mckenzieandjesus
  • mckenzieandjesus
Point P is the center of two concentric circles. PQ = 10.5 and PS = 20.9. RS is tangent to the smaller circle and a chord of the larger circle. What is length of RS to the nearest tenth? 36.1 41.4 31.4 41.8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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mckenzieandjesus
  • mckenzieandjesus
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mckenzieandjesus
  • mckenzieandjesus
I think it is C
campbell_st
  • campbell_st
well the tangent RS and the radius PQ are perpendicular to each other, so you need pythagoras \[PS^2 = QS^2 + PQ^2\] then you need to know the perpendicular from the centre to the chord bisects the chord so RQ = QS or RS = 2QS hope it helps

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campbell_st
  • campbell_st
and I don't think its C
mckenzieandjesus
  • mckenzieandjesus
i got 547 30/50
mckenzieandjesus
  • mckenzieandjesus
547 30/50^2 is 299274.6436
campbell_st
  • campbell_st
well using pythagoras' theorem you need to find QS manipulating the equation \[QS^2 = PS^2 - PQ^2\] so \[QS^2 = 20.9^2 - 10.5^2\] so start by finding QS then double it to get RS
mckenzieandjesus
  • mckenzieandjesus
20.9^2-10.5^2= 326 14/25
mckenzieandjesus
  • mckenzieandjesus
326 14/25^2 is 106641.4336
phi
  • phi
you don't want to square 326.56 you want to (or should want to) take the *square root*
mckenzieandjesus
  • mckenzieandjesus
so 18.0709712
phi
  • phi
and as campbell posted (see up above), that is 1/2 of the length you want.
mckenzieandjesus
  • mckenzieandjesus
so do i add 1/2?
mckenzieandjesus
  • mckenzieandjesus
or do 18.0709712^2?
phi
  • phi
multiply by 2
campbell_st
  • campbell_st
so that is QS now the perpendicular line from the centre, PQ, bisects RS. so RS = 2 x QS so you know QS, just double it... to find RS
phi
  • phi
or add to itself
mckenzieandjesus
  • mckenzieandjesus
36.1419424
phi
  • phi
now round to the nearest tenth
mckenzieandjesus
  • mckenzieandjesus
so 36.1?
phi
  • phi
yes
campbell_st
  • campbell_st
that's correct... now round it to 1 decimal place
mckenzieandjesus
  • mckenzieandjesus
36.1?
campbell_st
  • campbell_st
correct
mckenzieandjesus
  • mckenzieandjesus
what message?

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