To find the value of $x$ for each rectangle, we use the formula for the area of a rectangle:
$$ \text{Area} = \text{length} \times \text{width} $$
We will set up an equation for each problem using the numbers given in the diagram and solve for $x$.
1)
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Given: Width = 5, Length = $x$, Area = 70
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Equation: $5 \cdot x = 70$
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Solve: Divide both sides by 5.
$$ x = \frac{70}{5} $$
$$ x = 14 $$
2)
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Given: Width = $x$, Length = 16, Area = 80
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Equation: $16 \cdot x = 80$
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Solve: Divide both sides by 16.
$$ x = \frac{80}{16} $$
$$ x = 5 $$
3)
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Given: Width = $\frac{3}{4}$, Length = $x$, Area = $\frac{9}{8}$
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Equation: $\frac{3}{4} \cdot x = \frac{9}{8}$
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Solve: Multiply both sides by the reciprocal of $\frac{3}{4}$, which is $\frac{4}{3}$.
$$ x = \frac{9}{8} \cdot \frac{4}{3} $$
$$ x = \frac{36}{24} $$
Simplify the fraction by dividing top and bottom by 12:
$$ x = \frac{3}{2} \text{ or } 1.5 $$
4)
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Given: Width = 1.2, Length = $x$, Area = 11.04
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Equation: $1.2 \cdot x = 11.04$
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Solve: Divide both sides by 1.2.
$$ x = \frac{11.04}{1.2} $$
To make it easier, multiply top and bottom by 100 to remove decimals: $\frac{1104}{120}$.
$$ 1104 \div 120 = 9.2 $$
$$ x = 9.2 $$
5)
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Given: Width = $\frac{2}{3}$, Length = $x$, Area = $\frac{8}{15}$
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Equation: $\frac{2}{3} \cdot x = \frac{8}{15}$
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Solve: Multiply both sides by the reciprocal of $\frac{2}{3}$, which is $\frac{3}{2}$.
$$ x = \frac{8}{15} \cdot \frac{3}{2} $$
$$ x = \frac{24}{30} $$
Simplify the fraction by dividing top and bottom by 6:
$$ x = \frac{4}{5} \text{ or } 0.8 $$
6)
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Given: Width = $x$, Length = $x + 2$, Area = 24
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Equation: $x(x + 2) = 24$
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Solve: Expand the left side: $x^2 + 2x = 24$.
Subtract 24 from both sides to form a quadratic equation: $x^2 + 2x - 24 = 0$.
Factor the quadratic equation. We need two numbers that multiply to -24 and add to 2. These numbers are +6 and -4.
$$ (x + 6)(x - 4) = 0 $$
So, $x = -6$ or $x = 4$. Since length cannot be negative, $x$ must be positive.
$$ x = 4 $$
*(Check: If $x=4$, width is 4 and length is $4+2=6$. Area $4 \cdot 6 = 24$. Correct.)*
7)
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Given: Width = $x$, Length = $x$, Area = 169
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Equation: $x \cdot x = 169$
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Solve: $x^2 = 169$. Take the square root of both sides.
$$ x = \sqrt{169} $$
$$ x = 13 $$
8)
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Given: Width = $x$, Length = $x + 3$, Area = 40
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Equation: $x(x + 3) = 40$
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Solve: Expand the left side: $x^2 + 3x = 40$.
Subtract 40 from both sides: $x^2 + 3x - 40 = 0$.
Factor the quadratic equation. We need two numbers that multiply to -40 and add to 3. These numbers are +8 and -5.
$$ (x + 8)(x - 5) = 0 $$
So, $x = -8$ or $x = 5$. Since length cannot be negative, $x$ must be positive.
$$ x = 5 $$
*(Check: If $x=5$, width is 5 and length is $5+3=8$. Area $5 \cdot 8 = 40$. Correct.)*
Final Answer:
1) x = 14
2) x = 5
3) x = 3/2 (or 1.5)
4) x = 9.2
5) x = 4/5 (or 0.8)
6) x = 4
7) x = 13
8) x = 5
Parent Tip: Review the logic above to help your child master the concept of 7th grade printables.
