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
Two possibilities: (I)
Explanation:
The length of the given side is
From the formula of the triangle's area:
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 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 Excel
I 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
Orh=bsin gammah=bsinγ
Since the highest value ofsin betasinβ orsin gammasinγ is11 , the highest value ofhh , in Case 2, must bebb .
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)),
b^2=h^2+(a/2)^2b2=h2+(a2)2
b^2=(30/sqrt(85))^2+(sqrt(85)/2)^2b2=(30√85)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=√216568≅5.643
Solution considering Case 2 (shape of Fig. (b)),
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=b2−h2=(√85)2−(30√85)2=85−90085=85−18017=1445−18017 =>m=sqrt(1265/17)m=√126517
m+n=bm+n=b =>n=b-mn=b−m =>n=sqrt(85)-sqrt(1265/17)n=√85−√126517
a^2=h^2+n^2=(30/sqrt(85))^2+(sqrt(85)-sqrt(1265/17))^2a2=h2+n2=(30√85)2+(√85−√126517)2
a^2=900/85+85+1265/17-2sqrt((85*1265)/17)a2=90085+85+126517−2√85⋅126517
a^2=180/17+85+1265/17-2*sqrt(5*1265)a2=18017+85+126517−2⋅√5⋅1265
a^2=1445/17+85-2*5sqrt(253)a2=144517+85−2⋅5√253
a^2=85+85-10sqrt(253)a2=85+85−10√253
a=sqrt(170-10sqrt(253))~=3.308a=√170−10√253≅3.308