In the figure shown above, lengths AB, BD, and CD are all units. Angles A and C are both 45°. What is the perimeter of ABCD? What is the area of ABCD? i am going to attach the figure

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In the figure shown above, lengths AB, BD, and CD are all units. Angles A and C are both 45°. What is the perimeter of ABCD? What is the area of ABCD? i am going to attach the figure

Mathematics
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1 Attachment
12 +6sqrt(2) is perimiter, 18 is area
fisrt find the high \[\sin45=h/3\sqrt{2}\]

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Other answers:

\[\sin45=\sqrt{2}/2\]
do anothe way , this take to long
AD= \[(3\sqrt{2})^{2}+(3\sqrt{2})^{2}\]
AD=(9*2)+(9*2)=36
i tried using the special right triangle 45-45-90. for the perimenter i got 6 sqart(2)
\[p=( 36+3\sqrt{2})*2\]
Area: Use special triangles to notice that they are right triangles, then b*h/2 = 3sqrt(2)*3sqrt(2)/2 for each, and there are two, so 18 Perimeter: 3sqrt(2) *sqrt(2) =6 for the top and bottom, then add em up
sorry square 36=6
base =6
i tought the base is 6
yes base =6 I forget after square36=6
so the area is 18sqt(2). is it
now you have the base you can solve for Perimeter and area
\[P=(6+3\sqrt{2})*2\]
A=(b*h)/2
A=(3*6)/2 =9 2 triangle than (9*2)=18 A=18
thank u
P=20.49

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