# Two corners of an isosceles triangle are at #(6 ,6 )# and #(2 ,7 )#. If the triangle's area is #36 #, what are the lengths of the triangle's sides?

Lengths of the isosceles triangle are 4.1231, 17.5839, 17.5839

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To find the lengths of the sides of the isosceles triangle, first, determine the length of the base by finding the distance between the given points using the distance formula:

[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

Given points: ( (6, 6) ) and ( (2, 7) )

Distance = ( \sqrt{(2 - 6)^2 + (7 - 6)^2} )

[ = \sqrt{(-4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} ]

Now, since it's an isosceles triangle, the other two sides are congruent. Let's call the length of one of these congruent sides ( s ). The area of a triangle can be calculated using the formula:

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

We have the area (36) and the base ((\sqrt{17})), so we can solve for the height:

[ 36 = \frac{1}{2} \times \sqrt{17} \times \text{height} ]

[ \text{height} = \frac{36 \times 2}{\sqrt{17}} = \frac{72}{\sqrt{17}} ]

Now, using the Pythagorean theorem, we can find the length of one of the congruent sides:

[ s = \sqrt{\text{height}^2 + \left(\frac{\text{base}}{2}\right)^2} ]

[ s = \sqrt{\left(\frac{72}{\sqrt{17}}\right)^2 + \left(\frac{\sqrt{17}}{2}\right)^2} ]

[ s = \sqrt{\frac{72^2}{17} + \frac{17}{4}} ]

[ s = \sqrt{\frac{5184}{17} + \frac{17}{4}} ]

[ s = \sqrt{\frac{5184 + 17 \times 4}{17}} ]

[ s = \sqrt{\frac{5184 + 68}{17}} ]

[ s = \sqrt{\frac{5252}{17}} ]

[ s \approx \sqrt{308.941} ]

[ s \approx 17.57 ]

Therefore, the lengths of the triangle's sides are approximately (17.57), (17.57), and (\sqrt{17}).

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