Two corners of an isosceles triangle are at $(7, 5)$ $(7 ,5 )$ and $(3, 6)$ $(3 ,6 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(7, 5)$ $(7 ,5 )$ and $(3, 6)$ $(3 ,6 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-11-18T20:17:06

There's a couple ways to do it; the way with the fewest steps is explained below.

The question is ambiguous about which two sides are the same length. In this explanation, we will assume the two sides of equal length are the ones yet to be found.

Explanation:

One side length we can figure out just from the coordinates we've been given.

$a = \sqrt{{(7 - 3)}^{2} + {(5 - 6)}^{2}}$ $a=sqrt((7-3)^2+(5-6)^2)$
$a = \sqrt{4^{2} + {(- 1)}^{2}}$ $a=sqrt(4^2+(-1)^2)$
$a = \sqrt{16 + 1}$ $a=sqrt(16+1)$
$a = \sqrt{17}$ $a=sqrt17$

Then we can use the formula for area of a triangle in terms of its side lengths to figure out $b$ $b$ and $c$ $c$ .

$A = \sqrt{s (s - a) (s - b) (s - c)}$ $A=sqrt(s(s-a)(s-b)(s-c))$

where $s = \frac{a + b + c}{2}$ $s=(a+b+c)/2$ (called the semiperimeter)

Since $a = \sqrt{17}$ $a=sqrt(17)$ is known, and we assume $b = c$ $b=c$ , we have

$s = \frac{\sqrt{17} + b + b}{2}$ $s=(sqrt17+b+b)/2$
$s = \frac{\sqrt{17}}{2} + b$ $color(red)(s=sqrt17/2+b)$

Substituting this into the area formula above, as well as $A = 6$ $A=6$ and $a = \sqrt{17}$ $a=sqrt17$ , we get

$6 = \sqrt{(\frac{\sqrt{17}}{2} + b) (\frac{\sqrt{17}}{2} + b - \sqrt{17}) (\frac{\sqrt{17}}{2} + b - b) (\frac{\sqrt{17}}{2} + b - b)}$ $6=sqrt((color(red)(sqrt(17)/2+b))(color(red)(sqrt(17)/2+b)-sqrt17)(color(red)(sqrt(17)/2+b)-b)(color(red)(sqrt(17)/2+b)-b))$
$6 = \sqrt{(\frac{\sqrt{17}}{2} + b) (- \frac{\sqrt{17}}{2} + b) (\frac{\sqrt{17}}{2}) (\frac{\sqrt{17}}{2})}$ $6=sqrt((sqrt(17)/2+b)(-sqrt(17)/2+b)(sqrt(17)/2)(sqrt(17)/2))$

$6 = (\frac{\sqrt{17}}{2}) \sqrt{(b + \frac{\sqrt{17}}{2}) (b - \frac{\sqrt{17}}{2})}$ $6=(sqrt(17)/2)sqrt((b+sqrt(17)/2)(b-sqrt(17)/2))$

$\frac{12}{\sqrt{17}} = \sqrt{b^{2} - {(\frac{\sqrt{17}}{2})}^{2}}$ $12/sqrt17=sqrt(b^2-(sqrt17/2)^2)$

$\frac{144}{17} = b^{2} - \frac{17}{4}$ $144/17=b^2-17/4$

$\frac{144}{17} + \frac{17}{4} = b^{2}$ $144/17+17/4=b^2$

$\frac{576}{68} + \frac{289}{68} = b^{2}$ $576/68+289/68=b^2$

$\frac{865}{68} = b^{2}$ $865/68=b^2$

$b = \sqrt{\frac{865}{68}} = c$ $b=sqrt(865/68)=c$

Our solution is $a = \sqrt{17}, b = c = \sqrt{\frac{865}{68}}$ $a=sqrt(17), b=c=sqrt(865/68)$ .

Footnote 1:

It is possible to have a triangle with two sides of length $\sqrt{17}$ $sqrt(17)$ and area $A = 6$ $A=6$ (that is, to have $a = b = \sqrt{17}$ $a=b=sqrt(17)$ instead of $b = c$ $b=c$ ). This will lead to a different solution.

Footnote 2:

We could also have solved this question by finding the coordinates of the 3rd point. This would have involved:
a) finding the length of the known side $a$ $a$
b) finding the slope $m$ $m$ between the two given points
c) finding the midpoint $(x_{1}, y_{1})$ $(x_1,y_1)$ between the two given points
d) finding the "height" $h$ $h$ of this triangle using $A = \frac{1}{2} a h$ $A=1/2 ah$
e) finding the slope of the height using $m_{h} = \frac{- 1}{m}$ $m_h=(-1)/m$
f) using both the slope-point formula $m_{h} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $m_h=(y_2-y_1)/(x_2-x_1)$ and the height formula $h = \sqrt{{(y_{2} - y_{1})}_{2}^{2} + {(x_{2} - x_{1})}_{2}^{2}}$ $h=sqrt((y_2-y_1)^2+(x_2-x_1)^2)$ to solve for one of the coordinates of the 3rd point $(x_{2}, y_{2})$ $(x_2,y_2)$
g) after combining these two equations, simplifying yields
$x_{2} = \frac{h}{\sqrt{m_{h}^{2} + 1}} + x_{1}$ $x_2=h/(sqrt(m_h^2+1))+x_1$
h) plugging in the known values for $h$ $h$ , $m_{h}$ $m_h$ , and $x_{1}$ $x_1$ to get $x_{2}$ $x_2$
i) using one of the two equations in (f) to find $y_{2}$ $y_2$
j) using the distance formula to find the remaining (identical) side lengths
$b = c = \sqrt{{(x_{2} - 3)}_{2}^{2} + {(y_{2} - 6)}_{2}^{2}} = \sqrt{{(x_{2} - 7)}_{2}^{2} + {(y_{2} - 5)}_{2}^{2}}$ $b=c=sqrt((x_2-3)^2+(y_2-6)^2)=sqrt((x_2-7)^2+(y_2-5)^2)$

You can see why the first method is easier.

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Two corners of an isosceles triangle are at $(7, 5)$ $(7 ,5 )$ and $(3, 6)$ $(3 ,6 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

Footnote 1:

Footnote 2:

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Two corners of an isosceles triangle are at (7,5)(7 ,5 ) and (3,6)(3 ,6 ). If the triangle's area is 66 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

Footnote 1:

Footnote 2:

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Impact of this question

Two corners of an isosceles triangle are at $(7, 5)$ $(7 ,5 )$ and $(3, 6)$ $(3 ,6 )$ . If the triangle's area is $6$ $6$ , what are the lengths of the triangle's sides?