Ask your own question, for FREE!
Mathematics
AsleepAndUnafraid21:

URGENT PLZ HELP NOW Part C: Find the distance from B to E and from P to E. Show your work. (4 points)

AsleepAndUnafraid21:

here is the image

AsleepAndUnafraid21:

@aqual @uhhhkhakis plz help

UhhhKhakis:

I'm sorry fam I'm bad at math.

AsleepAndUnafraid21:

oh okay

AsleepAndUnafraid21:

@fortish @hunter3506 plz help

Fortish:

@uhhhkhakis wrote:
I'm sorry fam I'm bad at math.
Same xd

Aqual:

I'm looking now

AsleepAndUnafraid21:

ty

Aqual:

Fist you're going to give the perimeter of cgrp

Aqual:

So use \[p = 2(a+b)\]

Aqual:

p + 2(side + base)

Aqual:

p = 2(400 + 600)

AsleepAndUnafraid21:

so 2,000

Aqual:

Yes

Aqual:

And I realized I'm dropping that won't, my bad

Aqual:

Perimeter isn't necessary

AsleepAndUnafraid21:

oh ok

Aqual:

@aqual wrote:
And I realized I'm dropping that won't, my bad
Doing*

Aqual:

Wrong*

AsleepAndUnafraid21:

I belive i'm supposed to be using proportions but idk where to start

Aqual:

Is angle cgr congruent to bpe

AsleepAndUnafraid21:

i think so

AsleepAndUnafraid21:

aren't they alternative interior angles?

Aqual:

so line segment cb is in the same position as line segment, or side, bp

AsleepAndUnafraid21:

yes

Aqual:

so to find the change, you take side cb and divide the length by that of side bp

AsleepAndUnafraid21:

I'm confused. Is the person above me right?

Aqual:

yes

Aqual:

hold on

AsleepAndUnafraid21:

but i dont understand how pbc(a line segmen) is similar to an angle @surjithayer

Aqual:

pbc is a triangle

Aqual:

or an angle

Aqual:

depending on how you write it

AsleepAndUnafraid21:

How? pbc is on the same line

surjithayer:

if we assume GRPC is a parallelogram. Then triangles GCB and PEB are similar. \[\frac{ GB }{ BE }=\frac{ GC}{ PE }=\frac{ CB }{ PB}\] \[\frac{ 450 }{ BE }=\frac{ 400 }{ PE }=\frac{ 350 }{ 250 }\] \[350BE=450 \times 250\] \[BE=\frac{ 450\times250 }{ 350 }=\frac{ 2250 }{ 7 }=321\frac{ 3 }{7 }\] \[350 PE=400\times 250=100,000\] \[PE=\frac{ 100,000 }{ 350}=\frac{ 2000 }{ 7}=285\frac{ 5 }{ 7 }\]

AsleepAndUnafraid21:

pbe is a triangle/angle

Aqual:

@asleepandunafraid21 wrote:
pbe is a triangle/angle
yes

Aqual:

so how i do it is i take the number the triangl;e is dilated by, here its 350/250, and i divide the other side lengths to find those of the triangle im trying to find

surjithayer:

it was a mis print ,now i have corrected. thanks

AsleepAndUnafraid21:

that makes more sense

Aqual:

so 350/250=1.4 so 450/1.4 should give you line be

Aqual:

400/1.4 should give you line pe

AsleepAndUnafraid21:

oh i think i get it

AsleepAndUnafraid21:

thank you @aqual @surjithayer

Aqual:

@surjithayer wrote:
if we assume GRPC is a parallelogram. Then triangles GCB and PEB are similar. \[\frac{ GB }{ BE }=\frac{ GC}{ PE }=\frac{ CB }{ PB}\] \[\frac{ 450 }{ BE }=\frac{ 400 }{ PE }=\frac{ 350 }{ 250 }\] \[350BE=450 \times 250\] \[BE=\frac{ 450\times250 }{ 350 }=\frac{ 2250 }{ 7 }=321\frac{ 3 }{7 }\] \[350 PE=400\times 250=100,000\] \[PE=\frac{ 100,000 }{ 350}=\frac{ 2000 }{ 7}=285\frac{ 5 }{ 7 }\]
these are correct

Aqual:

@asleepandunafraid21 wrote:
thank you @aqual @surjithayer
np

surjithayer:

yw

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!