The length of a rectangular floor is 1 meter less than twiceits width. If a diag

Question

The length of a rectangular floor is 1 meter less than twiceits width. If a diagonal of the rectangle is 17 meters, find thelenght and width of the floor. A 5-inch by 7-inch picture is surrounded by a frame of uniformwidth. The area of the picture and frame together is 80 squareinches. Find the width of the frame. The length of a rectangular floor is 1 meter less than twiceits width. If a diagonal of the rectangle is 17 meters, find thelenght and width of the floor. A 5-inch by 7-inch picture is surrounded by a frame of uniformwidth. The area of the picture and frame together is 80 squareinches. Find the width of the frame.

Explanation / Answer

1. Let L = the length of the rectangle and W = itswidth. L=2w-1 "The length (L) of a rectangle is 1m less than twiceits width (W)." From the Pythagoreantheorem,...(L^2+W^2)=17 (L^2+W^2)^2=17^2 L^2+W^2=289 substitute L=2w-1 and solve for w. (2w-1)^2+w^2=289 (4w^2-4w+1)+w^2=289 simplify and subtract 289 from both sides. 5w^2-4w-288=0 solve by using the quadratic formula w=8 disregard the other answer because it's negative and you can't havea negative number for length/width L=2w-1 L=2(8)-1=15 The length is 15 meters and the width is 8meters. 2. Width of the frame (x): (5 + x)(7 + x) = 80 35 + 5x + 7x + x² = 80 x² + 12x = 45 x² + 6x = 45 + 6² x² + 6x = 45 + 36 (x + 6)² = 81 x + 6 = 9 x = 3 Answer: 3 feet Proof (total area is 80): = (5 feet + 3 feet)(7 feet + 3 feet) = 8 feet(10 feet) = 80 square feet

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