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alexira [117]
3 years ago
14

The midsegment through the legs of an isosceles triangle is 4 in. Find the sides of the triangle if its perimeter is 20 in. 24 p

oints

Mathematics
1 answer:
konstantin123 [22]3 years ago
6 0

an isosceles triangle has two twin sides, stemming from a vertex.

now, the midsegment of any triangle, is half the length of the parallel base, or namely the base itself is twice as long as the midsegment, check the picture below.

so, we know the midsegment is 4, therefore the parallel base of it will be 8.

we know its perimeter is 20, and we also know the twin sides are, well twins, let's say they're "a" units long.


\bf \stackrel{base}{(8)}+\stackrel{twin~sides}{(a+a)}=20\implies 8+2a=20\implies 2a=12 \\\\\\ a=\cfrac{12}{2}\implies \boxed{a=6}

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(11/x^2-25)+(4/x+5)= 3/x-5
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\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
-------------------------------\\\\
\cfrac{11}{x^2-25}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\implies \cfrac{11}{x^2-5^2}+\cfrac{4}{x+5}=\cfrac{3}{x-5}

\bf \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5}=\cfrac{3}{x-5}\impliedby 
\begin{array}{llll}
\textit{notice, LCD is }(x-5)(x+5)\\
\textit{so let's multiply all by it}\\
\textit{to toss the denominators}
\end{array}

\bf (x-5)(x+5)\left( \cfrac{11}{(x-5)(x+5)}+\cfrac{4}{x+5} \right)=(x-5)(x+5)\left( \cfrac{3}{x-5} \right)
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