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A rope of length 40 meters is cut into pieces and two squares are made on the floor with them. The sum of the areas of the enclosed is 58 square meters.

A) If the length of one piece is taken as x, what is the length of the other piece?

B) What are the lengths of the sides of the squares?

C) Write the given fact about area as an algebraic equation.

D) What is the length of each piece? (let x is length of the piece)

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A-

35 - x,

B-

\frac{3 x}{4} and \frac{40 - x}{5}

,C-

x^{2} - 50 x - 336 = 0,

D-28 m or 10 m

A-

40 - x,

B-

\frac{x}{4} and \frac{40 - x}{4}

,C-

x^{2} - 40 x + 336 = 0,

D-28 m or 12 m

A-

45 - x,

B-

\frac{5 x}{4} and \frac{45 - x}{4}

,C-

x^{2} - 30 x + 330 = 0,

D-20 m or 14 m

None

answer is B.

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Detailed Solution

Given,
A rope of length is 40 cm
The sum of the areas of the enclosed is 58 square meters.
Suppose length of the piece is x.
(A) So, other piece's length

= 40 - x

(B) Dimensions of the square's sides

= \frac{x}{4} and \frac{40 - x}{4}

(since, there are 4 sides in square)
(C) The area sum in algebraic form

58 {cm}^{2}

.
i.e.,

a^{2} + a^{2} = 58 {cm}^{2}

(As area of square =

sid e^{2}

)

\Rightarrow {(\frac{x}{4})}^{2} + {(\frac{40 - x}{4})}^{2} = 58

\Rightarrow \frac{x^{2}}{16} + \frac{1600 - 80 x + x^{2}}{16} = 58

\Rightarrow 2 x^{2} - 80 x + 1600 = 58 \times 16

\Rightarrow 2 x^{2} - 80 x + 1600 = 928

\Rightarrow 2 (x^{2} - 40 x + 800) = \frac{928}{2}

\Rightarrow x^{2} - 40 x + 800 - 464 = 0

\Rightarrow x^{2} - 40 x + 800 - 464 = 0

\Rightarrow x^{2} - 40 x + 336 = 0

(D) Each piece's length,

x^{2} - 40 \times 336 = 0 (factorize)

\Rightarrow (x - 28) (x - 12) = 0

\Rightarrow x - 28 = 0 or x - 12 = 0

x = 28 or x = 12

Thus, each piece's length = 28 m or 12 m.
Option 2 is correct.

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