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

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