Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-08-20T18:05:32

$\sqrt{\frac{2473}{13}}$ $sqrt(2473/13)$

Explanation:

Let the distance between the given points be s.
then $s^{2}$ $s^2$ = ${(9 - 3)}^{2} + {(6 - 2)}^{2}$ $(9-3)^2 + (6-2)^2$
$s^{2}$ $s^2$ = 52
hence s = 2 $\sqrt{13}$ $sqrt13$
The perpendicular bisector of s, cuts s $\sqrt{13}$ $sqrt13$ units from (9;6).
Let the altitude of the triangle given be h units.
Area of triangle = $\frac{1}{2}$ $1/2$ $2 \sqrt{13} . h$ $2sqrt13.h$
hence $\sqrt{13}$ $sqrt13$ h = 48
so h = $\frac{48}{\sqrt{13}}$ $48/sqrt13$
Let t be the lengths of the equal sides of the given triangle.
Then by Pythagoras' theorem,
$t^{2}$ $t^2$ = ${(\frac{48}{\sqrt{13}})}^{2}$ $(48/sqrt13)^2$ + ${\sqrt{13}}^{2}$ $sqrt13^2$
= $\frac{2304}{13}$ $2304/13$ + $\frac{169}{13}$ $169/13$
= $\frac{2473}{13}$ $2473/13$
hence t = $\sqrt{\frac{2473}{13}}$ $sqrt(2473/13)$

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Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

Two corners of an isosceles triangle are at (9,6)(9 ,6 ) and (3,2)(3 ,2 ). If the triangle's area is 4848 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(3, 2)$ $(3 ,2 )$ . If the triangle's area is $48$ $48$ , what are the lengths of the triangle's sides?