The perimeter of a rectangle is 24 ft. Its length is five times its width Let x be the length and y be the width. What is the area of the rectangle?

2 Answers
Jan 30, 2016

Answer:

#x=10" " y= 2" so area "=20 ft^2#

Explanation:

#color(brown)("Build the equation by breaking down the question into parts")#

Perimeter is #24 ft#

Width #->yft#
Length #"-> xft#

So perimeter is #->(2y+2x) ft=24 ft# ......................(1)

But Length = #5xx# width

So #x=5y# ..................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find y")#

Substitute equation (2) into equation (1) giving:

#2y+2(5y)=24#

#12y=24#

#color(brown)("y=2")# .....................................(3)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("To find x")#

Substitute (3) into (2) giving:

#color(brown)("x=5(2) -> x=10)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check
#2x+2y = 24 " "=>" " 2(10)+2(2)=24 " "color(red)("True")#

Jan 31, 2016

Answer:

#x=10,y=2#

#Area=20ft^2#

Explanation:

In the statement length is said to be #x# which is #5# times the width which is #y#

So,

#5x=y#

We can assume #y# as #x#

Then perimeter=#(5x+5x)+(x+x)=24#

#rarr10x+2x=24#

#rarr12x=24#

#rarrx=24/12=2#

So,the width=#2#

Then length=#2(5)=10#

Area of rectangle =#Leng##th*width=10(2)=20ft^2#