Two corners of an isosceles triangle are at #(2 ,6 )# and #(3 ,8 )#. If the triangle's area is #48 #, what are the lengths of the triangle's sides?

Question

Two corners of an isosceles triangle are at #(2  ,6   )# and #(3   ,8    )#.  If the triangle's area is #48   #, what are the lengths of the triangle's sides?

Wyatt Anderson · Accepted Answer

Measure of the three sides are (2.2361, 49.1212, 49.1212)

Autumn Armstrong · Answer

undefined To find the lengths of the sides of the isosceles triangle, we first need to determine the length of the base, which is the distance between the given points (2, 6) and (3, 8).

Using the distance formula:
$$ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

We calculate the distance between (2, 6) and (3, 8):
$$ \text{Distance} = \sqrt{(3 - 2)^2 + (8 - 6)^2} $$
$$ = \sqrt{1^2 + 2^2} $$
$$ = \sqrt{5} $$

Since the triangle is isosceles, the length of the base is the same as the length of one of the equal sides. Let's denote the length of each equal side as $ s $. Now, we know the area of the triangle is 48, and the formula for the area of a triangle is:
$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} $$

Substituting the known values, we get:
$$ 48 = \frac{1}{2} \times \sqrt{5} \times \text{height} $$

To solve for the height, we multiply both sides by 2 and divide by $ \sqrt{5} $:
$$ \text{height} = \frac{2 \times 48}{\sqrt{5}} = \frac{96}{\sqrt{5}} $$

Now, we can use the Pythagorean theorem to find the length of the equal sides. Let $ s $ be the length of each equal side:
$$ s^2 = \left( \frac{\sqrt{5}}{2} \right)^2 + \left( \frac{96}{\sqrt{5}} \right)^2 $$

$$ s^2 = \frac{5}{4} + \frac{9216}{5} $$

$$ s^2 = \frac{5 \times 5 + 9216}{4 \times 5} $$

$$ s^2 = \frac{9216 + 25}{20} $$

$$ s^2 = \frac{9241}{20} $$

$$ s = \sqrt{\frac{9241}{20}} $$

$$ s = \frac{\sqrt{9241}}{\sqrt{20}} $$

$$ s = \frac{\sqrt{9241}}{2\sqrt{5}} $$

Therefore, the length of each equal side of the triangle is $ \frac{\sqrt{9241}}{2\sqrt{5}} $.