anonymous
  • anonymous
Geometry help please!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
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anonymous
  • anonymous
please help me
AccessDenied
  • AccessDenied
should we assume that 'x' is the length of the altitude to the hypotenuse? If so, we only need to take the geometric mean of the hypotenuse segments x 9 - = -; for x 5 x

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anonymous
  • anonymous
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anonymous
  • anonymous
I don't get it. x would =1.3
AccessDenied
  • AccessDenied
in the triangle, we have a 45-45-90 triangle. do you know how the legs of that triangle compare to the hypotenuse?
anonymous
  • anonymous
what would be the answers?
AccessDenied
  • AccessDenied
im not here to give you the answer, just help you get there. :(
anonymous
  • anonymous
what you are saying doesnt make sense
AccessDenied
  • AccessDenied
|dw:1333486260813:dw| If this is what we know in the triangle, the last angle is 45, right? So, since the two angles are congruent, the legs are also congruent (isosceles triangle th). Pythagorean theorem to find the hypotenuse would be: a^2 + a^2 = h^2 2a^2 = h^2 sqrt(2a^2) = sqrt(h^2) a*sqrt(2) = h this is true for all 45-45-90 triangles with a side length "a" and hypotenuse "h"
anonymous
  • anonymous
5(5)+5(5)=h(h)
AccessDenied
  • AccessDenied
yes, 5*5 + 5*5 = h^2 (although h is y in the context of our problem)
anonymous
  • anonymous
ok next it would be 2(5)*5=h*h
anonymous
  • anonymous
it is 50
AccessDenied
  • AccessDenied
yeah, you could do that 50 = h*h , then you take the positive square root of both sides (h*h = h^2)
anonymous
  • anonymous
the answer is b?
AccessDenied
  • AccessDenied
Yes. :)
anonymous
  • anonymous
can you help me on #1?
AccessDenied
  • AccessDenied
the first problem you posted? well, when we have an altitude that divides the opposite hypotenuse, it divides them in a way that we can set up certain ratios between the triangles. |dw:1333486731312:dw| Basically, we set it up as \[ \frac{5}{x} = \frac{x}{9} \] In words, we take the left hypotenuse segment (5) over the altitude (x) and set it equal to the altitude (x) over the right hypotenuse segment (9). that's not the only way to write it necessarily, but I'll go with this way. they all work. then solve for x
anonymous
  • anonymous
x=22.5?
AccessDenied
  • AccessDenied
how do you get that?
anonymous
  • anonymous
5x=9x x=1.8?
AccessDenied
  • AccessDenied
to solve it, we cross-multiply, so we take the denominator of one side to the numerator of the other side |dw:1333487031640:dw|
anonymous
  • anonymous
2x=45
anonymous
  • anonymous
x*x=45
anonymous
  • anonymous
the answer is a?
AccessDenied
  • AccessDenied
Yes. :)
anonymous
  • anonymous
I have 2 more problems to ask
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anonymous
  • anonymous
thank you so much
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AccessDenied
  • AccessDenied
for the first of the two, we can just look at b and 120 as supplementary b + 120 = 180 the second one is exterior angle theorem, where the sum of nonadjacent interior angles are congruent to the opposite exterior angle
anonymous
  • anonymous
1=60?
anonymous
  • anonymous
2) 65+x=3x-5?
anonymous
  • anonymous
2)35?
AccessDenied
  • AccessDenied
Yes, #1 is 60 and #2 sounds correct as well
anonymous
  • anonymous
how do I do 3?
AccessDenied
  • AccessDenied
oh, there are two questions on the second thing! sec i gotta reorganize now. :P
AccessDenied
  • AccessDenied
for the first image, the b = 60 is correct for the second image's first question, we'd have to find x and substitute it back into 3x - 5 to find the angle measure of the exterior angle (which conveniently answers the last question). Then since the exterior angle and b are supplementary, we add the b to the value and set it equal to 180 / solve for b.
AccessDenied
  • AccessDenied
the value of x was correctly 35, so we evaluate 3(35) - 5 for the exterior angle measure
anonymous
  • anonymous
how do I set up it up? I am kind of confused
anonymous
  • anonymous
answer = 100
AccessDenied
  • AccessDenied
yeah, that answers the very last question then, we do the same thing as your first question with b and 120 they're supplementary, so we add them together and set = 180 to solve for b
anonymous
  • anonymous
do you me couple more. Please
AccessDenied
  • AccessDenied
b + 100 = 180 b = 80
anonymous
  • anonymous
do you mind helping me with a couple more?
AccessDenied
  • AccessDenied
i dont mind. :)
anonymous
  • anonymous
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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
5 problems total
anonymous
  • anonymous
thanks
AccessDenied
  • AccessDenied
For the first one, we're just looking at what we know. we know two angles that are congruent, and we also have this bit: |dw:1333488368342:dw| But, since that segment between the two congruent segments is shared by both triangles, it is also congruent to itself, so we can see that those sides are congruent.
AccessDenied
  • AccessDenied
#2 we just use corresponding angles postulate / other angle theorems to trace the relation between the known angle 30* to angle 5 i.e. m<1 = 30 because of vertical angles can we show the relation between m<1 and m<5?
AccessDenied
  • AccessDenied
for #3, we can use any of the three trig functions (sine, cosine, tangent) to find it since we know all three sides just have to keep it straight that sin t = opp/hyp, cos t = adj/hyp, and tan t = opp/adj evaluate, round
AccessDenied
  • AccessDenied
well, not evaluate necessarily, solve for the angle by using inverse functions
AccessDenied
  • AccessDenied
#4: BD is the perpendicular bisector of the base, so its an isosceles triangle. we can use that information to find an equation with x. #5: just take top of first triangle over top of second triangle intuitively, we can see that by going from 10 to 5, we're halving the number, so the scale factor would be 1/2 10(1/2) = 5

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