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Mathematics 8 Online
OpenStudy (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

OpenStudy (mckenzieandjesus):

OpenStudy (mckenzieandjesus):

I think it is C

OpenStudy (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

OpenStudy (campbell_st):

and I don't think its C

OpenStudy (mckenzieandjesus):

i got 547 30/50

OpenStudy (mckenzieandjesus):

547 30/50^2 is 299274.6436

OpenStudy (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

OpenStudy (mckenzieandjesus):

20.9^2-10.5^2= 326 14/25

OpenStudy (mckenzieandjesus):

326 14/25^2 is 106641.4336

OpenStudy (phi):

you don't want to square 326.56 you want to (or should want to) take the *square root*

OpenStudy (mckenzieandjesus):

so 18.0709712

OpenStudy (phi):

and as campbell posted (see up above), that is 1/2 of the length you want.

OpenStudy (mckenzieandjesus):

so do i add 1/2?

OpenStudy (mckenzieandjesus):

or do 18.0709712^2?

OpenStudy (phi):

multiply by 2

OpenStudy (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

OpenStudy (phi):

or add to itself

OpenStudy (mckenzieandjesus):

36.1419424

OpenStudy (phi):

now round to the nearest tenth

OpenStudy (mckenzieandjesus):

so 36.1?

OpenStudy (phi):

yes

OpenStudy (campbell_st):

that's correct... now round it to 1 decimal place

OpenStudy (mckenzieandjesus):

36.1?

OpenStudy (campbell_st):

correct

OpenStudy (mckenzieandjesus):

what message?

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