the length of a rectangle exceeds the width by 2cm. if the diagonal is 10cm long, what would the width be?? can you explain the answer
You need to come up with the following substitution for the length of rectangle such that:
length = `x`
The problem provides the information that the length exceeds the width by 2 such that:
width =`x + ` `2`
You should know that one diagonal of rectangle divides the rectangle into two right angle triangles.
Considering one right angle triangle, notice that diagonal represents the hypotenuse and the width and length represent the legs of triangle, hence, you may use Pythagorean theorem such that:
`10^2 = x^2 + (x+2)^2`
Expanding the square yields:
`100 = x^2 + x^2 + 4x + 4 => 2x^2 + 4x + 4 - 100 = 0`
`2x^2 + 4x - 96 = 0`
You may divide by 2 the equation such that:
`x^2 + 2x - 48 = 0`
Using quadratic formula yields:
`x_(1,2) = (-2+-sqrt(2^2 - 4*1*(-48)))/2`
`x_(1,2) = (-2+-sqrt(4 + 192))/2 => x_(1,2) = (-2+-sqrt196)/2`
`x_(1,2) = (-2+-14)/2 => x_1 = 6 ; x_2 = -8`
Since the length cannot be negative, you need to consider the length = x = 6, hence, the width is width = 6 - 2 = 4.
Hence evaluating the width of rectangle, under the given conditions, yields width = 4.