Find the lengths of the sides of a triangle with vertices: (0, 0), (5, 0), (5, 12).
graph
(1 point)
Responses
5, 12, 12
5, 12, 12
5, 5, 12
5, 5, 12
5, 12, 5
5, 12, 5
5, 12, 13
Let us name the vertices of the triangle as points A, B, and C:
A = (0, 0)
B = (5, 0)
C = (5, 12)
We will find the lengths of the segments AB, BC, and CA using the distance formula for points in a 2D space. The distance \(d\) between two points \((x_1, y_1)\) and \((x_2, y_2)\) is calculated by the formula:
\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
For AB, which is a horizontal line segment from A to B:
AB = distance from (0,0) to (5,0)
AB = \(\sqrt{(5 - 0)^2 + (0 - 0)^2}\)
AB = \(\sqrt{5^2 + 0}\)
AB = \(\sqrt{25}\)
AB = 5
For BC, which is a vertical line segment from B to C:
BC = distance from (5,0) to (5,12)
BC = \(\sqrt{(5 - 5)^2 + (12 - 0)^2}\)
BC = \(\sqrt{0 + 12^2}\)
BC = \(\sqrt{144}\)
BC = 12
For CA, the hypotenuse of this right-angle triangle from C to A:
CA = distance from (0,0) to (5,12)
CA = \(\sqrt{(5 - 0)^2 + (12 - 0)^2}\)
CA = \(\sqrt{5^2 + 12^2}\)
CA = \(\sqrt{25 + 144}\)
CA = \(\sqrt{169}\)
CA = 13
Thus, the lengths of the sides are 5, 12, and 13. The correct answer is:
5, 12, 13