Use 1.7 as the value of square root of 3. What is the area of equilateral triangle ABC? ____ sq yrds.
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OpenStudy (anonymous):
|dw:1360032001947:dw|
OpenStudy (anonymous):
@jim_thompson5910
jimthompson5910 (jim_thompson5910):
Area of Equilateral Triangle
A = (sqrt(3)/4)*s^2
A = (1.7/4)*s^2... replace sqrt(3) with 1.7 as instructed
A = (1.7/4)*(10)^2 ... plug in the given side length s = 10
A = ???
OpenStudy (anonymous):
what does s^2 mean ?
jimthompson5910 (jim_thompson5910):
s^2 = s*s = s times s
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jimthompson5910 (jim_thompson5910):
s^2 = s squared
jimthompson5910 (jim_thompson5910):
x^2 = x squared
OpenStudy (anonymous):
im confused /:
OpenStudy (anonymous):
can you explain step by step please
jimthompson5910 (jim_thompson5910):
(10)^2 is the same as 10*10
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jimthompson5910 (jim_thompson5910):
A = (1.7/4)*(10)^2
A = (1.7/4)*(10)*(10)
A = ???
OpenStudy (anonymous):
42.5 ?
jimthompson5910 (jim_thompson5910):
you got ti
jimthompson5910 (jim_thompson5910):
it*
OpenStudy (anonymous):
yay! ok next one :D
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OpenStudy (anonymous):
|dw:1360032977873:dw|
OpenStudy (anonymous):
is it 435.2 ? @jim_thompson5910
OpenStudy (anonymous):
@karatechopper am i right ?
jimthompson5910 (jim_thompson5910):
this is an equilateral triangle right?
OpenStudy (anonymous):
yes
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jimthompson5910 (jim_thompson5910):
A = (1.7/4)*s^2
A = (1.7/4)*(16)^2
A = (1.7/4)*(16)*(16)
A = ???
OpenStudy (anonymous):
108.8 right?
jimthompson5910 (jim_thompson5910):
good
OpenStudy (anonymous):
|dw:1360034618821:dw|
jimthompson5910 (jim_thompson5910):
A = (1.7/4)*(8)^2
A = (1.7/4)*(8)*(8)
A = ???
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jimthompson5910 (jim_thompson5910):
notice how all I'm doing is changing the last two numbers
the 1.7/4 part stays the same
OpenStudy (anonymous):
yea i got it haha this next one is different though
OpenStudy (anonymous):
|dw:1360034814628:dw|
OpenStudy (anonymous):
|dw:1360034893872:dw|
jimthompson5910 (jim_thompson5910):
multiply the base and the height
divide that result by 2
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