

Our base was 12? I think

- anonymous

P = 1/2 x 54

- anonymous

A =*

- amistre64

|dw:1320881588115:dw|

- amistre64

if you just divided the measures in half, then your not matching this problem.
It wants the measures divided by 3

- amistre64

12/3 = 4
18/3 = 6

- anonymous

My bad lol

- amistre64

i just checked out your other post; it was 6 for the bottom and 9 for the height ....

- amistre64

the normal area, is then: 1/2 x 54 as you stated

- amistre64

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

which is also:
1/2 * 6/3 * 9/3
1/2 * 2 * 3
1/2 x 6

- anonymous

A = 3

- amistre64

yep

- anonymous

Man thanks your real good at this math.

- amistre64

we can say that if the sides are divided by 3, then the area is divided by 3^2

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