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?___ feet by ___ feet
Perimeter Formula: Let l be the length and w be the width of the map.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.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:1. 10=l+w2. 21=lwWe 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−l2.
Rearrange Equation: Rearrange the equation to form a quadratic equation: l2−10l+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.For l=7 and w=3: P=2(7)+2(3)=14+6=20 and A=7×3=21.For l=3 and w=7: P=2(3)+2(7)=6+14=20 and A=3×7=21.Both solutions satisfy the original equations.
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