## anonymous 5 years ago 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

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

$\sin45=\sqrt{2}/2$

5. anonymous

do anothe way , this take to long

6. anonymous

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

7. anonymous

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

sorry square 36=6

12. anonymous

base =6

13. anonymous

i tought the base is 6

14. anonymous

yes base =6 I forget after square36=6

15. anonymous

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

16. anonymous

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

17. anonymous

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

18. anonymous

A=(b*h)/2

19. anonymous

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

20. anonymous

thank u

21. anonymous

P=20.49