Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2007-04-23 22:48:16

Neela2
Guest

Algebra

The perimeter of a rectangle is 34 cm. Given that the diagonal is of length 13 cm, and that the width is x cm, derive the equation x¨2 - 17x + 60 =0. Hence find the dimensions.

Thanks smile

#2 2007-04-23 23:23:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Algebra

The perimeter is 34cm, and as the perimeter is made of 2 lengths and 2 widths, that means that length+width = 17.

We're also told that the diagonal is 13cm, and so by Pythagoras, length² + width² = 13² = 169.

We now have 2 simultaneous equations, so we need to eliminate one of the variables.
Rearranging the first equation shows that length = 17-width.

Substituting this into the second gives width² + (17-width)² = 169.

Expanding and replacing width by x turns this into x² + 289 - 34x + x² = 169.

Simplify: 2x² - 34x + 120 = 0
--> x² - 17x + 60 = 0.

From here, it's just a case of solving the quadratic equation that we now have.
x² - 17x + 60 = 0
(x-5)(x-12) = 0
x = 5 or 12.

By convention, length is generally larger than width, so the length is 12cm and the width is 5cm.


Why did the vector cross the road?
It wanted to be normal.

Offline

Board footer

Powered by FluxBB