Two corners of an isosceles triangle are at (9 ,4 )(9,4) and (3 ,2 )(3,2). If the triangle's area is 48 48, what are the lengths of the triangle's sides?

1 Answer
Feb 9, 2016

2sqrt(10),sqrt(1202/5),sqrt(1202/5)~=6.325,15.505,15.505210,12025,120256.325,15.505,15.505

Explanation:

The length of the given side is
s=sqrt((9-3)^2+(4-2)^2)=sqrt(40)=2sqrt(10)~=6.325s=(93)2+(42)2=40=2106.325

From the formula of the triangle's area:
S=(b*h)/2S=bh2 => 48=(cancel(2)sqrt(10)*h)/cancel(2) => h=48/sqrt(10)~=15.179

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)/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=(48/sqrt(10))^2+(cancel(2)sqrt(10)/cancel(2))^2
b^2=2304/10+10=2404/10=1202/5 => b=sqrt(1202/5)~=15.505