imehs18
Find the number of sides of each of the two polygons if the total numbers of sides of the polygon is 15, and the sum of the number of diagonals is 36.



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Decart
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the number of sides is 30

imehs18
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i was able to get the number of sides but i don't know how to show the solution

imehs18
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the number of side is 9 and 6
i just need a solution how to get this

Decart
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then the numbers of sides of the polygon should be polygons

kc_kennylau
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dw:1384944170133:dw


kc_kennylau
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So the number of diagonals of a polygon with \(n\) sides is \(\dfrac{n(n3)}2\)

kc_kennylau
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Why don't you set up equations :)

imehs18
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number of diagonals i already given. i need a solution.

imehs18
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can you help me?

kc_kennylau
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You could try letting the numbers of sides of the two polygons be \(m\) and \(n\) respectively :)

imehs18
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can you pls show me the solution. i got this m(m3)/2 + n(n3)/2 = 36

kc_kennylau
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Ok I'll set up the equations for you but you'll work out the steps yourself :)
\[m+n=15\]\[\frac{m(m+3)}2+\frac{n(n+3)}2=36\]

kc_kennylau
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ok here's a hint: \(m=15n\)

imehs18
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m^23m+n^23n = 72
m^2+n^23(m+n) = 72
we know that m+n = 15
m^2+n^23(15) = 72
m^2+n^2 = 27
after that i dont know

Decart
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substitute 15n for m

imehs18
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huh?

kc_kennylau
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substitute \(m=15n\) to \(\dfrac{m(m3)}2+\dfrac{n(n3)}2=36\) :)

kc_kennylau
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into*

Decart
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\[\left( 15n ^{} \right)^{2}3\left( 15n \right)+n ^{2}3n=72\]

kc_kennylau
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yep, and this will be a quadratic :)

imehs18
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im stuck

imehs18
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i got 108 = 27n+n^2

kc_kennylau
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Have you learnt quadratic? :)

kc_kennylau
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Use quadratics pl0x

imehs18
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can you just pls provide me the whole solution
im stuck. sorry if im stupid about this..

imehs18
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i stop studying for 5years and i need to recall everything about geometry

Decart
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2n^230n+108=0

Decart
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(_n+_)(_n_) factor

Decart
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you did it right you just got 27 instead of 30

imehs18
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how did you get the 30?

imehs18
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got it

Decart
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from the first equation
22530n+n^245+3n+n^23n=72

Decart
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combine like terms

imehs18
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after that. what is next?

Decart
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(15n)^2 gave me the thirty

imehs18
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how do i get the n now?

Decart
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then you factor the 2n^230n+108
you can divide both sides by two to make it easier
n^215n+54

Decart
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two factors of 54 that total to 15

Decart
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(x_)*(x_)=0

imehs18
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(n9)(n6)

Decart
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yes
so those are your sides

imehs18
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thank you so much for your patience with me. i really appreciate it.