Answer:
The area of rectangle BEFD is 180 square units.
Step-by-step explanation:
After checking the figure given, we have the following information:
[tex]AD = 15[/tex], [tex]AB = 12[/tex], [tex]BC = 15[/tex], [tex]CD = 12[/tex]
By Pythagorean Theorem, we determine the length of the line segment BD:
[tex]BD = \sqrt{AB^{2}+AD^{2}}[/tex] (1)
[tex]BD = \sqrt{12^{2}+15^{2}}[/tex]
[tex]BD = 3\sqrt{41}[/tex]
In addition, we know the following characteristics of the rectangle BEFD:
[tex]BD = EF[/tex], [tex]EF = EC + CF[/tex], [tex]BE = FD[/tex] (2), (3), (4)
By Pythagorean Theorem:
[tex]BC^{2} = BE^{2}+EC^{2}[/tex] (5)
[tex]CD^{2} = CF^{2}+DF^{2}[/tex] (6)
By (3), (4), (5) and (6):
[tex]BC^{2} = BE^{2} + EC^{2}[/tex] (7)
[tex]CD^{2} = (EF-EC)^{2} + BE^{2}[/tex] (8)
By (7) in (8):
[tex]CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})[/tex]
[tex]CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}[/tex]
[tex]CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}[/tex]
Then, we clear [tex]EC[/tex]:
[tex]2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}[/tex]
[tex]EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}[/tex]
If we know that [tex]EF = 3\sqrt{41}[/tex], [tex]BC = 15[/tex] and [tex]CD = 12[/tex], then the length of the segment [tex]EC[/tex] is:
[tex]EC = \frac{75\sqrt{41}}{41}[/tex]
And the length of the line segment [tex]CF[/tex] is:
[tex]CF = EF - EC[/tex]
[tex]CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}[/tex]
[tex]CF = \frac{48\sqrt{41}}{41}[/tex]
And the length of the line segment [tex]DF[/tex] is determined by Pythagorean Theorem:
[tex]FD = \sqrt{CD^{2}-CF^{2}}[/tex]
[tex]FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}[/tex]
[tex]FD = \frac{60\sqrt{41}}{41}[/tex]
And the area of the rectangle is determined by the following formula:
[tex]A = FD\cdot EF[/tex]
[tex]A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})[/tex]
[tex]A = 180[/tex]
The area of rectangle BEFD is 180 square units.
