## elisichi 2 years ago Could someone help me with the following question?

1. elisichi

Here's the question.. I have no idea on what to do!

2. zordoloom

Do you need help?

3. elisichi

I kind of do..

4. amistre64

|dw:1353507196703:dw| i might be thinking this to hard, but id say we would need to define the volume of water that can be collected in a year

5. amistre64

|dw:1353507466106:dw|

6. elisichi

I don't get what youre doing :/

7. amistre64

im trying to determine the volume of water that the roof collects by finding the area along the edge of the roof that would collect 50cm of water a year

8. amistre64

then using similar triangles i can add up all the areas along the roof line to cover 7 meters and then multiply it by 12 to get the total volume on one side of the roof

9. amistre64

|dw:1353507706506:dw| but it loks like the area im looking for is 1/4 the area of the bigger triangle

10. amistre64

and 4 of those rake down the roof, which means the area of the big triangle equals the area of the rake .... area of a triangle is base times height divided by 2

11. amistre64

Area = sqrt(48)/2 times 12 is: 6 sqrt(48) 6 sqrt(4.4.3) 24 sqrt(3) is the total volume of water that one side of the roof collects during the year

12. amistre64

how many cubed meters are in a liter?

13. phi

Imagine all the rain fell at once |dw:1353508016540:dw|

14. amistre64

the rest of it seems pretty basic, except for the last one

15. amistre64

phi, should we assume its asking for total roof, or just the side thats showing?

16. phi

good question. I initially focused on 1/2 the roof (due to the picture), but the roof is both sides... I hate trying to be a mind-reader for people that make up these questions

17. amistre64

indeed, but then i got thinking (since a gable end) that the only rain that they barrel catches is the part that falls down the one side ....

18. phi

sounds reasonable. 1/2 the roof it is!

19. elisichi

Im confused..

20. amistre64

|dw:1353508980599:dw| $V=12\int_{0}^{\sqrt{48}}{(1.5-\frac{x}{\sqrt{48}})-(1-\frac{x}{\sqrt{48}})}~dx$ $12\int_{0}^{\sqrt{48}}{.5}~dx=12(.5\sqrt{48})$ $V=6\sqrt{48}$ lol :)