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

1 Answer
Feb 9, 2016

Two possibilities: (I) sqrt(85),sqrt(2165/68),sqrt(2165/68)~=9.220,5.643,5.643 or (II) 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

From the formula of the triangle's area:
S=(b*h)/2 => 15=(sqrt(85)*h)/2 => 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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I created this figure using MS ExcelI created this figure using MS Excel

For this problem Case 1 always applies, because:

tan(alpha/2)=(a/2)/h => h=(1/2)a/tan(alpha/2)

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

sin(beta)=h/b => h=bsin beta
Or h=bsin gamma
Since the highest value of sin beta or sin gamma is 1, the highest value of h, in Case 2, must be 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)

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

Solution considering Case 2 (shape of Fig. (b)), b=sqrt(85)

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

a^2=h^2+n^2=(30/sqrt(85))^2+(sqrt(85)-sqrt(1265/17))^2
a^2=900/85+85+1265/17-2sqrt((85*1265)/17)
a^2=180/17+85+1265/17-2*sqrt(5*1265)
a^2=1445/17+85-2*5sqrt(253)
a^2=85+85-10sqrt(253)
a=sqrt(170-10sqrt(253))~=3.308