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
Feb 9, 2016

Two possibilities: (I) sqrt(85),sqrt(2165/68),sqrt(2165/68)~=9.220,5.643,5.64385,216568,2165689.220,5.643,5.643 or (II) sqrt(170-10sqrt(253)),sqrt(85),sqrt(85)~=3.308,9.220,9.22017010253,85,853.308,9.220,9.220

Explanation:

The length of the given side is
s=sqrt((1-8)^2+(7-1)^2)=sqrt(49+36)=sqrt(85)~=9.220s=(18)2+(71)2=49+36=859.220

From the formula of the triangle's area:
S=(b*h)/2S=bh2 => 15=(sqrt(85)*h)/215=85h2 => h=30/sqrt(85)~=3.254h=30853.254

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

I created this figure using MS ExcelI created this figure using MS Excel

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

I created this figure using MS ExcelI created this figure using MS Excel
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)/htan(α2)=a2h => h=(1/2)a/tan(alpha/2)h=(12)atan(α2)

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

sin(beta)=h/bsin(β)=hb => h=bsin betah=bsinβ
Or h=bsin gammah=bsinγ
Since the highest value of sin betasinβ or sin gammasinγ is 11, the highest value of hh, in Case 2, must be bb.

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=85

b^2=h^2+(a/2)^2b2=h2+(a2)2
b^2=(30/sqrt(85))^2+(sqrt(85)/2)^2b2=(3085)2+(852)2
b^2=900/85+85/4=180/17+85/4=(720+1445)/68=2165/68b2=90085+854=18017+854=720+144568=216568 => b=sqrt(2165/68)~=5.643b=2165685.643

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

b^2=m^2+h^2b2=m2+h2
m^2=b^2-h^2=(sqrt(85))^2-(30/sqrt(85))^2=85-900/85=85-180/17=(1445-180)/17m2=b2h2=(85)2(3085)2=8590085=8518017=144518017 => m=sqrt(1265/17)m=126517
m+n=bm+n=b => n=b-mn=bm => n=sqrt(85)-sqrt(1265/17)n=85126517

a^2=h^2+n^2=(30/sqrt(85))^2+(sqrt(85)-sqrt(1265/17))^2a2=h2+n2=(3085)2+(85126517)2
a^2=900/85+85+1265/17-2sqrt((85*1265)/17)a2=90085+85+126517285126517
a^2=180/17+85+1265/17-2*sqrt(5*1265)a2=18017+85+126517251265
a^2=1445/17+85-2*5sqrt(253)a2=144517+8525253
a^2=85+85-10sqrt(253)a2=85+8510253
a=sqrt(170-10sqrt(253))~=3.308a=170102533.308