Question

the area of a rectangle is 54m square whose length is 3m more than its width. Find the dimension of the rectangle

Answer 1

Let width be x units and length be (x + 3) units

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Area = 54

Length×breadth = 54

x(x + 3) = 54

x² + 3x = 54

x² + 3x - 54 = 0

x² + (9 - 6)x - 54 = 0

x² + 9x - 6x - 54 = 0

x(x + 9) - 6(x + 9) = 0

(x + 9)(x - 6) = 0

x= -9 or x = 6

Taking positive value, x= 6

×××××××××××××××××××××


then,

breadth = x => 6
Length = x + 3 = 6 + 3 = 9




I hope this will help you

(-:

Answer 2

$hi \: ..$
$here's \: \: your \: \: answer$

Let the width of the rectangle be x.

So,
By the condition ,
Length = x + 3

Area of rectangle = 54 sq .m

$we \: \: \: know \: \: \: that \: ....$

$area \: of \: rectangle \: = l \times b$

$54 = x(x + 3)$

$54 = {x}^{2} + 3x$

${x}^{2} + 3x - 54 = 0$

${x}^{2} + 9x - 6x - 54 = 0$

$x(x + 9) - 6(x + 9) = 0$

$(x + 9)(x - 6) = 0$

So,

$x + 9 = 0 \: \: or \: \: x - 6 = 0$

$x = - 9 \: \: \: or \: \: \: x = 6$

But , x = -9 is not acceptable .

Thus ,
x = 6
Breadth = 6m

Length = x + 3
Length = 6 + 3
Length = 9

Therefore ,
The dimensions of the rectangle are 9m & 6m respectively .

$hope \: \: \: this \: \: \: helps.$