Question

The area of the rectangle on the right is (x2 + x - 12). If it’s length is x + 4, what is it’s width?
X-3
X+6
X+1
X+2

The legs of one right triangle are 9 and 12, while those of another triangle are 12 and 16. How much longer is the perimeter of the larger than the perimeter of the smaller triangle?
12
7
84
14

Answer 1

To find the width of the rectangle, you can use the formula for the area of a rectangle:

\[
\text{Area} = \text{Length} \times \text{Width}
\]

Given that the area is \(x^2 + x - 12\) and the length is \(x + 4\), you can set up the equation:

\[
x^2 + x - 12 = (x + 4) \times \text{Width}
\]

To find the width, divide the area by the length:

\[
\text{Width} = \frac{x^2 + x - 12}{x + 4}
\]

Now, we can perform polynomial long division or simplify it:

1. Factor \(x^2 + x - 12\):
\[
x^2 + x - 12 = (x - 3)(x + 4)
\]

2. Therefore, we can substitute back in:
\[
\frac{(x - 3)(x + 4)}{x + 4}
\]

3. This simplifies to:
\[
x - 3
\]

So, the width of the rectangle is \(x - 3\).

---

Next, for the triangles:

1. **Perimeter of the first triangle (9, 12)**:
- Use the Pythagorean theorem to find the hypotenuse \(c_1\):
\[
c_1 = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15
\]
- So, the perimeter \(P_1\) is:
\[
P_1 = 9 + 12 + 15 = 36
\]

2. **Perimeter of the second triangle (12, 16)**:
- Again using the Pythagorean theorem for hypotenuse \(c_2\):
\[
c_2 = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20
\]
- Therefore, the perimeter \(P_2\) is:
\[
P_2 = 12 + 16 + 20 = 48
\]

3. Now, find how much longer the perimeter of the larger triangle is than that of the smaller triangle:
\[
P_2 - P_1 = 48 - 36 = 12
\]

Thus, the perimeter of the larger triangle is **12** units longer than that of the smaller triangle.

**Final Answers:**
- Width of the rectangle: \(x - 3\)
- Difference in perimeter lengths: \(12\)