## anonymous 4 years ago the length of a rectangle box is one inch greater than its width. the height is three times the length. the diagonal box exceeds the height by one inch. Find the volume of the box.

1. Mertsj

|dw:1328539986535:dw|

2. Mertsj

$AB=\sqrt{w^2+(w+1)^2}=\sqrt{2w^2+2w+1)}$

3. Mertsj

$AC=\sqrt{2w^2+2w+1+9(w+1)^2}=\sqrt{11w^2+20w+10}$

4. Mertsj

Now AC is one more than the height which is 3w+3 so AC = 3w+4

5. Mertsj

$\sqrt{11w^2+20w+10}=3w+4$

6. Mertsj

Let's square both sides.

7. anonymous

now im really confused lol

8. Mertsj

9. anonymous

uhhh everything lol

10. anonymous

like idk what the main answer is lol

11. Mertsj

That might be because we haven't found the answer yet. We have to find w based on the information they gave us. Then we will be able to find the dimensions of the box and then the volume.

12. Mertsj

As i said, let's square both sides of the equation.

13. Mertsj

$11w^2+20w+10=9w^2+24w+16$

14. anonymous

is 16 the square root ?

15. Mertsj

$2w^2-4w-6=0$

16. Mertsj

Divide by 2

17. Mertsj

$w^2-2w-3=0$

18. Mertsj

Factor

19. Mertsj

$(w-3)(w+1)=0$

20. Mertsj

$w=3 or w=-1$

21. Mertsj

Discard -1 since the width cannot be negative. So w = 3. That means : |dw:1328541450706:dw|

22. Mertsj

so the box is 3 by 4 by 12 and the volume is 3(4)(12) or 144 cubic inches