# Two corners of an isosceles triangle are at (5 ,2 ) and (2 ,3 ). If the triangle's area is 6 , what are the lengths of the triangle's sides?

Aug 11, 2017

If the base is $\sqrt{10}$, then the two sides are $\sqrt{\frac{29}{2}}$

#### Explanation:

It depends on whether or not these points form the base or the sides.

First, find the length between the two points.
This is done by finding the lenght of the vector between the two points:

$\sqrt{{\left(5 - 2\right)}^{2} + {\left(2 - 3\right)}^{2}} = \sqrt{10}$

If this is the lenght of the base, then:
Start by finding the height of the triangle.
Area of a triangle is given by: $A = \frac{1}{2} \cdot h \cdot b$ , where (b) is the base and (h) is the height.

Therefore:
$6 = \frac{1}{2} \cdot \sqrt{10} \cdot h \iff$ $\frac{12}{\sqrt{10}} = h$

Because the height cuts a isosceles triangle into two similar right-angeled triangles, we can use pythagoras.
The two sides will then be:
$\sqrt{{\left(\frac{1}{2} \cdot \sqrt{10}\right)}^{2} + {\left(\frac{12}{\sqrt{12}}\right)}^{2}} = \sqrt{\frac{1}{4} \cdot 10 + 12} = \sqrt{\frac{58}{4}} = \sqrt{\frac{29}{2}}$

If it was the lenght of the two sides, then:
Use the area formula for triangles in generel, $A = \frac{1}{2} \cdot a \cdot b \cdot \sin \left(C\right)$, because (a) and (b) are the same, we get; $A = \frac{1}{2} \cdot {a}^{2} \cdot \sin \left(C\right)$, where (a) is the side we calculated.

$6 = \frac{1}{2} \cdot 10 \cdot \sin \left(C\right) \iff$ $\sin \left(C\right) = \frac{6}{5}$

But that isn't possible for a real triangle, so we must asume the two coordinates formed the base.