Two corners of an isosceles triangle are at #(9 ,6 )# and #(7 ,2 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(9   ,6   )# and #(7  ,2    )#.  If the triangle's area is #64   #, what are the lengths of the triangle's sides?

Violet Aldrich · Accepted Answer

#"sides "a = c = 28.7" units"# and #"side " b = 2sqrt5" units"#

let #b = # the distance between the two points:

#b = sqrt((9-7)^2+(6-2)^2)#

#b = 2sqrt5" units"#

We are given that the #"Area" = 64" units"^2#

Let "a" and "c" be the other two sides.

For a triangle, #"Area " = 1/2bh#

Substituting in the values for "b" and the Area:

#64" units"^2 = 1/2(2sqrt5" units")h#

Solve for the height:

#h = 64/sqrt5 = 64/5sqrt5" units"#

Let #C = # the angle between side "a" and side "b", then we may use the right triangle formed by side "b" and the height to write the following equation:

#tan(C) = h/(1/2b)#

#tan(C) = (64/5sqrt5" units")/(1/2(2sqrt5" units"))#

#C = tan^-1(64/5)#

We can find the length of side "a", using the following equation:

#h = (a)sin(C)#

#a = h/sin(C)#

Substitute in the values for "h" and "C":

#a = (64/5sqrt5" units")/sin(tan^-1(64/5))#

#a = 28.7" units"#

Intuition tells me that side "c" is the same length as side "a" but we can prove this using the Law of Cosines:

#c^2 = a^2 + b^2 - 2(a)(b)cos(C)#

Substitute in the values for a, b, and C:

#c^2 = (28.7" units")^2 + (2sqrt5" units")^2 - 2(28.7" units")(2sqrt5" units")cos(tan^-1(64/5))#

#c = 28.7" units"#

Theodore Abad · Answer

undefined First, find the length of the base of the triangle, which is the distance between the two given corners. Then, use the formula for the area of a triangle to find the height. Finally, use the Pythagorean theorem to find the length of the equal sides.

1. Find the length of the base:
Distance formula: 
$ \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $

$ \sqrt{(9 - 7)^2 + (6 - 2)^2} $  
$ \sqrt{(2)^2 + (4)^2} $  
$ \sqrt{4 + 16} $  
$ \sqrt{20} $  
$ 2\sqrt{5} $

2. Use the area formula for a triangle: 
$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $
$ 64 = \frac{1}{2} \times 2\sqrt{5} \times \text{height} $
$ 128 = 2\sqrt{5} \times \text{height} $
$ \text{height} = \frac{128}{2\sqrt{5}} $
$ \text{height} = \frac{64}{\sqrt{5}} $
$ \text{height} = \frac{64\sqrt{5}}{5} $

3. Use the Pythagorean theorem to find the length of the equal sides:
$ \text{Side}^2 = \text{height}^2 + \left(\frac{\text{base}}{2}\right)^2 $
$ \text{Side}^2 = \left(\frac{64\sqrt{5}}{5}\right)^2 + \left(\frac{2\sqrt{5}}{2}\right)^2 $
$ \text{Side}^2 = \left(\frac{64\sqrt{5}}{5}\right)^2 + \left(\sqrt{5}\right)^2 $
$ \text{Side}^2 = \frac{64^2 \times 5}{5^2} + 5 $
$ \text{Side}^2 = \frac{64^2 \times 5 + 5^2 \times 5}{5^2} $
$ \text{Side}^2 = \frac{64^2 \times 5 + 5^2 \times 5}{5^2} $
$ \text{Side}^2 = \frac{64^2 \times 5 + 5^2 \times 5}{5^2} $
$ \text{Side}^2 = \frac{5(64^2 + 5^2)}{5^2} $
$ \text{Side}^2 = \frac{5(4096 + 25)}{25} $
$ \text{Side}^2 = \frac{5 \times 4121}{25} $
$ \text{Side}^2 = \frac{20605}{25} $
$ \text{Side}^2 = 824.2 $
$ \text{Side} \approx \sqrt{824.2} $
$ \text{Side} \approx 28.71 $

Therefore, the lengths of the triangle's sides are approximately 28.71, 28.71, and $2\sqrt{5}$.