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The perimeter of a rectangle is 70 cm. If its length is decreased by 5 cm and its width is increased by 5 cm, its area will increase by 50 squared cm. Find the length and the width of the original rectangle.

User Zaara
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Answer:

Length = 25cm; Width = 10cm.

Explanation:

My method is to use simultaneous equations.

Let x be the length of the rectangle and y be the width.

The perimeter of a rectangle is (2 x length) + (2 x width).

Substituting our values in we can say that 70 = 2x + 2y.

This is our first of two equations.

Next, we find an equation using the area of the rectangle.

The length decreases by 5, so we can call this x - 5.

The width increases by 5, so so we can call this y + 5.

The area of a rectangle is simply length x width, so in our original rectangle, the area would be xy.

However, with our new lengths the area increases by 50, allowing us to set up an equation where xy + 50 = (x - 5)(y + 5)

We now have two simultaneous equations:

1) 70 = 2x + 2y

2) xy + 50 = (x - 5)(y + 5)

We rearrange equation 1 to make y the subject.

This gives us y = 35 - x.

We can now substitute this into equation 2.

x(35 - x) + 50 = (x - 5)(35 - x + 5)

-x^2 + 35x + 50 = (x - 5)(40 - x) [expand the left and side and collect like terms in the second bracket on the RH]

-x^2 + 35x + 50 = -x^2 + 45x - 200 [expand the RH]

35x + 50 = 45x - 200 [The -x^2’s cancel out each other]

250 = 10x

x = 25cm

Substitute this into equation 1.

70 = 2(25) + 2y

70 = 50 + 2y

2y = 20

y = 10cm

User Jeff Trull
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