Two corners of an isosceles triangle are at $(1, 6)$ $(1 ,6 )$ and $(2, 9)$ $(2 ,9 )$ . 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 $(1, 6)$ $(1 ,6 )$ and $(2, 9)$ $(2 ,9 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-02-09T19:53:34

$\sqrt{10}, \sqrt{520.9}, \sqrt{520.9} ≅ 3.162, 22.823, 22.823$ $sqrt(10),sqrt(520.9), sqrt(520.9)~=3.162,22.823,22.823$

Explanation:

The length of the given side is
$s = \sqrt{{(2 - 1)}^{2} + {(9 - 6)}^{2}} = \sqrt{1 + 9} = \sqrt{10} ≅ 3.162$ $s=sqrt((2-1)^2+(9-6)^2)=sqrt(1+9)=sqrt(10)~=3.162$

From the formula of the triangle's area:
$S = \frac{b \cdot h}{2}$ $S=(b*h)/2$ => $36 = \frac{\sqrt{10} \cdot h}{2}$ $36=(sqrt(10)*h)/2$ => $h = \frac{72}{\sqrt{10}} ≅ 22.768$ $h=72/sqrt(10)~=22.768$

Since the figure is an isosceles triangle we could have Case 1 , where the base is the singular side, ilustrated by Fig. (a) below

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Or we could have Case 2 , where the base is one of the equal sides, ilustrated by Figs. (b) and (c) below

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For this problem Case 1 always applies, because:

$tan (\frac{α}{2}) = \frac{\frac{a}{2}}{h}$ $tan(alpha/2)=(a/2)/h$ => $h = (\frac{1}{2}) \frac{a}{tan (\frac{α}{2})}$ $h=(1/2)a/tan(alpha/2)$

But there's a condition so that Case 2 apllies:

$sin (β) = \frac{h}{b}$ $sin(beta)=h/b$ => $h = b sin β$ $h=bsin beta$
Or $h = b sin γ$ $h=bsin gamma$
Since the highest value of $sin β$ $sin beta$ or $sin γ$ $sin gamma$ is $1$ $1$ , the highest value of $h$ $h$ , in Case 2, must be $b$ $b$ .

In the present problem h is longer than the side to which it is perpendicular, so for this problem only the Case 1 applies.

Solution considering Case 1 (Fig. (a))

$b^{2} = h^{2} + {(\frac{a}{2})}^{2}$ $b^2=h^2+(a/2)^2$
$b^{2} = {(\frac{72}{\sqrt{10}})}^{2} + {(\frac{\sqrt{10}}{2})}^{2}$ $b^2=(72/sqrt(10))^2+(sqrt(10)/2)^2$
$b^{2} = \frac{5184}{10} + \frac{10}{4} = \frac{5184 + 25}{10} = \frac{5209}{10}$ $b^2=5184/10+10/4=(5184+25)/10=5209/10$ => $b = \sqrt{520.9} ≅ 22.823$ $b=sqrt(520.9)~=22.823$

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Two corners of an isosceles triangle are at $(1, 6)$ $(1 ,6 )$ and $(2, 9)$ $(2 ,9 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

Two corners of an isosceles triangle are at (1,6)(1 ,6 ) and (2,9)(2 ,9 ). If the triangle's area is 3636 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(1, 6)$ $(1 ,6 )$ and $(2, 9)$ $(2 ,9 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?