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Geometry 16 Online
OpenStudy (vayacondios1):

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

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

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

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

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

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?

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?

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

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

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

OpenStudy (vayacondios1):

it does thanks

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