Consider four triangles with side lengths (5, | vedclass

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(C) For a triangle with sides $a, b, c$,the area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.For all tr

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$(5, 12, 13)$

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(C) For a triangle with sides $a, b, c$,the area $A$ is given by Heron's formula: $A = \sqrt{s(s-a)(s-b)(s-c)}$,where $s = \frac{a+b+c}{2}$.For all triangles,two sides are fixed at $5$ and $12$. Let the third side be $x$. Then $s = \frac{5+12+x}{2} = \frac{17+x}{2}$.The area $A(x) = \sqrt{\frac{17+x}{2} \cdot \frac{17-x}{2} \cdot \frac{x+7}{2} \cdot \frac{x-7}{2}} = \frac{1}{4} \sqrt{(17^2 - x^2)(x^2 - 7^2)}$.Let $u = x^2$. The function $f(u) = (289 - u)(u - 49) = -u^2 + 338u - 14161$. This is a downward parabola with a maximum at $u = \frac{-338}{2(-1)} = 169$.Thus,$x^2 = 169$,which means $x = 13$.For $x = 13$,the sides are $(5, 12, 13)$,which satisfy $5^2 + 12^2 = 13^2$,forming a right-angled triangle with area $\frac{1}{2} 	imes 5 	imes 12 = 30$.Comparing areas for $x \in \{9, 11, 13, 15\}$,the maximum area occurs at $x = 13$.

$i$. $P$	$A$. $(0,0)$
$ii$. $Q$	$B$. $(6,0)$
$iii$. $R$	$C$. $(-2,1)$
$iv$. $S$	$D$. $(-6,0)$
	$E$. $(-6,-3)$
	$F$. $(-6,3)$

Consider four triangles with side lengths $(5, 12, 9)$,$(5, 12, 11)$,$(5, 12, 13)$,and $(5, 12, 15)$. Among these,the triangle with the maximum area has sides:

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