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