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Mathematics
Geometry
Perimeter and Sides Length
The perimeter of a quadrilateral is 56. The lengths of three consecutive sides are 9, 13, and 16. What is the length of the fourth side?
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The perimeter of a quadrilateral is 56. The lengths of three consecutive sides are 9, 13, and 16. What is the length of the
Top answer:
9+13+16 = 38 56 - 38 = ?
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The consecutive sides of a quadrilateral measure (x-17), (24-x), (3x-40) and (x+1). The perimeter is 42 cm . is the
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let's see: x-17 + 24-x + 3x-40 + x+1 = 42 4x=74 x = 74/4 = 18.5 so the sides are: 1.5, 5.5 , 15.5 ,
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the sides of a triangle are consecutive integers.If its perimeter is 126cm.find the lengths of the sides.
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x + x+1 + x+2 = 126 ...
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Let x be the length of the shortest side. x+ (x+2) + (x+4) = 195 3x + 6 = 195 3x = 189 You finish
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Quadrilateral ABCD has a perimeter of 26 and sides of integer lengths. If AB=m and BC=CD=DA=n, then what is the difference
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The perimeter is: 3n + m = 26 So: n = (26-m)/3 Thus n must be an integer value such that: 1/3 < n <
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The perimeter of a triangle is 195 mm.If the lengths of the sides are consecutive odd integers,find the length of each side.
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3x+6=195 -6 -6 3x=189 divide by 3 on both sides and you get x=63
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One leg of a right triangle has a length of 15 m. The other sides have lengths that are consecutive integers. Find the number of
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15^2 + x^2 = (x+1)^2 225 + x^2 = x^2 + 2n + 1 225 = 2n + 1 224 = 2n 112 = n So the three sides have
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One leg of a right triangle has a length of 15 m. The other sides have lengths that are consecutive integers. Find the number of
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x^2+15^2 = (x+1)^2 x = 112 p = 15+112+113
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One leg of a right triangle has a length of 15 m. The other sides have lengths that are consecutive integers. Find the number of
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225 + x^2= (x+1)^2 225 + x^2 = x^2 + 2 x + 1 2 x = 224 x = 112 x+1 = 113 perimeter = 15 + 112 + 113
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