Two corners of an isosceles triangle are at $(5, 4)$ $(5 ,4 )$ and $(9, 2)$ $(9 ,2 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(5, 4)$ $(5 ,4 )$ and $(9, 2)$ $(9 ,2 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-03-11T21:23:02

Tony B The length of the sides are both: $s \approx 16.254$ $s~~16.254$ to 3 dp

Explanation:

It usually helps to draw a diagram:

$Method$ $color(blue)("Method")$
Find base width $w$ $w$
Use in conjunction with area to find $h$ $h$
Using $h$ $h$ and $\frac{w}{2}$ $w/2$ in Pythagoras find $s$ $s$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$To determine the value of w$ $color(blue)("To determine the value of "w)$

Consider the green line in the diagram (base as would be plotted)

Using Pythagoras:
$w = \sqrt{{(9 - 5)}^{2} + {(2 - 4)}^{2}}$ $w=sqrt((9-5)^2+(2-4)^2)$

$w = \sqrt{4^{2} + {(- 2)}^{2}} = \sqrt{20} = 2 \sqrt{5}$ $color(blue)(w=sqrt(4^2+(-2)^2) =sqrt(20) =2sqrt(5))$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
$To determine the value of h$ $color(blue)("To determine the value of "h)$

$A r e a = \frac{w}{2} \times h$ $"Area = w/2xxh$

$36 = \frac{2 \sqrt{5}}{2} \times h$ $36=(2sqrt(5))/2xxh$

$36 = \frac{2}{2} \times \sqrt{5} \times h$ $36=2/2xxsqrt(5)xxh$

$h = \frac{36}{\sqrt{5}}$ $color(blue)(h=36/sqrt(5))$
'~~~~~~~~~~~~~~~~~~~~~~~~~
$To determine the value of s$ $color(blue)("To determine the value of "s)$

Using Pythagoras

${(\frac{w}{2})}^{2} + h^{2} = s^{2}$ $(w/2)^2+h^2=s^2$

$\Rightarrow s = \sqrt{{(\frac{2 \sqrt{5}}{2})}^{2} + {(\frac{36}{\sqrt{5}})}^{2}}$ $=>s=sqrt( ( (2sqrt(5))/2)^2 +(36/sqrt(5))^2$

$\Rightarrow s = \sqrt{(5 + \frac{36^{2}}{5})}$ $=>s=sqrt( ( 5 +(36^2)/5)$

$s = \sqrt{\frac{25 + 36^{2}}{5}} = \sqrt{\frac{1321}{5}}$ $s=sqrt( (25+36^2)/5) = sqrt(1321/5)$

$s \approx 16.254$ $s~~16.254$

Answer 2 · 2016-03-19T11:35:35

Support of the decision that the given points are for the base of the triangle.

Explanation:

Suppose the coordinates give were not for the base of an Isosceles triangle but for one of the other two sides. Then we would have:
Tony B

Where

$x = 2 \sqrt{5} \times sin (θ)$ $x=2sqrt(5)xxsin(theta)$
$h = 2 \sqrt{5} \times cos (θ)$ $h=2sqrt(5)xxcos(theta)$

Given that Area $= 36 = x \times h$ $= 36 = x xx h$

Thus we have:

$36 = {(2 \sqrt{5} .)}^{2} (sin (θ) cos (θ))$ $" "color(blue)(36=(2sqrt(5)color(white)(.))^2( sin(theta)cos(theta)))$

$Using the Trig Identity of sin (2 θ) = 2 sin (θ) cos (θ)$ $color(brown)("Using the Trig Identity of "sin(2theta)=2sin(theta)cos(theta))$

$36 = 20 (sin (θ) cos (θ)) \to 36 = \frac{20}{2} sin (2 θ)$ $" "color(brown)(36=20( sin(theta)cos(theta)) ->)color(blue)(36=20/2sin(2theta))$

$\Rightarrow sin (2 θ) = \frac{72}{20}$ $=> sin(2theta)=72/20$

But $- 1 \leq sin (2 θ) \leq + 1$ $" "-1<=sin(2theta)<=+1$

and $\frac{72}{20} > + 1$ $72/20>+1$ so there is a $" "color(red)(underline("contradiction"))$

Implying that this scenario of $2sqrt(5)$ not being the base is false.

$color(magenta)("The length of "2sqrt(5)" applies to the base of the triangle")$

Tony B

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Two corners of an isosceles triangle are at $(5, 4)$ $(5 ,4 )$ and $(9, 2)$ $(9 ,2 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

2 Answers

Explanation:

Explanation:

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Two corners of an isosceles triangle are at (5 ,4 )(5,4)(5 ,4 ) and (9 ,2 )(9,2)(9 ,2 ). If the triangle's area is 36 3636 , what are the lengths of the triangle's sides?

2 Answers

Explanation:

Explanation:

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Two corners of an isosceles triangle are at $(5, 4)$ $(5 ,4 )$ and $(9, 2)$ $(9 ,2 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?