Question

PLEASE ILL MAKE BRAINIEST!!!

Answer 1

Answer:

The area of rectangle BEFD is 180 square units.

Step-by-step explanation:

After checking the figure given, we have the following information:

$AD = 15$, $AB = 12$, $BC = 15$, $CD = 12$

By Pythagorean Theorem, we determine the length of the line segment BD:

$BD = \sqrt{AB^{2}+AD^{2}}$ (1)

$BD = \sqrt{12^{2}+15^{2}}$

$BD = 3\sqrt{41}$

In addition, we know the following characteristics of the rectangle BEFD:

$BD = EF$, $EF = EC + CF$, $BE = FD$ (2), (3), (4)

By Pythagorean Theorem:

$BC^{2} = BE^{2}+EC^{2}$ (5)

$CD^{2} = CF^{2}+DF^{2}$ (6)

By (3), (4), (5) and (6):

$BC^{2} = BE^{2} + EC^{2}$ (7)

$CD^{2} = (EF-EC)^{2} + BE^{2}$ (8)

By (7) in (8):

$CD^{2} = (EF-EC)^{2}+ (BC^{2}-EC^{2})$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC + EC^{2}+BC^{2}-EC^{2}$

$CD^{2} = EF^{2}-2\cdot EF\cdot EC +BC^{2}$

Then, we clear $EC$:

$2\cdot EF\cdot EC = EF^{2} + BC^{2} - CD^{2}$

$EC = \frac{EF^{2}+BC^{2}-CD^{2}}{2\cdot EF}$

If we know that $EF = 3\sqrt{41}$, $BC = 15$ and $CD = 12$, then the length of the segment $EC$ is:

$EC = \frac{75\sqrt{41}}{41}$

And the length of the line segment $CF$ is:

$CF = EF - EC$

$CF = 3\sqrt{41}-\frac{75\sqrt{41}}{41}$

$CF = \frac{48\sqrt{41}}{41}$

And the length of the line segment $DF$ is determined by Pythagorean Theorem:

$FD = \sqrt{CD^{2}-CF^{2}}$

$FD = \sqrt{12^{2}-\left(\frac{48\sqrt{41}}{41} \right)^{2}}$

$FD = \frac{60\sqrt{41}}{41}$

And the area of the rectangle is determined by the following formula:

$A = FD\cdot EF$

$A = \left(\frac{60\sqrt{41}}{41} \right)\cdot (3\sqrt{41})$

$A = 180$

The area of rectangle BEFD is 180 square units.