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?

Question

8838 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-30T15:39:55

$x = 10 y = 2 so area = 20 f t^{2}$ $x=10" " y= 2" so area "=20 ft^2$

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

Perimeter is $24 f t$ $24 ft$

Width $\to y f t$ $->yft$
Length $\to x f t$ $"-> xft$

So perimeter is $\to (2 y + 2 x) f t = 24 f t$ $->(2y+2x) ft=24 ft$ ......................(1)

But Length = $5 \times$ $5xx$ width

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

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

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

$12 y = 24$ $12y=24$

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

Substitute (3) into (2) giving:

$x = 5 (2) \to x = 10$ $color(brown)("x=5(2) -> x=10)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Check
$2 x + 2 y = 24 \Rightarrow 2 (10) + 2 (2) = 24 True$ $2x+2y = 24 " "=>" " 2(10)+2(2)=24 " "color(red)("True")$

Answer 2 · 2016-01-31T08:52:12

$x = 10, y = 2$ $x=10,y=2$

$A r e a = 20 f t^{2}$ $Area=20ft^2$

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

So,

$5 x = y$ $5x=y$

We can assume $y$ $y$ as $x$ $x$

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

$\to 10 x + 2 x = 24$ $rarr10x+2x=24$

$\to 12 x = 24$ $rarr12x=24$

$\to x = \frac{24}{12} = 2$ $rarrx=24/12=2$

So,the width= $2$ $2$

Then length= $2 (5) = 10$ $2(5)=10$

Area of rectangle = $L e n g$ $Leng$ $t h \cdot w i d t h = 10 (2) = 20 f t^{2}$ $th*width=10(2)=20ft^2$