Question

Solve the problem by applying the sum and product of roots
of quadratic equations.

The perimeter of a rectangular metal plate is 36 dm and its area is 80
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 product of its length and width. Such that,
Perimeter = 2(L + w) and Area = Lºw

ineed answer plss asap.​

Answer 1

SOLUTION :

GIVEN

The perimeter of a rectangular metal plate is 36 dm and its area is 80 dm2

TO DETERMINE

The dimensions by relating the measures to the sum

and product of a quadratic equation

EVALUATION

Let L = Length of the rectangular metal plate

W = Width of the rectangular metal plate

Then perimeter = 2 ( L + W )

Area = L. W

So by the given condition

$\sf{2(L + W) = 36 \: }$

$\implies \: \sf{L + W = 18 \: } \: \: \: ....(1)$

Again

$\sf{L. W \: } = 80 \: \: \: \: \: ....(2)$

Equation (1) & Equation (2) shows that L & W are the roots of the quadratic equation

$\sf{ {x}^{2} - ( L + W)x +L W = 0\: }$

$\implies \sf{ {x}^{2} - 18x + 80 = 0 \: }$

$\implies \sf{ {x}^{2} - 10x - 8x + 80 = 0 \: }$

$\implies \sf{x(x - 10) - 8(x - 10) = 0 \: }$

$\implies \sf{(x - 10) (x - 8) = 0 \: }$

Which gives either x - 10 = 0 or x - 8 = 0

Now x -10 = 0 gives x = 10

Again x - 8 = gives x = 8

Since for a rectangle length is greater than width

So

Length of the rectangular

metal plate = 10 dm

Width of the rectangular

metal plate = 8 dm

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