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The area of a rectangle is 45f * t ^ 2 , and the length of the rectangle is 1 ft more than twice the width. Find the dimensions of the rectangle.

The area of a rectangle is 45f * t ^ 2 , and the length of the rectangle is 1 ft more-example-1
User Kaa
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5 votes

ANSWER

The width of the rectangle is 4.5ft

The length of the rectangle is 10ft

Step-by-step explanation:

Given information


\begin{gathered} \text{ Area of a rectangle = 45ft}^2 \\ \text{ Length of a rectangle = 1 more than twice the width} \end{gathered}

To find the dimension of the rectangle, follow the steps below

Step 1: Write the formula for calculating the area of a rectangle


\text{ Area of a rectangle = length }*\text{ width}

Step 2: Express the length of the rectangle in algebraic form

The length of the rectangle is 1 more than twice the width

Algebraically,

l = 2w + 1

l = 2w + 1

Step 3: Substitute the given data into the formula in step 1


\begin{gathered} \text{ Area of the rectangle = length }*\text{ width} \\ \text{ Area = 45ft}^2 \\ \text{ 45 = \lparen2w + 1\rparen}*\text{ w} \\ \text{ open the parenthesis} \\ \text{ 45 = 2w}^2\text{+ w} \\ \text{ 2w}^2\text{ + w - 45 = 0} \end{gathered}

Step 4: Solve the quadratic function in step 3 using the general method


\begin{gathered} \text{ The general form of the quadratic method is written below as} \\ \text{ ax}^2\text{ + bx + c = 0} \\ \text{ Since the quadratic function is 2w}^2\text{ + w - 45} \\ \text{ Relating the two functions, we have} \\ \text{ a= 2} \\ b\text{ = 1} \\ \text{ c =- 45} \\ \text{ } \end{gathered}
\begin{gathered} \text{ The general formula is } \\ \text{ x= }\frac{-b\text{ }\pm\sqrt{b^2\text{ - 4ac}}}{2a} \\ x\text{ = w} \\ \text{ w = }\frac{-(1)\pm\sqrt{1^2\text{ - 4}*2*(-45)}}{2*2} \\ w\text{ = }\frac{-1\text{ }\pm\sqrt{1^2\text{ -\lparen-360\rparen }}}{4} \\ w\text{ = }\frac{-1\pm\sqrt{1\text{ + 360}}}{4} \\ w\text{ = }(-1\pm√(361))/(4) \\ w\text{ = }(-1\pm19)/(4) \\ \text{ w =}\frac{-1\text{ + 19}}{4}\text{ or }\frac{-1\text{ - 19}}{4} \\ w\text{ = }(18)/(4)\text{ or }(-20)/(4) \\ \text{ w = 4.5 ft or -5ft} \end{gathered}

Hence, the width of the rectangle is 4.5ft

Step 5: Find the length of the width by putting w = 4.5


\begin{gathered} \text{ Recall, that } \\ \text{ l = 2w + 1} \\ \text{ w= 4.5} \\ \text{ l = 2\lparen4.5\rparen + 1} \\ l\text{ = 9 + 1} \\ \text{ l = 10 ft} \end{gathered}

Hence, the length of the rectangle is 10ft

User Inthu
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