We are tasked with finding the dimensions of a rectangle given:

1. Area = $105 \, \text{cm}^2$
2. Perimeter = $44 \, \text{cm}$
3. Length ($l$) is greater than Width ($w$).

Step 1: Express relations

1. Area Formula: $l \cdot w = 105$
2. Perimeter Formula: $2(l + w) = 44$, which simplifies to: $l + w = 22$

Step 2: Solve for dimensions

From $l + w = 22$, we get: $l = 22 - w$

Substitute $l = 22 - w$ into the area equation: $(22 - w) \cdot w = 105$ Simplify: $22w - w^2 = 105$ Rearrange into standard quadratic form: $w^2 - 22w + 105 = 0$

Step 3: Solve quadratic equation

Use the quadratic formula: $w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -22$, $c = 105$. Plugging in: $w = \frac{-(-22) \pm \sqrt{(-22)^2 - 4(1)(105)}}{2(1)}$ $w = \frac{22 \pm \sqrt{484 - 420}}{2}$ $w = \frac{22 \pm \sqrt{64}}{2}$ $w = \frac{22 \pm 8}{2}$

Thus: $w = \frac{22 + 8}{2} = 15 \quad \text{or} \quad w = \frac{22 - 8}{2} = 7$

Step 4: Find $l$

If $w = 7$, then: $l = 22 - 7 = 15$

If $w = 15$, then: $l = 22 - 15 = 7$

Since $l > w$, the dimensions are: $\boxed{l = 15 \, \text{cm}, \, w = 7 \, \text{cm}}$

Final Answer:

• Length: $15 \, \text{cm}$
• Width: $7 \, \text{cm}$

Would you like a detailed breakdown of any step or further explanations?

Here are five related questions for practice:

1. How do you derive the perimeter formula for a rectangle?
2. What are the steps to factorize a quadratic equation?
3. Can you use trial and error instead of the quadratic formula here?
4. How does knowing area and one dimension help find the other?
5. What happens if length equals width (square case)?

Tip: Always verify solutions by substituting back into the original equations!