Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(7, 5)$ $(7 ,5 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(7, 5)$ $(7 ,5 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-07-11T12:15:00

$5.099$ $5.099$ , $25.6155$ $25.6155$ & $25.6155$ $25.6155$

Explanation:

The length of side of isosceles triangle joining the vertices $(2, 6)$ $(2, 6)$ & $(7, 5)$ $(7, 5)$
$= \sqrt{{(2 - 7)}^{2} + {(6 - 5)}^{2}}$ $=\sqrt{(2-7)^2+(6-5)^2}$

$= \sqrt{26}$ $=\sqrt{26}$

If $h$ $h$ is the length of altitude dropped from third vertex to the side/base joining the vertices $(2, 6)$ $(2, 6)$ & $(7, 5)$ $(7, 5)$ then the area of isosceles triangle

$\frac{1}{2} (base) \times (altitude)$ $\frac{1}{2}(\text{base})\times \text{(altitude)}$

$\frac{1}{2} \sqrt{26} h = 64$ $1/2 \sqrt{26}h=64$

$h = \frac{128}{\sqrt{26}}$ $h=128/\sqrt{26}$

Now, using the Pythagorean theorem in a right triangle with legs $\sqrt{26}$ $\sqrt26$ & $\frac{128}{\sqrt{26}}$ $128\/sqrt{26}$ , length of each of equal sides of isosceles triangle

$\sqrt{{(\sqrt{26})}^{2} + {(\frac{128}{\sqrt{26}})}^{2}}$ $\sqrt{(\sqrt26)^2+(128/\sqrt{26})^2}$

$= 25.6155$ $=25.6155$

hence, the sides of given isosceles triangle are $\sqrt{26} = 5.099$ $\sqrt{26}=5.099$ , $25.6155$ $25.6155$ & $25.6155$ $25.6155$

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(7, 5)$ $(7 ,5 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Humanities

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Two corners of an isosceles triangle are at (2 ,6 )(2,6)(2 ,6 ) and (7 ,5 )(7,5)(7 ,5 ). If the triangle's area is 64 6464 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(2, 6)$ $(2 ,6 )$ and $(7, 5)$ $(7 ,5 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?