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robinfr93
Can someone help me solve this....
|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...
Can you use that information to set up the two equations you need?
Ive tried but Im getting two second degree equations!!!
Actually I'm down with a terrible cold so my brain aint working so well.. So need a lil help!!!
Woah. Well, "The length is thrice the width" --> L=3W.
"Perimeter is 56" --> 2L+2W=56.
yah got that part ryt!!
Sub the first equation into the second to solve for W.
length = 18 units and breadth 6
18 and 6 won't get you a perimeter of 56.
ryt!! Sorry!!! 21 and 7!!
Excellent. Now you can just count over and down from the given point to find all the other coordinates.
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
I think i made some mistake somewhere. Would you be kind enough to help me out!! Its the darn cold!!
Forget about the distance formula! Just count.
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.
then?? Should I just put in simple increment formula and solve??
I know its suppose to be darn simple but my heads all frozen up!! @CliffSedge
|dw:1348872917396:dw|
|dw:1348872954175:dw| etc.
No formulas, no solving - elementary arithmetic.
k got it!! :D Thank man!!! Thanx a ton!! :P hahahahha!! Ryt!!