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Mathematics 10 Online
OpenStudy (anonymous):

Can someone help me solve this....

OpenStudy (anonymous):

|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...

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

yah got that part ryt!!

OpenStudy (anonymous):

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

OpenStudy (anonymous):

length = 18 units and breadth 6

OpenStudy (anonymous):

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

OpenStudy (anonymous):

:P

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Forget about the distance formula! Just count.

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

|dw:1348872917396:dw|

OpenStudy (anonymous):

|dw:1348872954175:dw| etc.

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