## Dallasb22 3 years ago AD=4√3 in.; AB = 8 in. Area =? times the square root of ? sq. in.

1. Dallasb22

2. Dallasb22

@timo86m Help with mine?

3. timo86m

do you know trig? or does this only require pythogoras theorem?

4. Dallasb22

It should only require Pythagoras theorem. It is merely Geometry.

5. Dallasb22

But the Sq Rt is what confuses me.

6. Dallasb22

I tried using the 30-60-90 ratio and i got the answer: 2 times sqrt of 1.5 sq in which is wrong.

7. timo86m

|dw:1346873004579:dw|

8. Dallasb22

Don't i take half of 4√3 since its the hypotunese?

9. timo86m

|dw:1346873167383:dw| once you have a then a*2 would be height

10. Dallasb22

And how do i find A? A is what?

11. timo86m

a is 4 i think the height is 8

12. Dallasb22

SO it would be area= 8 times the sqrt of 3?

13. timo86m

then i think it is just 8*8 the area Something seems odd tho. THe figure is not to scale or i am wrong.

14. timo86m

i did it wrong

15. timo86m

all you have to do is find height then it is simply 8*h

16. Dallasb22

But how will that get me a sq rt?

17. timo86m

i found where i gone wrong lol

18. Dallasb22

Lol

19. timo86m

|dw:1346874388605:dw|

20. timo86m

a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?

21. Dallasb22

The height is √3 correct?

22. timo86m

a=2*sqrt(3) So h = 2*sqrt(3)*sqrt(3) and A=h*b can you do that?

23. Dallasb22

19-3= 16 times √3

24. Dallasb22

Thanks man!

25. Dallasb22

I got it.

26. timo86m

:)