Q. The perimeter of an envelope is 28 inches. The area is 40 square inches. What are the dimensions of the envelope?____ inches by ____ inches
Define Variables: Let's denote the length of the envelope as l inches and the width as w inches.The perimeter (P) of a rectangle is given by the formula P=2l+2w.
Perimeter Equation: Given the perimeter of the envelope is 28 inches, we can write the equation 2l+2w=28.
Simplify Equation: We can simplify this equation by dividing all terms by 2, which gives us l+w=14.
Area Formula: The area A of a rectangle is given by the formula A=lw.
Area Equation: Given the area of the envelope is 40 square inches, we can write the equation lw=40.
Solve System of Equations: Now we have a system of two equations with two variables:1. l+w=142. lw=40We 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 w in terms of l: Let's express w in terms of l from the first equation: w=14−l.
Substitute into Equation: Substitute w=14−l into the second equation lw=40:l(14−l)=40
Expand Equation: Expand the equation: 14l−l2=40.
Rearrange Equation: Rearrange the equation to form a quadratic equation: l2−14l+40=0.
Factor Quadratic Equation: Factor the quadratic equation: l - \(10)(l - 4) = 0\
Solve for l: Set each factor equal to zero and solve for l:1.l−10=0 gives l=102.l−4=0 gives l=4
Find Solutions: Since l and w are interchangeable as length and width, we can have two solutions:1. l=10 inches and w=4 inches2. l=4 inches and w=10 inchesBoth sets of dimensions satisfy the given perimeter and area.
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