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Question
the length of a rectangle is 5 inches longer than it is wide. if the area is 66 square inches dimensions of the rectangle? the width, or shorter side is inches the length, or longer side is inches add work > next question
Step1: Define variables
Let the width of the rectangle be \( x \) inches. Then the length is \( x + 5 \) inches (since length is 5 inches longer than width).
Step2: Use area formula
The area of a rectangle is given by \( A=\text{length}\times\text{width} \). We know the area \( A = 66 \) square inches. So we set up the equation:
\( x(x + 5)=66 \)
Step3: Expand and simplify
Expand the left - hand side: \( x^{2}+5x = 66 \)
Rearrange it to a quadratic equation: \( x^{2}+5x - 66=0 \)
Step4: Factor the quadratic equation
We need to find two numbers that multiply to \( - 66 \) and add up to \( 5 \). The numbers are \( 11 \) and \( - 6 \) (since \( 11\times(-6)=-66 \) and \( 11+( - 6)=5 \)).
So the factored form is \( (x + 11)(x - 6)=0 \)
Step5: Solve for x
Set each factor equal to zero:
- \( x+11 = 0\) gives \( x=-11 \). But the width cannot be negative, so we discard this solution.
- \( x - 6=0\) gives \( x = 6 \)
Step6: Find the length
Since the length is \( x + 5 \), substitute \( x = 6 \) into it. Then length \(=6 + 5=11 \) inches.
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The width, or shorter side is \( \boldsymbol{6} \) inches.
The length, or longer side is \( \boldsymbol{11} \) inches.