The diagram shows a section of a bridge between the points A and K. The length of line segment is 640 meters. ∆ABC, ∆CDF, and ∆FJK are similar, and 2AC = CF = 2FK. The first pillar,--BG , is 20 meters tall.
The area of ∆CDF is___________ square meters.
link-https://cdn.ple.platoweb.com/EdAssets/1ede0da051bd4a47b13ca7013501ae97?ts=635405481637200000
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OpenStudy (aleahani):
Try using the pythagorean theorem. So, plug in what you have for ABC
OpenStudy (vayacondios1):
idk how to do that lol i bombed the class
OpenStudy (homeworkhelp):
The Pythagorean theorem states that the square of the hypotenuse in right triangle is equal to the sum of the squares of the other two sides. This is expressed as a^2 + b^2 = c^2.
OpenStudy (aleahani):
Okay well the pythagorean theorem is where you have an triangleand either of the angles are A B or C. |dw:1465410780055:dw|
OpenStudy (vayacondios1):
oh i remember that formula now lol u jogged my memory
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OpenStudy (alekos):
No I dont think you need pythagoras in this case
OpenStudy (vayacondios1):
i really need help on getting this im so close to passing this assighnment by this question
OpenStudy (alekos):
Area of CDF = 1/2 x CF x DH
OpenStudy (vayacondios1):
what is ---cf and ---dh equal it doesnt say
OpenStudy (vayacondios1):
is it basically 1/2 base times hight
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OpenStudy (alekos):
well AC+CF+AK = 640m correct?
OpenStudy (vayacondios1):
it says --ak is 640 all together
OpenStudy (alekos):
i meant AC+CF+FK = 640
OpenStudy (alekos):
happy with that?
OpenStudy (vayacondios1):
yep
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OpenStudy (alekos):
OK now AC = CF/2 & FK = CF/2 right?
OpenStudy (vayacondios1):
yes
OpenStudy (alekos):
So now we have CF/2 + CF + CF/2 = 640
and from here you can work out CF!
OpenStudy (vayacondios1):
so would you divide to find cf
OpenStudy (vayacondios1):
by 3 maybe
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OpenStudy (alekos):
no. what does the LHS of the equation work out to?
OpenStudy (vayacondios1):
213?
OpenStudy (alekos):
no thats way off. What does CF/2 + CF + CF/2 become?
OpenStudy (vayacondios1):
idk im sorry
OpenStudy (vayacondios1):
640?
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OpenStudy (vayacondios1):
in total
OpenStudy (alekos):
Thats the RHS. Im asking you what the LHS simplifies to?
OpenStudy (alekos):
one half of CF + CF + one half of CF = what in terms of CF
OpenStudy (vayacondios1):
320?
OpenStudy (alekos):
320 is what?
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OpenStudy (vayacondios1):
half of segment cf plus cf/2
OpenStudy (alekos):
so CF/2 + CF/2 = 320 iS THAT WHAT YOU MEAN?
OpenStudy (vayacondios1):
yes
OpenStudy (alekos):
Very Good. So we have CF/2 + CF/2 = CF = 320 m
Now we have to find DH
OpenStudy (vayacondios1):
is says ---bg is 20
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OpenStudy (alekos):
If CF = 2AC then DH would equal 2BG because they are similar triangles, correct?
OpenStudy (vayacondios1):
so 40? for height
OpenStudy (alekos):
Brilliant, so now you can work out the area!!
OpenStudy (vayacondios1):
ight 1 sec
OpenStudy (vayacondios1):
i got 6400
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OpenStudy (alekos):
Fabulous, well done
OpenStudy (vayacondios1):
is it in square meters?
OpenStudy (alekos):
Yep
OpenStudy (vayacondios1):
thank you so much
OpenStudy (alekos):
no problem, it was a pleasure. i hope that helps you to understand things
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