The Quadratic Relationships (length vs area of rectangles with fixed perimeter)
Quiz yourself by thinking what should be in
each of the black spaces below before clicking
on it to display the answer.
Help!
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| A rectangle has a perimeter of 14. Determine the dimensions of ALL of the possible rectangles. | 1x6 4x3
2x5 5x2
3x4 6x1
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| If the perimeter of a rectangle is equal to 50, then the L plus the width is equal to . . . ? | L + W + L + W = 50
so, L + W = 25 (half of the perimeter)
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| L + W = 45. Solve the equation for L, in terms of W. | L = 45 - W
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| L + W = 250. Solve the equation for W, in terms of L. | W = 250 - L
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| If the area of rectangles with a fixed perimeter is A = L(26 - L), determine the fixed perimeter. | 52, the fixed perimeter is double the constant in the formula.
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| If the area of rectangles with a fixed perimeter is A = L(78 - L) and the length is 40, determine the width. | In the equation, A = L(78 - L), the quantity (78-L) represents the width so plug in 40 to get 78 - 40 = 38 as the width.
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| If the area of rectangles with a fixed perimeter is A = L(100 - L) and the length is 60, determine the area. | A = 60(100 - 60)
A = 60(40)
A = 2400
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| A rectangle has a fixed perimeter of 8, determine the rectangle with the maximum area. | 1x3 = 3
2x2 = 4
3x1 = 3 , so 2x2 has maximum area
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| If a rectangle has a fixed perimeter, the rectangle with the maximum area is always a _____________________ . | Square, L = W
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| A rectangle has a perimeter of 32. Determine the dimensions of ALL of the possible rectangles. | 1x15, 2x14, 3x13, 4x12, 5x11, 6x10, 7x9, 8x8
plus all of the rectangles in reverse.
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| The following is a list of dimensions of rectangles with a fixed area. Which one represents the rectangle with the maximum area? Explain. 1x15, 2x14, 3x13, 4x12, 5x11, 6x10, 7x9, 8x8 | 8x8, because it is a square plus it's area is 64 which is the largest number.
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| Given a graph comparing length and area of rectangles with a fixed perimeter, how can you determine the maximum area? | By looking at the highest point on the graph.
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| L + W = 68. Solve the equation for L, in terms of W. | L = 68 - W
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| L + W = 190. Solve the equation for W, in terms of L. | W = 190 - L
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| Rectangles with a fixed perimeter of 80 have a length of L and a width of 40 - L. Write an equation for Area using only A and L as the variables. | A = L(W)
A = L(40 - L), must have parenthesis for full credit
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| The table displays rectangles with a fixed perimeter. Length 0 1 2 3 4 5 6 7 8 Area 0 7 12 15 16 15 12 7 0 Determine the fixed perimeter. | 16, Double the length that corresponds to an area of zero. Or determine the width and find the perimeter, ex: L=1 A=7 so W=7, so perimeter is 1 + 7 + 1 + 7.
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| The table displays rectangles with a fixed perimeter. Length 0 1 2 3 4 5 6 7 8 Area 0 7 12 15 16 15 12 7 0 Determine the maximum area and the dimensions of that rectangle with the maximum area. | 16 and 4x4, the rectangle with the maximum area is always a square.
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Created by:
gunningta
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