The length of a rectangle is 6.7 cm more than 2 times the width. If the perimeter of the rectangle is 57.2 cm, what are its dimensions? alength = 7.9 cm; width = 14 cm blength = 7.3 cm; width = 21.3 cm clength = 21.3 cm; width = 14 cm dlength = 21.3 cm; width = 7.3 cm
Question
The length of a rectangle is 6.7 cm more than 2 times the width. If the perimeter of the rectangle is 57.2 cm, what are its dimensions? alength = 7.9 cm; width = 14 cm blength = 7.3 cm; width = 21.3 cm clength = 21.3 cm; width = 14 cm dlength = 21.3 cm; width = 7.3 cm
Solution
Let's solve this step by step:
Step 1: Let's denote the width of the rectangle as w (in cm). According to the problem, the length of the rectangle is 6.7 cm more than 2 times the width. So, we can express the length as l = 2w + 6.7.
Step 2: The perimeter of a rectangle is given by the formula P = 2l + 2w. We know from the problem that P = 57.2 cm.
Step 3: Substituting the expressions for l and P into the perimeter formula gives us 57.2 = 2(2w + 6.7) + 2w.
Step 4: Simplifying the equation gives us 57.2 = 4w + 13.4 + 2w, which simplifies further to 57.2 = 6w + 13.4.
Step 5: Solving for w, we subtract 13.4 from both sides to get 43.8 = 6w, and then divide by 6 to get w = 7.3 cm.
Step 6: Substituting w = 7.3 cm into the expression for l gives us l = 2(7.3) + 6.7 = 21.3 cm.
So, the dimensions of the rectangle are length = 21.3 cm and width = 7.3 cm. The correct answer is (d) length = 21.3 cm; width = 7.3 cm.
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