Two corners of an isosceles triangle are at #(6 ,3 )# and #(5 ,8 )#. If the triangle's area is #8 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(6  ,3   )# and #(5   ,8    )#.  If the triangle's area is #8   #, what are the lengths of the triangle's sides?

Theodore Abad · Accepted Answer

case 1. Base#=sqrt26 and# leg#=sqrt(425/26)#
case 2. Leg #=sqrt26 and # base#=sqrt(52+-sqrt1680)#

Given Two corners of an isosceles triangle are at #(6,3) and (5,8)#.
Distance between the corners is given by the expression
#d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)#, inserting given values
#d=sqrt((5-6)^2+(8-3)^2)#
#d=sqrt((-1)^2+(5)^2)#
#d=sqrt26#

Now area of triangle is given by
#"Area"=1/2"base"xx"height"#

Case 1. The corners are base angles.
#:."base"=sqrt26#
#"height"=2xx"Area"/"base"# .....(1)
#=2xx8/sqrt26=16/sqrt26#
Now using the Pythagoras theorem
#"leg"=sqrt("height"^2+("base"/2)^2)#
#"leg"=sqrt((16/sqrt26)^2+(sqrt26/2)^2)#
#=sqrt(256/26+26/4#
#=sqrt(128/13+13/2)#
#=sqrt(425/26)#

Case 2. The corners are base angle and the vertex.
#"Leg"=sqrt26#
Let #"base"=b#
Also from (1) #"height"=2xx"Area"/"base"#
#"height"=2xx8/"base"#
#"height"=16/"base"#
Now using the Pythagoras theorem
#"leg"=sqrt("height"^2+("base"/2)^2)#
#sqrt26=sqrt("256/b^2+b^2/4)#, squaring both sides
#26="256/b^2+b^2/4#
#104b^2=1024+b^4#
#b^4-104b^2+1024=0#, solving for #b^2# using the quadratic formula
#b^2=(104+-sqrt((-104)^2-4xx1024xx1))/2#
#b^2=52+-sqrt1680#, taking square root
#b=sqrt(52+-sqrt1680)#, we have ignored the negative sign as length can not be negative.

Aurora Ayala · Answer

undefined To find the lengths of the sides of the triangle, we can use the formula for the area of a triangle, which is given by $ \frac{1}{2} \times \text{base} \times \text{height} $. Since the triangle is isosceles, we can determine the base by finding the distance between the two given points. Then, we can use the given area to find the height. Once we have the base and height, we can use the Pythagorean theorem to find the lengths of the other two sides, as the triangle is isosceles.

First, we find the distance between the two given points (6, 3) and (5, 8) to get the base:
$$ \text{Base} = \sqrt{(6 - 5)^2 + (3 - 8)^2} $$
$$ = \sqrt{(1)^2 + (-5)^2} $$
$$ = \sqrt{1 + 25} $$
$$ = \sqrt{26} $$

Now, we use the formula for the area of a triangle to find the height:
$$ 8 = \frac{1}{2} \times \sqrt{26} \times \text{height} $$
$$ \text{height} = \frac{8 \times 2}{\sqrt{26}} $$
$$ \text{height} = \frac{16}{\sqrt{26}} $$

Next, we use the Pythagorean theorem to find the lengths of the other two sides, which are equal since the triangle is isosceles:
$$ \text{Side} = \sqrt{\left(\frac{\text{Base}}{2}\right)^2 + \text{height}^2} $$
$$ = \sqrt{\left(\frac{\sqrt{26}}{2}\right)^2 + \left(\frac{16}{\sqrt{26}}\right)^2} $$
$$ = \sqrt{\frac{26}{4} + \frac{256}{26}} $$
$$ = \sqrt{\frac{13}{2} + \frac{256}{26}} $$
$$ = \sqrt{\frac{13}{2} + 10} $$
$$ = \sqrt{\frac{33}{2}} $$

Therefore, the lengths of the sides of the isosceles triangle are $ \sqrt{26} $ and $ \sqrt{\frac{33}{2}} $.