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tatiyna
3 years ago
14

In triangle ABC, P is a centroid on a median AD. If AD=33 and AP>PD find AP

Mathematics
1 answer:
jasenka [17]3 years ago
7 0

Answer:

AP = 22

Step-by-step explanation:

In a triangle, the centroid divides the median in the ratio 2:1.

It is given that AD is the median and AD = 33

It is also given that P is the centroid on the median AD.

Therefore, P divides AD in the ratio 2:1.

\frac{AP}{PD} =\frac{2}{1}

AP = 2k and PD = k

AD = 33

AP + PD = 33

2k + k = 33

3k = 33

k = 11

So, AP = 2k = 2(11) = 22

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According to the given question, we have –

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Let's solve it!

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\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

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