How do you find the dimensions of a rectangle if the area is $63 c m^{2}$ $63cm^2$ and the perimeter is $32 c m$ $32cm$ where $x$ $x$ represents the width of the rectangle?

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Answer 1 · 2017-04-03T08:06:31

Width = $7 c m$ $7cm$ ; Length = $9 c m$ $9cm$

let $x$ $color(red)x$ be the width and $y$ $color(red)y$ be the length of the rectangle.

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Given
$a r e a$ $area$ $= x \cdot y = 63$ $= x*y = 63$ ----------- $(1)$ $(1)$

$p e r i m e t e r$ $perimeter$ $= x + y + x + y = 2 x + 2 y = 32$ $=x+y+x+y = 2x+2y = 32$ ---------- $(2)$ $(2)$

from $(1), x = \frac{63}{y}$ $(1), x=63/y$

substituting this in $(2),$ $(2),$

$\frac{63}{y} + y = 16$ $63/y+y=16$

$\Rightarrow y^{2} - 16 y + 63 = 0$ $=>y^2-16y+63=0$

$\Rightarrow y^{2} - 7 \cdot y - 9 \cdot y + 63 = 0$ $=> y^2-7*y-9*y+63=0$

$\Rightarrow y (y - 7) - 9 (y - 7) = 0$ $=>y(y-7)-9(y-7)=0$

$\Rightarrow (y - 9) (y - 7) = 0$ $=>(y-9)(y-7)=0$

$\Rightarrow$ $=>$ either $y - 9 = 0$ $y-9=0$ i.e. $y = 9$ $y=9$

or $y - 7 = 0$ $y-7=0$ i.e. $y = 7$ $y=7$

therefore $x = \frac{63}{9} = 7$ $x=63/9=7$ or $x = \frac{63}{7} = 9$ $x=63/7=9$

Hence, width = $7 c m$ $7cm$ ; length = $9 c m$ $9cm$

How do you find the dimensions of a rectangle if the area is 63cm263cm^2 and the perimeter is 32cm32cm where xx represents the width of the rectangle?