An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of #8 # and the triangle has an area of #32 #. What are the lengths of sides A and B?

Question

An isosceles triangle has sides A, B, and C, such that sides A and B have the same length.  Side C has a length of #8 # and the triangle has an area of #32  #.  What are the lengths of sides A and B?

Sadie Adams · Accepted Answer

#sqrt80#

Area of a triangle #A=1/2*b*h#, where #b# is the base and #h# is the height.

Given that #A=B#, base #C=8#, and area of the triangle #=32#,
#=> 32=1/2*8*h#
#=> h=8#

#A^2=h^2+(C/2)^2#

#=> A=sqrt(8^2+(8/4)^2)=sqrt80#

Hence, #A=B=sqrt80#

Caroline Aucoin · Answer

undefined Using the formula for the area of a triangle, $ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $, where in the case of an isosceles triangle, the base is one of the equal sides (let's say $ A $) and the height is the perpendicular distance from the base to the opposite vertex (let's say $ h $), we can set up the equation:

$ 32 = \frac{1}{2} \times A \times h $

Given that side $ C $ is 8 units long, it can be seen as the base of two right triangles, each formed by $ A $, $ B $, and half of $ C $. Using the Pythagorean theorem, the height $ h $ can be expressed as:

$ h = \sqrt{C^2 - \left(\frac{C}{2}\right)^2} $

Substitute the given values:

$ h = \sqrt{8^2 - \left(\frac{8}{2}\right)^2} = \sqrt{64 - 16} = \sqrt{48} = 4\sqrt{3} $

Now, substitute the values of $ h $ and the given area into the area formula:

$ 32 = \frac{1}{2} \times A \times 4\sqrt{3} $

Solve for $ A $:

$ 32 = 2A\sqrt{3} $

$ A\sqrt{3} = 16 $

$ A = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} $

Since sides $ A $ and $ B $ are equal in length:

$ A = B = \frac{16\sqrt{3}}{3} $