Question

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

Answer 1

12 +6sqrt(2) is perimiter, 18 is area

Answer 2

fisrt find the high
\[\sin45=h/3\sqrt{2}\]

Answer 3

\[\sin45=\sqrt{2}/2\]

Answer 4

do anothe way , this take to long

Answer 5

AD=
\[(3\sqrt{2})^{2}+(3\sqrt{2})^{2}\]

Answer 6

AD=(9*2)+(9*2)=36

Answer 7

i tried using the special right triangle 45-45-90. for the perimenter i got 6 sqart(2)

Answer 8

\[p=( 36+3\sqrt{2})*2\]

Answer 9

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

Answer 10

sorry square 36=6

Answer 11

base =6

Answer 12

i tought the base is 6

Answer 13

yes base =6 I forget after square36=6

Answer 14

so the area is 18sqt(2). is it

Answer 15

now you have the base you can solve for Perimeter and area

Answer 16

\[P=(6+3\sqrt{2})*2\]

Answer 17

A=(b*h)/2

Answer 18

A=(3*6)/2 =9
2 triangle than (9*2)=18
A=18

Answer 19

thank u

Answer 20

P=20.49