Two corners of an isosceles triangle are at $(8, 1)$ $(8 ,1 )$ and $(1, 7)$ $(1 ,7 )$ . If the triangle's area is $15$ $15$ , what are the lengths of the triangle's sides?

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Two corners of an isosceles triangle are at $(8, 1)$ $(8 ,1 )$ and $(1, 7)$ $(1 ,7 )$ . If the triangle's area is $15$ $15$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-02-09T22:00:07

Two possibilities: (I) $\sqrt{85}, \sqrt{\frac{2165}{68}}, \sqrt{\frac{2165}{68}} ≅ 9.220, 5.643, 5.643$ $sqrt(85),sqrt(2165/68),sqrt(2165/68)~=9.220,5.643,5.643$ or (II) $\sqrt{170 - 10 \sqrt{253}}, \sqrt{85}, \sqrt{85} ≅ 3.308, 9.220, 9.220$ $sqrt(170-10sqrt(253)),sqrt(85),sqrt(85)~=3.308,9.220,9.220$

Explanation:

The length of the given side is
$s = \sqrt{{(1 - 8)}^{2} + {(7 - 1)}^{2}} = \sqrt{49 + 36} = \sqrt{85} ≅ 9.220$ $s=sqrt((1-8)^2+(7-1)^2)=sqrt(49+36)=sqrt(85)~=9.220$

From the formula of the triangle's area:
$S = \frac{b \cdot h}{2}$ $S=(b*h)/2$ => $15 = \frac{\sqrt{85} \cdot h}{2}$ $15=(sqrt(85)*h)/2$ => $h = \frac{30}{\sqrt{85}} ≅ 3.254$ $h=30/sqrt(85)~=3.254$

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 smaller than the side to which it is perpendicular, so for this problem besides the Case 1, also the Case 2 applies.

Solution considering Case 1 (Fig. (a)), $a = \sqrt{85}$ $a=sqrt(85)$

$b^{2} = h^{2} + {(\frac{a}{2})}^{2}$ $b^2=h^2+(a/2)^2$
$b^{2} = {(\frac{30}{\sqrt{85}})}^{2} + {(\frac{\sqrt{85}}{2})}^{2}$ $b^2=(30/sqrt(85))^2+(sqrt(85)/2)^2$
$b^{2} = \frac{900}{85} + \frac{85}{4} = \frac{180}{17} + \frac{85}{4} = \frac{720 + 1445}{68} = \frac{2165}{68}$ $b^2=900/85+85/4=180/17+85/4=(720+1445)/68=2165/68$ => $b = \sqrt{\frac{2165}{68}} ≅ 5.643$ $b=sqrt(2165/68)~=5.643$

Solution considering Case 2 (shape of Fig. (b)), $b = \sqrt{85}$ $b=sqrt(85)$

$b^{2} = m^{2} + h^{2}$ $b^2=m^2+h^2$
$m^{2} = b^{2} - h^{2} = {(\sqrt{85})}^{2} - {(\frac{30}{\sqrt{85}})}^{2} = 85 - \frac{900}{85} = 85 - \frac{180}{17} = \frac{1445 - 180}{17}$ $m^2=b^2-h^2=(sqrt(85))^2-(30/sqrt(85))^2=85-900/85=85-180/17=(1445-180)/17$ => $m = \sqrt{\frac{1265}{17}}$ $m=sqrt(1265/17)$
$m + n = b$ $m+n=b$ => $n = b - m$ $n=b-m$ => $n = \sqrt{85} - \sqrt{\frac{1265}{17}}$ $n=sqrt(85)-sqrt(1265/17)$

$a^{2} = h^{2} + n^{2} = {(\frac{30}{\sqrt{85}})}^{2} + {(\sqrt{85} - \sqrt{\frac{1265}{17}})}^{2}$ $a^2=h^2+n^2=(30/sqrt(85))^2+(sqrt(85)-sqrt(1265/17))^2$
$a^{2} = \frac{900}{85} + 85 + \frac{1265}{17} - 2 \sqrt{\frac{85 \cdot 1265}{17}}$ $a^2=900/85+85+1265/17-2sqrt((85*1265)/17)$
$a^{2} = \frac{180}{17} + 85 + \frac{1265}{17} - 2 \cdot \sqrt{5 \cdot 1265}$ $a^2=180/17+85+1265/17-2*sqrt(5*1265)$
$a^{2} = \frac{1445}{17} + 85 - 2 \cdot 5 \sqrt{253}$ $a^2=1445/17+85-2*5sqrt(253)$
$a^{2} = 85 + 85 - 10 \sqrt{253}$ $a^2=85+85-10sqrt(253)$
$a = \sqrt{170 - 10 \sqrt{253}} ≅ 3.308$ $a=sqrt(170-10sqrt(253))~=3.308$

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Two corners of an isosceles triangle are at $(8, 1)$ $(8 ,1 )$ and $(1, 7)$ $(1 ,7 )$ . If the triangle's area is $15$ $15$ , 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 (8,1)(8 ,1 ) and (1,7)(1 ,7 ). If the triangle's area is 1515 , what are the lengths of the triangle's sides?

1 Answer

Explanation:

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Two corners of an isosceles triangle are at $(8, 1)$ $(8 ,1 )$ and $(1, 7)$ $(1 ,7 )$ . If the triangle's area is $15$ $15$ , what are the lengths of the triangle's sides?