SOLUTION: There is a right-angled triangle (height=4x, base=x) and a rectangle (length=5x-2, width=x-1). all measurements are in centimeters. the area of the rectangle is greater than the ar

Algebra ->  Inequalities -> SOLUTION: There is a right-angled triangle (height=4x, base=x) and a rectangle (length=5x-2, width=x-1). all measurements are in centimeters. the area of the rectangle is greater than the ar      Log On


   

Question 1204211: There is a right-angled triangle (height=4x, base=x) and a rectangle (length=5x-2, width=x-1). all measurements are in centimeters. the area of the rectangle is greater than the area of the triangle. find the set of possible values of x. step-by-step explanation needed please
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
There is a right-angled triangle (height=4x, base=x) and a rectangle (length=5x-2, width=x-1). all measurements are in centimeters. the area of the rectangle is greater than the area of the triangle. find the set of possible values of x. step-by-step explanation needed please

There is a right-angled triangle (height=4x, base=x)
Area of triangle 1/2 * 2x*x = 2x^2
a rectangle (length=5x-2, width=x-1).
Area of rectangle = L*W =(5x-2)(x-1)
= 5x^2-7x+2
5x^2-7x+2 >2x^2
3x^2-7x+2>0
3x^2-6x-x+2>0
3x(x-2)-1(x-2)>0
(x-2)(3x-1)>0
either (x-2)>0 or 3x-1>0
x>2 or x>1/3
x>2



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


To start with, lengths must be positive numbers. So x, 4x, 5x-2, and x-1 must all be positive numbers. That means x has to be greater than 1.

The area of the triangle is one-half base times height: %281%2F2%29%284x%29%28x%29+=+2x%5E2

The area of the rectangle is base times height: %285x-2%29%28x-1%29

The area of the rectangle is greater than the area of the triangle:

%285x-2%29%28x-1%29%3E2x%5E2

Expand the expression on the left

5x%5E2-7x%2B2%3E2x%5E2

Put the quadratic in standard form

3x%5E2-7x%2B2%3E0

Factor the quadratic

%283x-1%29%28x-2%29%3E0

Find the values of x for which the product is equal to 0. Those are the values of x for which the areas of the triangle and rectangle are the same.

3x-1 = 0 --> x = 1/3

or

x-2 = 0 --> x = 2

Mathematically, the areas are equal for x=1/3 and for x=2; the area of the rectangle is less than the area of the triangle for values of x between 1/3 and 2; and the area of the rectangle is greater than the area of the triangle for x<1/3 and for x>2.

But we know x has to be greater than 1.

The requirement that the area of the rectangle be greater than the area of the triangle tells us x is less than 1/3 or greater than 2; the requirement that all the lengths in the problem be positive tells us x has to be greater than 1.

Since both requirements must be met...

ANSWER: The area of the rectangle is greater than the area of the triangle for all values of x greater than 2

-------------------------------------------------------------

Note in the response from tutor @mananth her reasoning is deficient and the work she shows contains errors. It is only by chance that she shows the correct final answer.