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The length of a rectangle is 8 more than twice it's width.If the area of the rectangle is 24 sq. meters,find its dimensions​

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KAntoineDoixidn

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[tex] \large \bold{\blue{PROBLEM:}} [/tex]The length of a rectangle is 8 more than twice it's width.If the area of the rectangle is 24 sq. meters, find its dimensions.

[tex] \large \bold{\blue{SOLUTION:}} [/tex] Represent 'l' and 'w' as the length and the width respectively. Make equations on the given statements and use the formula on how to find the area of a rectangle.

  • [tex] \begin{cases} l = 2w + 8 \\ 24 = l \cdot w \end{cases} \: \begin{align} \red{(eq. \: 1)} \\ \red{(eq. \: 2)} \end{align} [/tex]

» Substitute the value of 'l' in terms of 'w' into the second equation.

  • [tex] \begin{cases} l = 2w + 8 \\ 24 = (2w + 8) \cdot w \end{cases} [/tex]

  • [tex] \begin{cases} l = 2w + 8 \\ 24 = 2w^2 + 8w \end{cases} [/tex]

  • [tex] \begin{cases} l = 2w + 8 \\ 2w^2 + 8w - 24 = 0 \end{cases} [/tex]

» Solve the second equation using the quadratic formula. Make sure that the solution has the positive result.

  • [tex] w = \frac{-8 \pm \sqrt{8^2 - 4(2)(-24)}}{2(2)} \\ [/tex]

  • [tex] w = \frac{-8 \pm \sqrt{64 + 192}}{4} \\ [/tex]

  • [tex] w = \frac{-8 \pm \sqrt{256}}{4} \\ [/tex]

  • [tex] w = \frac{-8 \pm 16}{4} \\ [/tex]

  • [tex] w = \frac{-8 + 16}{4} \\ [/tex]

  • [tex] w = \frac{8}{4} \\ [/tex]

  • [tex] w = 2 [/tex]

  • [tex] \begin{cases} l = 2w + 8 \\ w = 2 \end{cases} [/tex]

» After finding the length of the width, substitute it to the first equation to find the length.

  • [tex] \begin{cases} l = 2(2) + 8 \\ w = 2\end{cases} [/tex]

  • [tex] \begin{cases} l = 4 + 8 \\ w = 2 \end{cases} [/tex]

  • [tex] \begin{cases} l = 12 \\ w = 2 \end{cases} [/tex]

[tex] \large \therefore \underline{\boxed{\tt \purple{l = 12 \: meters, \: w = 2 \: meters}}} [/tex]

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[tex] \huge{\bold{ Answer: }}[/tex]

[tex] \bold{length = 12m} \\ \bold{width = 2m}[/tex]

Solution:

We know that :

  • Area of rectangle = length × width

Given :

  • Area of the rectangle = 24 m[tex]^{2}[/tex]
  • The length of a rectangle is 8 more than twice it's width.

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Let x be the width of rectangle

Therefore :

[tex] Length = 2x + 8 \\ x(2x + 8) = 24 \\ 2x^{2} + 8x = 24 \\ 2x^{2} + 8x - 24 = 0 \\ x^{2} + 4x - 12 = 0 \\ (x+6)(x-2)=0 \\ (x+6)=0 \: or \: (x=2)=0 \\ x=-6 \: or \: x=2[/tex]

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Hence width = 2m

Length = 2 × 2 + 8 = 4 + 8 = 12m

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