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The perimeter of a playing card is 3030 centimeters. The area is 5454 square centimeters. What are the dimensions of the playing card?\newline___\_\_\_ centimeters by ___\_\_\_ centimeters

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Q. The perimeter of a playing card is 3030 centimeters. The area is 5454 square centimeters. What are the dimensions of the playing card?\newline___\_\_\_ centimeters by ___\_\_\_ centimeters
  1. Define Card Dimensions: Let the length of the playing card be ll centimeters and the width be ww centimeters.\newlineThe perimeter of a rectangle is given by the formula P=2l+2wP = 2l + 2w.
  2. Perimeter Equation: Given the perimeter of the playing card is 3030 centimeters, we can write the equation:\newline30=2l+2w30 = 2l + 2w
  3. Simplify Perimeter: Simplify the perimeter equation by dividing all terms by 22 to make it easier to work with:\newline15=l+w15 = l + w
  4. Area Calculation: The area of a rectangle is given by the formula A=lwA = lw.\newlineGiven the area of the playing card is 5454 square centimeters, we can write the equation:\newline54=lw54 = lw
  5. System of Equations: We now have a system of two equations with two variables:\newline15=l+w15 = l + w (Equation 11)\newline54=lw54 = lw (Equation 22)\newlineWe can solve this system by expressing one variable in terms of the other using Equation 11 and then substituting into Equation 22.
  6. Express Width in Terms of Length: From Equation 11, express ww in terms of ll:w=15lw = 15 - l
  7. Substitute Width into Area Equation: Substitute w=15lw = 15 - l into Equation 22:\newline54=l(15l)54 = l(15 - l)
  8. Expand and Rearrange Equation: Expand the equation and rearrange it into a quadratic equation:\newline54=15ll254 = 15l - l^2\newline0=l215l+540 = l^2 - 15l + 54
  9. Factor Quadratic Equation: Factor the quadratic equation: 0=(l9)(l6)0 = (l - 9)(l - 6)
  10. Solve for Length: Solve for ll by setting each factor equal to zero:\newlinel9=0l - 9 = 0 or l6=0l - 6 = 0\newlinel=9l = 9 or l=6l = 6
  11. Find Possible Solutions: Since ll and ww are interchangeable in a rectangle (it doesn't matter which one is length and which one is width), we can have two solutions:\newlineIf l=9l = 9, then w=159=6w = 15 - 9 = 6.\newlineIf l=6l = 6, then w=156=9w = 15 - 6 = 9.
  12. Check Validity of Solutions: Check the solutions by verifying that they satisfy both the perimeter and area equations:\newlineFor l=9l = 9 and w=6w = 6:\newlinePerimeter check: 2l+2w=2(9)+2(6)=18+12=302l + 2w = 2(9) + 2(6) = 18 + 12 = 30 (matches given perimeter)\newlineArea check: lw=9×6=54lw = 9 \times 6 = 54 (matches given area)

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