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Question
determining characteristics of a rectangle
which statement is true of a rectangle that has an area of 4x² + 39x - 10 square units and a width of (x + 10) units?
the perimeter of the rectangle is (10x + 18) units.
the rectangle is a square.
the area of the rectangle can be represented by(4x² + 20x - 2x - 10) square units.
the rectangle has a length of (2x - 5) units.
Step1: Recall area formula for rectangle
Area \(A = l\times w\), where \(A = 4x^{2}+39x - 10\) and \(w=x + 10\). We need to find \(l\) by dividing \(A\) by \(w\), i.e., \(l=\frac{4x^{2}+39x - 10}{x + 10}\).
Step2: Factor the numerator
We factor \(4x^{2}+39x - 10\). We need two numbers that multiply to \(4\times(- 10)=-40\) and add up to \(39\). The numbers are \(40\) and \(-1\). So \(4x^{2}+39x - 10=4x^{2}+40x - x - 10=4x(x + 10)-(x + 10)=(4x - 1)(x + 10)\).
Step3: Calculate the length
\(l=\frac{(4x - 1)(x + 10)}{x + 10}=4x-1\). Now let's check each statement:
- Perimeter \(P = 2(l + w)=2((4x-1)+(x + 10))=2(5x + 9)=10x+18\).
- For the rectangle to be a square, \(l = w\), but \(4x-1
eq x + 10\) (solving \(4x-1=x + 10\) gives \(3x=11\), \(x=\frac{11}{3}\), generally not true always).
- The area is given as \(4x^{2}+39x - 10\) not \(4x^{2}+20x-2x - 10\).
- The length is not \(2x - 5\) as we found it to be \(4x-1\).
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The perimeter of the rectangle is \((10x + 18)\) units.