Two corners of an isosceles triangle are at (8 ,3 ) and (5 ,4 ). If the triangle's area is 15 , what are the lengths of the triangle's sides?

1 Answer
Feb 9, 2016

sqrt(10),5sqrt(3.7), 5sqrt(3.7)~=3.162,9.618,9.618

Explanation:

The length of the given side is
s=sqrt((5-8)^2+(4-3)^2)=sqrt(9+1)=sqrt(10)~=3.162

From the formula of the triangle's area:
S=(b*h)/2 => 15=(sqrt(10)*h)/2 => h=30/sqrt(10)~=9.487

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

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

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 longer than the side to which it is perpendicular, so for this problem only the Case 1 applies.

Solution considering Case 1 (Fig. (a))

b^2=h^2+(a/2)^2
b^2=(30/sqrt(10))^2+(sqrt(10)/2)^2
b^2=900/10+10/4=(900+25)/10=925/10 => b=sqrt(92.5)=5sqrt(3.7)~=9.618