Learning Task 3.
Solve the problem by applying the sum and product of roots of quadratic equations.
(refer to supplementary materials for more details)
The perimeter of a rectangular metal plate is 38 dm and its area is 90 dm2. Find its dimensions.
(Relate the measures to the sum and product of a quadratic equation.)
The perimeter of a rectangle is twice the sum of its length and width while its area is the produc
of its length and width. Such that,
Perimeter = 2(L + w) and Area = Lºw
Solve the problem by applying the sum and product of roots of quadratic equations.
(refer to supplementary materials for more details)
The perimeter of a rectangular metal plate is 38 dm and its area is 90 dm2. Find its dimensions.
(Relate the measures to the sum and product of a quadratic equation.)
The perimeter of a rectangle is twice the sum of its length and width while its area is the produc
of its length and width. Such that,
Perimeter = 2(L + w) and Area = Lºw
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Answer:
9dm and 10dm
Step-by-step explanation:
2L + 2W = 38
LW = 90
Find for L in the first equation:
2L = 38 - 2W
L = 19 - W
Substitute into the second equation:
(19 - W)W = 90
19W - W^2 = 90
Transpose to the other side:
W^2 - 19W + 90
Factor:
(W - 9)(W - 10)
W = 9, 10
Therefore the width and the heights are: 9dm and 10dm