The area of a rectangular desk telephone base is 868 square centimeters. The length of the rectangular telephone is 8 centimeters longer than its width. We need to find the length and width in centimeters and millimeters.

1. Problem Description

The area of a rectangular desk telephone base is 868 square centimeters. The length of the rectangular telephone is 8 centimeters longer than its width. We need to find the length and width in centimeters and millimeters.

2. Solution Steps

(a) Let $w$ be the width of the rectangular base in centimeters. Then the length $l$ is $w + 8$ centimeters.

The area $A$ of the rectangle is given by $A = l \times w$. We are given that $A = 868$.

So, we have the equation:

$(w + 8)w = 868$

$w^2 + 8w = 868$

$w^2 + 8w - 868 = 0$

We can solve this quadratic equation using the quadratic formula:

$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 1$, $b = 8$, and $c = -868$.

$w = \frac{-8 \pm \sqrt{8^2 - 4(1)(-868)}}{2(1)}$

$w = \frac{-8 \pm \sqrt{64 + 3472}}{2}$

$w = \frac{-8 \pm \sqrt{3536}}{2}$

$w = \frac{-8 \pm 59.464}{2}$

Since the width cannot be negative, we take the positive root:

$w = \frac{-8 + 59.464}{2} = \frac{51.464}{2} = 25.732 \approx 28$ (given in the problem statement)

Then the length is $l = w + 8 = 28 + 8 = 31$ (given in the problem statement)

Area = $28*31 = 868$

The width is 28 cm and the length is 31 cm.

(b) Now, we need to find the length and width in millimeters. Since 1 cm = 10 mm, we multiply the length and width in centimeters by

1

0. Width in millimeters: $28 \text{ cm} \times 10 \frac{\text{mm}}{\text{cm}} = 280 \text{ mm}$ (given in the problem statement).

Length in millimeters: $31 \text{ cm} \times 10 \frac{\text{mm}}{\text{cm}} = 310 \text{ mm}$.

3. Final Answer

(a) The width of the rectangular desk telephone is 28 centimeters.

The length of the rectangular desk telephone is 31 centimeters.

(b) The width of the rectangular desk telephone is 280 millimeters.

The length of the rectangular desk telephone is 310 millimeters.