A map of the world has a perimeter of $20$ feet. Its area is $21$ square feet. What are the dimensions of the map? $\newline$ $\_\_\_$ feet by $\_\_\_$ feet

View step-by-step help

Home

Math Problems

Algebra 1

Area and perimeter: word problems

Full solution

Q. A map of the world has a perimeter of

20

feet. Its area is

21

square feet. What are the dimensions of the map?

\newline

\_\_\_

feet by

\_\_\_

feet

Perimeter Formula: Let $l$ be the length and $w$ be the width of the map. $\newline$ The perimeter of a rectangle is given by the formula $P = 2l + 2w$ .
Perimeter Equation: Given the perimeter of the map is $20$ feet, we can write the equation $20 = 2l + 2w$ .
Simplify Equation: Simplify the perimeter equation by dividing all terms by $2$ to get $10 = l + w$ .
Area Formula: The area of a rectangle is given by the formula $A = lw$ . $\newline$ Given the area of the map is $21$ square feet, we can write the equation $21 = lw$ .
System of Equations: We now have a system of two equations with two variables: $\newline$ $1$ . $10 = l + w$ $\newline$ $2$ . $21 = lw$ $\newline$ We can solve this system by expressing one variable in terms of the other using the first equation and then substituting into the second equation.
Express Width in Terms of Length: From the first equation, express $w$ in terms of $l$ : $w = 10 - l$ .
Substitute Width into Equation: Substitute $w = 10 - l$ into the second equation: $21 = l(10 - l)$ .
Expand Equation: Expand the equation: $21 = 10l - l^2$ .
Rearrange Equation: Rearrange the equation to form a quadratic equation: $l^2 - 10l + 21 = 0$ .
Factor Quadratic Equation: Factor the quadratic equation: $(l - 7)(l - 3) = 0$ .
Solve for Length: Solve for $l$ : $l = 7$ or $l = 3$ . Since $l$ and $w$ are interchangeable as length and width, we can have two solutions: $l = 7$ and $w = 3$ , or $l = 3$ and $w = 7$ .
Check Solutions: Check the solutions with the original equations to ensure they satisfy both the perimeter and area. $\newline$ For $l = 7$ and $w = 3$ : $P = 2(7) + 2(3) = 14 + 6 = 20$ and $A = 7 \times 3 = 21$ . $\newline$ For $l = 3$ and $w = 7$ : $P = 2(3) + 2(7) = 6 + 14 = 20$ and $A = 3 \times 7 = 21$ . $\newline$ Both solutions satisfy the original equations.