Triangle D E F is shown with point P at the center. Lines are drawn from each point of the triangle to point P. Line segments are drawn from point P to the sides of the triangle to form right angles and line segments P H, P J, and P G. The length of F J is 3 x minus 1, the length of J E is x + 3, the length of H E is 4 y minus 3, and the length of H D is 9. Given that point P is equidistant from the vertices of ΔDEF, what is EF? EF =

d

gimmee one of those gold crowns!

~coco

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !

18 units²

step-by-step explanation:

first let's find altitude bd using the distance formula

d = √((y₂ - y₁)² + (x₂ - x₁)²)

bd = √((0 - -3)² + (0 - 3)²)

bd = √18

given the placement of points a and c, we can denote that point d bisects segment ac.

we will call ac the base.

using the distance formula

ac = √((0 - 6)² + (-6 - 0)²)

ac = √72

knowing our base and our height, we can plug these into our triangle area formula

area = (b * h)/2

area = (√18 * √72) / 2

area = (√1296) / 2

area = 36 / 2

area = 18 units²