Question

The width of a rectangle is 3 cm longer than it is wide. The area of the rectangle is 550 cm2. Find the dimensions of the rectangle.

Answer 1

Answer:

Step-by-step explanation:

Length rectangle = 3cm.  longer than its width.

Rectangle perimeter = 56 cm.

find its dimensions?

P = 2L +2w

L= w + 3

P= 56

P  = 2L +2w

56 = 2(w+3) +2(w)

56 = 2w +6  + 2w

4w= 56  - 6  

4w = 50        

w = 50/4  

w = 12.5  

L= 12.5 +3 = 15.5

L=  15.5

 Solution:

Length= 15.5 cm

Width = 12.5 cm

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Answer 2

AnswEr :

$\bf{\pink{\underline{\underline{\bf{Given\::}}}}}$

The length of a rectangle is 3 cm longer than it's wide. The area of the rectangle is 550 cm².

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The dimensions of the rectangle.

$\bf{\purple{\underline{\underline{\bf{Explanation\::}}}}}$

Let the breadth of rectangle be r cm

Let the Length of rectangle be r+3 cm

Formula use :

$\bf{\large{\boxed{\bf{Area\:of\:rectangle\:=\:Length \times breadth}}}}}}$

A/q

$\hookrightarrow\tt{(r+3)(r)=550}\\\\\\\hookrightarrow\tt{r^{2} +3r=550}\\\\\\\hookrightarrow\tt{\red{r^{2} +3r-550=0}}$

$\bf{\large{\underline{\sf{\dag{Using\:Quadratic\:Formula\::}}}}}$

Above, given quadratic equation as we compared with ax² + bx + c = 0.

• a = 1
• b = 3
• c = -550

Then;

$\mapsto\tt{x=\dfrac{-b\pm\sqrt{b^{2}-4ac } }{2a} }}}\\\\\\\\\mapsto\tt{x=\dfrac{-3\pm\sqrt{(3)^{2} -4*1*(-550)} }{2*1} }}\\\\\\\\\mapsto\tt{x=\dfrac{-3\pm\sqrt{9-4*(-550)} }{2} }}\\\\\\\\\mapsto\tt{x=\dfrac{-3\pm\sqrt{9+2200} }{2} }}\\\\\\\\\mapsto\tt{x=\dfrac{-3\pm\sqrt{2209} }{2} }}\\\\\\\\\mapsto\tt{x=\dfrac{-3\pm47}{2} }}\\\\\\\\\mapsto\tt{x=\dfrac{-3+47}{2} \:\:\:Or\:\:\:\dfrac{-3-47}{2} }}\\\\\\\\\mapsto\tt{x=\cancel{\dfrac{44}{2}} \:\:\:Or\:\:\:\cancel{\dfrac{-50}{2}} }}}$

$\mapsto\tt{\purple{x=22\:\:\:\:\:Or\:\:\:\:x=-25}}$

We know that negative value isn't acceptable.

Thus,

$\underbrace{\sf{The\:breadth\:of\:rectangle\:(b)=r=22\:cm}}}\\\\\underbrace{\sf{The\:length\:of\:rectangle\:(l)=3+r=3+22=25\:cm}}}$