(a) A rectangular corner lot, originally twice as long as itwas wide, lost a 2-m
ID: 3092362 • Letter: #
Question
(a) A rectangular corner lot, originally twice as long as itwas wide, lost a 2-meter strip along two adjacent sides due tostreet widening. Its new area is 684 m2. Find its newdimensions. (b) A rancher plans to use 160 yd of fencing to enclose arectangular corral and to divide it into two parts by a fenceparallel to the shorter sides of the corral. Find the dimensions ofthe corral if its area is 1000 yd2. (a) A rectangular corner lot, originally twice as long as itwas wide, lost a 2-meter strip along two adjacent sides due tostreet widening. Its new area is 684 m2. Find its newdimensions. (b) A rancher plans to use 160 yd of fencing to enclose arectangular corral and to divide it into two parts by a fenceparallel to the shorter sides of the corral. Find the dimensions ofthe corral if its area is 1000 yd2.Explanation / Answer
(a) Let w be the lot's new width after the street was widened. Then the old width was w + 2, the old length was 2(w + 2) because it's twice the old width,and the new length is 2(w + 2) - 2 = (2w + 2). The new length times the new width gives the new area: (2w + 2) w = 684 m^2 Solve that for w and calculate the length from that. ---------------------------------------------------------------------------------------------------------------------------------- (b)the Area is length (l) x width (w) Area = 1000 = l * w the fencing is 160 yards and there are 2 lengths but 3 widths (penis divided into 2 smaller pens) then 160 = 2l + 3w. solve the area for l = 1000/w then we have 160 = 2(1000/w) + 3w or 2000/w + 3w = 160, multiplying by w 2000 + 3w^2 = 160w, rearranging into a quadratic 3w2 -160w + 2000 = 0, factoring (3w - 100)(w -20) = 0 implies that w = 100/3, 20. then l = 1000/w yield l = 3000/100 = 30, 1000/20 = 50. then two possible ways 100/3 (33.3) by 30 or 20 by 50.
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