Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(9, 4)$ $(9 ,4 )$ . 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 $(1, 3)$ $(1 ,3 )$ and $(9, 4)$ $(9 ,4 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-09-16T08:27:29

The lengths of the triangle's sides are:

$\sqrt{65}, \sqrt{\frac{266369}{260}}, \sqrt{\frac{266369}{260}}$ $sqrt(65), sqrt(266369/260), sqrt(266369/260)$

Explanation:

The distance between two points $(x_{1}, y_{1})$ $(x_1, y_1)$ and $(x_{2}, y_{2})$ $(x_2, y_2)$ is given by the distance formula:

$d = \sqrt{{(x_{2} - x_{1})}_{2}^{2} + {(y_{2} - y_{1})}_{2}^{2}}$ $d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)$

So the distance between $(x_{1}, y_{1}) = (1, 3)$ $(x_1, y_1) = (1, 3)$ and $(x_{2}, y_{2}) = (9, 4)$ $(x_2, y_2) = (9, 4)$ is:

$\sqrt{{(9 - 1)}^{2} + {(4 - 3)}^{2}} = \sqrt{64 + 1} = \sqrt{65}$ $sqrt((9-1)^2+(4-3)^2) = sqrt(64+1) = sqrt(65)$

which is an irrational number a little larger than $8$ $8$ .

If one of the other sides of the triangle was the same length, then the maximum possible area of the triangle would be:

$\frac{1}{2} \cdot {\sqrt{65}}^{2} = \frac{65}{2} < 64$ $1/2*sqrt(65)^2 = 65/2 < 64$

So that cannot be the case. Instead, the other two sides must be the same length.

Given a triangle with sides $a = \sqrt{65}, b = t, c = t$ $a=sqrt(65), b=t, c=t$ , we can use Heron's formula to find its area.

Herons formula tells us that the area of a triangle with sides $a, b, c$ $a, b, c$ and semi perimeter $s = \frac{1}{2} (a + b + c)$ $s = 1/2(a+b+c)$ is given by:

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

In our case the semi perimeter is:

$s = \frac{1}{2} (\sqrt{65} + t + t) = t + \frac{\sqrt{65}}{2}$ $s = 1/2(sqrt(65)+t+t) = t+sqrt(65)/2$

and Heron's formula tells us that:

$64 = \frac{1}{2} \sqrt{(t + \frac{\sqrt{65}}{2}) (t - \frac{\sqrt{65}}{2}) (\frac{\sqrt{65}}{2}) (\frac{\sqrt{65}}{2})}$ $64 = 1/2sqrt((t+sqrt(65)/2)(t-sqrt(65)/2)(sqrt(65)/2)(sqrt(65)/2))$

$64 = \frac{1}{2} \sqrt{\frac{65}{4} (t^{2} - \frac{65}{4})}$ $color(white)(64) = 1/2sqrt(65/4(t^2-65/4))$

Multiply both ends by $2$ $2$ to get:

$128 = \sqrt{\frac{65}{4} (t^{2} - \frac{65}{4})}$ $128 = sqrt(65/4(t^2-65/4))$

Square both sides to get:

$16384 = \frac{65}{4} (t^{2} - \frac{65}{4})$ $16384 = 65/4(t^2-65/4)$

Multiply both sides by $\frac{4}{65}$ $4/65$ to get:

$\frac{65536}{65} = t^{2} - \frac{65}{4}$ $65536/65 = t^2-65/4$

Transpose and add $\frac{65}{4}$ $65/4$ to both sides to get:

$t^{2} = \frac{65536}{65} + \frac{65}{4} = \frac{262144}{260} + \frac{4225}{260} = \frac{266369}{260}$ $t^2 = 65536/65+65/4 = 262144/260+4225/260 = 266369/260$

Take the positive square root of both sides to get:

$t = \sqrt{\frac{266369}{260}}$ $t = sqrt(266369/260)$

So the lengths of the triangle's sides are:

$\sqrt{65}, \sqrt{\frac{266369}{260}}, \sqrt{\frac{266369}{260}}$ $sqrt(65), sqrt(266369/260), sqrt(266369/260)$

Alternative method

Instead of using Heron's formula, we can reason as follows:

Given that the base of the isosceles triangle is of length:

$\sqrt{8^{2} + 1^{2}} = \sqrt{65}$ $sqrt(8^2+1^2) = sqrt(65)$

The area is $64 = \frac{1}{2} base \times height$ $64 = 1/2 "base" xx "height"$

So the height of the triangle is:

$\frac{64}{\frac{1}{2} \sqrt{65}} = \frac{128}{\sqrt{65}} = \frac{128 \sqrt{65}}{65}$ $64/(1/2 sqrt(65)) = 128/sqrt(65) = (128sqrt(65))/65$

This is the length of the perpendicular bisector of the triangle, which passes through the midpoint of the base.

So the other two sides form the hypotenuses of two right angled triangles with legs $\frac{\sqrt{65}}{2}$ $sqrt(65)/2$ and $\frac{128 \sqrt{65}}{65}$ $(128sqrt(65))/65$

So by Pythagoras, each of those sides is of length:

$\sqrt{{(\frac{\sqrt{65}}{2})}^{2} + {(\frac{128 \sqrt{65}}{65})}^{2}} = \sqrt{\frac{65}{4} + \frac{65536}{65}} = \sqrt{\frac{266369}{260}}$ $sqrt((sqrt(65)/2)^2+((128sqrt(65))/65)^2) = sqrt(65/4+65536/65) = sqrt(266369/260)$

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Two corners of an isosceles triangle are at $(1, 3)$ $(1 ,3 )$ and $(9, 4)$ $(9 ,4 )$ . 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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Two corners of an isosceles triangle are at (1,3)(1 ,3 ) and (9,4)(9 ,4 ). If the triangle's area is 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 $(1, 3)$ $(1 ,3 )$ and $(9, 4)$ $(9 ,4 )$ . If the triangle's area is $64$ $64$ , what are the lengths of the triangle's sides?