28Tylerr
Find one area of each triangle. Round intermediate values of the nearest tenth. Ust the rounded values to calcuate the next value. Round your final answer to the nearest tenth. Please help and show work! Drawing Below -->!
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28Tylerr
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|dw:1337137612703:dw|
GOODMAN
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Hmm...lol, the drawing is a bit unclear..
28Tylerr
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Lol, I know it's kind of hard to read.
GOODMAN
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|dw:1337041518669:dw|
GOODMAN
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I dont know where the 2 goes, lol.
28Tylerr
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There was no 2 and it was 5.2
GOODMAN
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|dw:1337041626863:dw|
GOODMAN
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Like that right?
28Tylerr
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Yepp
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|dw:1337041674389:dw|
GOODMAN
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Thers a more clearer version..
As you can see, this looks like an isosceles.
28Tylerr
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Ahh, Yes it kinda looks like a triangle split in two
GOODMAN
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Now..because of this one rule..lol, sorry forgot the rule, but remember its function,
but it stated that that line in the middle, i think altitude its called, not sure, that separates the triangle, makes the two halves equal.
GOODMAN
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So, basically, that means:
|dw:1337041876443:dw|
28Tylerr
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Ok, Would I use the a2 + b2 = c2 and the a=1/2bh?
GOODMAN
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Yes..I have found all the necessary lengths, all you need to do is plug it in to your formula.
GOODMAN
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So..
What is our base?
28Tylerr
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Ok, Would it be like this: 7^2 + 7^2 = 5.2^2?
GOODMAN
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No..wait.
28Tylerr
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Or would you add the 2 5.2's to make 10.4?
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|dw:1337042106132:dw|
GOODMAN
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Use Pythagorean Theorem.
Our hypotenuse, or "c" is 7
our leg 1 or "a" is 5.2
we need to find leg2 or "h"
28Tylerr
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Ok, So how would I set that up?
GOODMAN
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\[a ^{2}+b ^{2}=c ^{2}\]
\[(5.2)^{2}+b ^{2}=7^{2}\]
\[27.04+b ^{2}=49\]
\[27.04-Subtract \]
\[b ^{2}=49-27.04\]
\[b ^{2}=21.96\]
\[b=4.7\]
GOODMAN
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Our "b" will substitute as our "h" or height.
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|dw:1337042542370:dw|
GOODMAN
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So, now we have all the components of our triangle.
A=1/2bh
A=1/2(10.4)(4.7)
A=1/2(48.88)
A=24.44
28Tylerr
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Would that be my answer and my answer for my area
GOODMAN
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Yes..that would be your area.
28Tylerr
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Mhmmk, can you help me with 1 more like this?
GOODMAN
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Okay..but I gotta hurry. I have some trig functions, lol.
28Tylerr
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|dw:1337140075755:dw|
GOODMAN
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|dw:1337042863061:dw|
GOODMAN
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Like that? Is it isosceles as well?
28Tylerr
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Yes
GOODMAN
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Okay, because of the altitude, we can also determine this..|dw:1337042961699:dw|
28Tylerr
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Ok
GOODMAN
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Now..
We have our formula.
A=1/2bh
Our Base: 16
And our height: 5
28Tylerr
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Can you show me some work for that plz on both? (:
GOODMAN
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A=1/2bh
A=1/2(16)(5)
A=1/2(80)
A=40
28Tylerr
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What about for a2 + b2 = c2
GOODMAN
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We didn't need the Pythagorean Theorem, because we had all the necessary lengths to find the Area.
28Tylerr
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Oh ok so all I needed was the Area because
28Tylerr
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I had all of the lengths
GOODMAN
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Yes..but the first one, we needed the height, and we used the Pythagorean Theorem to find it. In this problem, we already had everything we needed to find the Area.
28Tylerr
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Ok, I thank you for all of the help Goodman. Hopefully you can help me some more some other time. Thank You (:
GOODMAN
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No Problem :D Iam glad I helped