Question

a rectangle the area and one dimension is given(a-3).find other dimensions. area=12+5a-3a²​

Answer 1

Solution :

We know that,

area of a rectangle = length × width

Given, area = 12 + 5a - 3a²

& one dimension = a - 3

Now, 12 + 5a - 3a²

= 12 + 9a - 4a - 3a²

= 3 (4 + 3a) - a (4 + 3a)

= (4 + 3a) (3 - a)

= (- 4 - 3a) (a - 3)

So the other dimension be

= (- 4 - 3a), must be > 0

or, we simply can take (3a + 4) as its modulus value, where we will find such values of a for which 3a + 4 > 0 always.

Answer 2

Answer:

(-3a^2 + 5a + 12) / (a - 3)

Step-by-step explanation:

We know that the area of a rectangle is given by the formula A = L x W, where A is the area, L is the length, and W is the width.

In this case, we are given the area as 12 + 5a - 3a^2 and one dimension as (a - 3), which we can assume to be the length. Let's use these values to find the other dimension, which will be the width.

A = L x W

12 + 5a - 3a^2 = (a - 3) x W (substituting L with (a-3))

To find W, we can simplify the equation above by expanding the product on the right-hand side:

12 + 5a - 3a^2 = aW - 3W

Now, we can rearrange the terms to isolate W on one side of the equation:

aW - 3W = 12 + 5a - 3a^2

W(a - 3) = -3a^2 + 5a + 12

W = (-3a^2 + 5a + 12) / (a - 3)

Therefore, the width of the rectangle is (-3a^2 + 5a + 12) / (a - 3).