robinfr93 3 years ago Can someone help me solve this....

1. robinfr93

|dw:1348871903875:dw| If length of this rectangle be thrice its width and its perimeter be 56, then find the co ordinates of the rest of the sides...

2. CliffSedge

Can you use that information to set up the two equations you need?

3. robinfr93

Ive tried but Im getting two second degree equations!!!

4. robinfr93

Actually I'm down with a terrible cold so my brain aint working so well.. So need a lil help!!!

5. CliffSedge

Woah. Well, "The length is thrice the width" --> L=3W.

6. CliffSedge

"Perimeter is 56" --> 2L+2W=56.

7. robinfr93

yah got that part ryt!!

8. CliffSedge

Sub the first equation into the second to solve for W.

9. robinfr93

length = 18 units and breadth 6

10. CliffSedge

18 and 6 won't get you a perimeter of 56.

11. robinfr93

:P

12. robinfr93

ryt!! Sorry!!! 21 and 7!!

13. CliffSedge

Excellent. Now you can just count over and down from the given point to find all the other coordinates.

14. robinfr93

yeah!! But when I try to use the distance formula!!! Im getting two second degree equations!! How m I suppose to solve them when ive got two equation something like ax^2+b^y2+cx+dy+e=0

15. robinfr93

I think i made some mistake somewhere. Would you be kind enough to help me out!! Its the darn cold!!

16. CliffSedge

Forget about the distance formula! Just count.

17. CliffSedge

The distance formula is so overrated. It is merely the pythagorean theorem, but when you have horizontal and vertical lines, you don't need to solve for a hypotenuse.

18. robinfr93

then?? Should I just put in simple increment formula and solve??

19. robinfr93

I know its suppose to be darn simple but my heads all frozen up!! @CliffSedge

20. CliffSedge

|dw:1348872917396:dw|

21. CliffSedge

|dw:1348872954175:dw| etc.

22. CliffSedge

No formulas, no solving - elementary arithmetic.

23. robinfr93

k got it!! :D Thank man!!! Thanx a ton!! :P hahahahha!! Ryt!!