## anonymous one year ago In trapezoid PQRS, PQ is parallel to SR. What is the area of PQRS in simplest radical form?

1. anonymous

2. anonymous

i got the height as 16 ?? idk how to get the bottom base can someone help

3. jim_thompson5910

|dw:1433379207636:dw|

4. jim_thompson5910

|dw:1433379358414:dw| you have the correct height, TQ = 16

5. jim_thompson5910

what is the distance from T to R?

6. anonymous

i got 27.71281292 ??

7. jim_thompson5910

triangle TRQ is a 30-60-90 triangle, so you'll find that TQ = 16 $$\large TR = 16\sqrt{3}$$ |dw:1433380082720:dw|

8. jim_thompson5910

now construct a perpendicular segment to PQ that goes through point S |dw:1433380133685:dw| mark the new point point U as shown above

9. jim_thompson5910

angle UST = 90 degrees since I made SU perpendicular to PQ |dw:1433380192075:dw| so that means angle PSU must be 45 degrees (45+90 = 135)

10. jim_thompson5910

|dw:1433380223132:dw|

11. jim_thompson5910

agreed so far?

12. anonymous

ya

13. jim_thompson5910

TQ = SU since SUQT is a rectangle so SU = 16 SU = PU due to the fact that PSU is a 45-45-90 triangle PU = 16 as well

14. jim_thompson5910

|dw:1433380378598:dw|

15. jim_thompson5910

|dw:1433380396688:dw|

16. jim_thompson5910

UQ = 8 and ST = 8 because UQ = ST |dw:1433380435649:dw|

17. jim_thompson5910

So all of that allows us to conclude that $\Large SR = 8+16\sqrt{3} \approx 35.71281$

18. anonymous

for the area i got $A \approx 477.7$

19. anonymous

in the simplest radical form it would be 256+128sqrt3 ?

20. jim_thompson5910

yep $\Large 256+128\sqrt{3}$ is the answer they want

21. jim_thompson5910

and $\Large 256+128\sqrt{3} \approx 477.7025034$ is the approximate area

22. anonymous

okay!! thank you so much :D

23. jim_thompson5910

yw