We have a rectangle and a right-angled triangle. All measurements are in centimetres.
The area of the rectangle is greater than the
area of the triangle.
Find the set of possible values of x.

x > 2

Step-by-step explanation:

Step 1: Find area of the rectangle.

Length of rectangle = 5x - 6

Width of rectangle = x - 1

Area of rectangle = length*width

Area of rectangle = (5x - 6)(x - 1)

= 5x(x - 1) -6(x - 1)

= 5x² - 5x - 6x + 6

Area of rectangle = 5x² - 11x + 6

Step 2: Find are of the triangle

Area of triangle = ½*base*height

Base of triangle = x

Height of triangle = 2x

Area = ½*x*2x

Area = x*x

Area of triangle = x²

Step 3: Write the inequality statement for "the area of the rectangle is greater than the area of the triangle", and solve for possible values of x.

Area of rectangle = 5x² - 11x + 6

Area of triangle = x²

Therefore:

5x² - 11x + 6 > x²

Subtract x² from both sides

5x² - 11x + 6 - x² > x² - x²

5x² - x² - 11x + 6 > 0

4x² - 11x + 6 > 0

Factorise (4x² - 11x + 6)

(The 2 possible factors when multiplied together will give us "24", and when added together will give us "-11" are -8 and -3).

Therefore, we factorise as follows:

4x² - 8x - 3x + 6 > 0

(4x² - 8x) - (3x + 6) > 0

4x(x - 2) -3(x - 2) > 0

(4x - 3)(x - 2) > 0

Find the solution of each factor of the inequality

(4x - 3) > 0

Or

(x - 2) > 0

Thus, solving for each for each, we have:

4x - 3 > 0

Add 3 to both sides

4x - 3 + 3 > 0 + 3

4x > 3

Divide both sides by 4

4x/4 > 3/4

x > ¾

Or

x - 2 > 0

Add 2 to both sides

x - 2 + 2 > 0 + 2

x > 2

x > ¾ or x > 2

The possible values of x that fits the inequality given is x > 2, rather than x > ¾

Rationale:

We are told that the area of rectangle > area of triangle. Thus, the inequality given is:

5x² - 11x + 6 > x²

If x > ¾, that is the first possible value of x is 1, plug in the value of x into the inequality let's find out if it is true for the statement.

Thus:

5x² - 11x + 6 > x²

5(1)² - 11(1) + 6 > 1²

5 - 11 + 6 > 1

0 > 1 (this is very untrue)

Let's try out x > 2, meaning the first possible value of x = 3

Thus,

5x² - 11x + 6 > x²

5(3)² - 11(3) + 6 > 3²

5(9) - 33 + 6 > 9

45 - 33 + 6 > 9

18 > 9 (very true)

Therefore, the possible value of x would be values of x that are greater than 2.

Our answer is x > 2.