5. If the area of a rectangle is $$ x^{2}-x-6 $$ sq. units and its length is $$ (x-3) $$ units, then its breadth is $$ \begin{array}{llll} ext { a. }(x-2) ext { units } & ext { b. }(x+3) ext { units } & ext { c. }(x+2) ext { units } & ext { d. }(x-3) ext { units }\end{array} $$

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5. If the area of a rectangle is $$ x^{2}-x-6 $$ sq. units and its length is $$ (x-3) $$ units, then its breadth is $$ \begin{array}{llll}	ext { a. }(x-2) 	ext { units } & 	ext { b. }(x+3) 	ext { units } & 	ext { c. }(x+2) 	ext { units } & 	ext { d. }(x-3) 	ext { units }\end{array} $$

UpStudy AI · Accepted Answer

To find the breadth of the rectangle, we'll use the relationship between area, length, and breadth:

$$
\text{Area} = \text{Length} \times \text{Breadth}
$$

Given:
- **Area** = $$ x^{2} - x - 6 $$ square units
- **Length** = $$ x - 3 $$ units

We need to find the **Breadth**. Let's denote the breadth by $$ B $$.

$$
B = \frac{\text{Area}}{\text{Length}} = \frac{x^{2} - x - 6}{x - 3}
$$

**Step 1: Factor the Numerator**

Factor the quadratic expression in the numerator:

$$
x^{2} - x - 6 = (x - 3)(x + 2)
$$

**Step 2: Simplify the Expression**

Now, substitute the factored form back into the equation:

$$
B = \frac{(x - 3)(x + 2)}{x - 3}
$$

Cancel out the $$ (x - 3) $$ terms (assuming $$ x \neq 3 $$):

$$
B = x + 2
$$

**Therefore, the breadth of the rectangle is $$ x + 2 $$ units.**

**Answer:**  
c. $$ (x+2) $$ units
The breadth of the rectangle is $$ (x+2) $$ units.

If the area of a rectangle is sq. units and its length is units, then its
breadth is

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If the area of a rectangle is sq. units and its length is units, then its breadth is