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Rectangle problem solving
Question:
Given: ABCD is a rectangle. The area of the rectangle is 132 square feet and the perimeter of the rectangle is 46 feet. AB > BC. Find the length of AB and BC.
find the length and width of a rectangle.
Solution:
since ABCD is a rectangle, so AB = DC = x and AD = BC = y
area of the rectangle is 132 square feet
x y = 132 ... equation1
perimeter of the rectangle is 46 feet
2(x + y) = 46 ...equation2
from equation2, (x + y) = 46/2 = 23
express x in terms of y
x = 23 - y
substitute the expression of x into equation1
(23 - y) y = 132
remove parenthesis
23y - y2 = 132
move the number to the left side of the equation since the degree of the variable is 2
23y - y2 - 132 = 0
multiply (-1) in each term of the equation and write them in degree decrease order
y2 - 23y + 132 = 0
left side of the equation is a two degree polynomial, factoring it
(y - 11)(y - 12) = 0
now we get the solution of y
y1 = 11
y2 = 12
when y1 = 11, x1 = 23 - y = 23 - 11 = 12
when y2 = 12, x2 = 23 - y = 23 - 12 = 11
Given: AB > BC, so x > y
so the solution is: x = 12 and y = 11
AB = 12 feet and AD = 11 feet.