Answer: The segment EF of the isoceles triangle ΔEFP is 10 units
Solution
If point P is equidistant from the vertices of ΔDEF, the segment PE must measure the same that the segment PF, then these two segments are congruent:
PE=PF
Then ΔEFP is an isisceles triangle, because it has two congruent sides (PE=PF), and the height of the side EF (PJ perpendicular to EF) divides ΔEFP into two congruent right triangles (ΔEJP and ΔFJP). The legs EJ and FJ must be congruent:
FJ=EJ
3x-1=x+3
Solving for x: Subtracting x and adding 1 both sides of the equation:
3x-1-x+1=x+3-x+1
2x=4
Dividing both sides of the equation by 2:
2x/2=4/2
x=2
Then:
FJ=3x-1→FJ=3(2)-1→FJ=6-1→FJ=5
EJ=x+3→EJ=2+3→EJ=5
EF=EJ+FJ→EF=5+5→EF=10
first of all you have to draw a rhombus
area (d+d)/2
(4+6)/2
5
the first one
100x² + 20x +1
its a factor of different 2 squares
(10x - 1 )²
