# Two corners of an isosceles triangle are at #(9 ,2 )# and #(4 ,7 )#. If the triangle's area is #64 #, what are the lengths of the triangle's sides?

Solution.

By signing up, you agree to our Terms of Service and Privacy Policy

Let's denote the coordinates of the two corners of the isosceles triangle as A(9, 2) and B(4, 7). The third vertex of the triangle can be denoted as C.

We know that the area of a triangle can be calculated using the formula:

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

Since the triangle is isosceles, we can find the length of the base as the distance between points A and B. Using the distance formula:

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

Let's find the length of AB first:

[ AB = \sqrt{(4 - 9)^2 + (7 - 2)^2} = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} ]

Now, let's use the area formula and the fact that the area is given as 64 to find the height of the triangle:

[ 64 = \frac{1}{2} \times 5\sqrt{2} \times \text{height} ] [ 128 = 5\sqrt{2} \times \text{height} ] [ \text{height} = \frac{128}{5\sqrt{2}} = \frac{128\sqrt{2}}{10} = \frac{64\sqrt{2}}{5} ]

Now that we have the height and the base, we can use the Pythagorean theorem to find the length of the sides of the triangle. Let's denote the length of the sides as AC and BC.

[ AC^2 = AB^2 - (\frac{1}{2} \times \text{height})^2 ] [ AC^2 = (5\sqrt{2})^2 - \left(\frac{64\sqrt{2}}{10}\right)^2 ] [ AC^2 = 50 - \frac{4096}{100} ] [ AC^2 = 50 - 40.96 ] [ AC^2 = 9.04 ] [ AC = \sqrt{9.04} = 3.008 ]

Similarly, for BC: [ BC^2 = AB^2 - (\frac{1}{2} \times \text{height})^2 ] [ BC^2 = (5\sqrt{2})^2 - \left(\frac{64\sqrt{2}}{10}\right)^2 ] [ BC^2 = 50 - \frac{4096}{100} ] [ BC^2 = 50 - 40.96 ] [ BC^2 = 9.04 ] [ BC = \sqrt{9.04} = 3.008 ]

Therefore, the lengths of the sides AC and BC of the triangle are approximately 3.008 units each.

By signing up, you agree to our Terms of Service and Privacy Policy

- Two corners of a triangle have angles of #pi / 3 # and # pi / 2 #. If one side of the triangle has a length of #2 #, what is the longest possible perimeter of the triangle?
- Two corners of a triangle have angles of # (7 pi )/ 12 # and # pi / 4 #. If one side of the triangle has a length of 1, what is the longest possible perimeter of the triangle?
- Two corners of a triangle have angles of #pi / 12 # and # pi / 3 #. If one side of the triangle has a length of #6 #, what is the longest possible perimeter of the triangle?
- A pyramid has a base in the shape of a rhombus and a peak directly above the base's center. The pyramid's height is #3 #, its base has sides of length #4 #, and its base has a corner with an angle of #(2 pi)/3 #. What is the pyramid's surface area?
- An ellipsoid has radii with lengths of #4 #, #5 #, and #7 #. A portion the size of a hemisphere with a radius of #5 # is removed form the ellipsoid. What is the remaining volume of the ellipsoid?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7