Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(7, 2)$ $(7 ,2 )$ . 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 $(9, 6)$ $(9 ,6 )$ and $(7, 2)$ $(7 ,2 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-09-21T02:11:38

$sides a = c = 28.7 units$ $"sides "a = c = 28.7" units"$ and $side b = 2 \sqrt{5} units$ $"side " b = 2sqrt5" units"$

Explanation:

let $b =$ $b =$ the distance between the two points:

$b = \sqrt{{(9 - 7)}^{2} + {(6 - 2)}^{2}}$ $b = sqrt((9-7)^2+(6-2)^2)$

$b = 2 \sqrt{5} units$ $b = 2sqrt5" units"$

We are given that the $Area = 64 {units}^{2}$ $"Area" = 64" units"^2$

Let "a" and "c" be the other two sides.

For a triangle, $Area = \frac{1}{2} b h$ $"Area " = 1/2bh$

Substituting in the values for "b" and the Area:

$64 {units}^{2} = \frac{1}{2} (2 \sqrt{5} units) h$ $64" units"^2 = 1/2(2sqrt5" units")h$

Solve for the height:

$h = \frac{64}{\sqrt{5}} = \frac{64}{5} \sqrt{5} units$ $h = 64/sqrt5 = 64/5sqrt5" units"$

Let $C =$ $C =$ the angle between side "a" and side "b", then we may use the right triangle formed by side "b" and the height to write the following equation:

$tan (C) = \frac{h}{\frac{1}{2} b}$ $tan(C) = h/(1/2b)$

$tan (C) = \frac{\frac{64}{5} \sqrt{5} units}{\frac{1}{2} (2 \sqrt{5} units)}$ $tan(C) = (64/5sqrt5" units")/(1/2(2sqrt5" units"))$

$C = {tan}^{- 1} (\frac{64}{5})$ $C = tan^-1(64/5)$

We can find the length of side "a", using the following equation:

$h = (a) sin (C)$ $h = (a)sin(C)$

$a = \frac{h}{sin (C)}$ $a = h/sin(C)$

Substitute in the values for "h" and "C":

$a = \frac{\frac{64}{5} \sqrt{5} units}{sin ({tan}^{- 1} (\frac{64}{5}))}$ $a = (64/5sqrt5" units")/sin(tan^-1(64/5))$

$a = 28.7 units$ $a = 28.7" units"$

Intuition tells me that side "c" is the same length as side "a" but we can prove this using the Law of Cosines:

$c^{2} = a^{2} + b^{2} - 2 (a) (b) cos (C)$ $c^2 = a^2 + b^2 - 2(a)(b)cos(C)$

Substitute in the values for a, b, and C:

$c^{2} = {(28.7 units)}^{2} + {(2 \sqrt{5} units)}^{2} - 2 (28.7 units) (2 \sqrt{5} units) cos ({tan}^{- 1} (\frac{64}{5}))$ $c^2 = (28.7" units")^2 + (2sqrt5" units")^2 - 2(28.7" units")(2sqrt5" units")cos(tan^-1(64/5))$

$c = 28.7 units$ $c = 28.7" units"$

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Two corners of an isosceles triangle are at $(9, 6)$ $(9 ,6 )$ and $(7, 2)$ $(7 ,2 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Science

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Humanities

... and beyond

Two corners of an isosceles triangle are at (9 ,6 )(9,6)(9 ,6 ) and (7 ,2 )(7,2)(7 ,2 ). 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 $(9, 6)$ $(9 ,6 )$ and $(7, 2)$ $(7 ,2 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?