The problem has two parts. (a) Solve the inequality: $4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1$. (b) The diagram shows a rectangle PQRS from which a square of side $x$ cm has been cut. If the area of the shaded portion is 484 $cm^2$, find the values of $x$. The dimensions of the rectangle are 20 cm by (10 cm + 10 cm) = 20 cm.

1. Problem Description

The problem has two parts.

(a) Solve the inequality: $4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1$.

(b) The diagram shows a rectangle PQRS from which a square of side $x$ cm has been cut. If the area of the shaded portion is 484 $cm^2$, find the values of $x$. The dimensions of the rectangle are 20 cm by (10 cm + 10 cm) = 20 cm.

2. Solution Steps

(a) Solve the inequality:

$4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1$

$4 + \frac{3}{4}x + \frac{6}{4} \le \frac{3}{8}x + 1$

Multiply by 8 to eliminate fractions:

$8(4 + \frac{3}{4}x + \frac{3}{2}) \le 8(\frac{3}{8}x + 1)$

$32 + 6x + 12 \le 3x + 8$

$6x + 44 \le 3x + 8$

$6x - 3x \le 8 - 44$

$3x \le -36$

$x \le -12$

(b)

The area of the rectangle PQRS is $20 \times (10 + 10) = 20 \times 20 = 400$ $cm^2$.

The area of the square is $x^2$ $cm^2$.

Two such squares have been removed. So the total area removed is $2x^2$ $cm^2$.

The shaded area is the area of the rectangle minus the area of the two squares, which is given as 484 $cm^2$. Thus, this statement is incorrect.

Let us suppose the area of the shaded region is actually 384 $cm^2$.

$400 - 2x^2 = 384$

$2x^2 = 400 - 384$

$2x^2 = 16$

$x^2 = 8$

$x = \sqrt{8} = 2\sqrt{2}$

However, it appears the shaded area is $484$ $cm^2$, which doesn't make sense. Let us assume that the two sections cut out are actually rectangles each of length $x$ and width

1

0. Then the area of the shaded area should be $20 \times 20 - 2 \times (10 \times x) = 484$.

$400 - 20x = 484$

$-20x = 84$

$x = -\frac{84}{20} = -\frac{21}{5}$, which also does not make sense.

Let us assume the width of the rectangle is unknown, call it $w$. Then, $(w)(20) - 2x^2 = 484$. We are also given that segments of length 10 are included in $w$ on both sides, so $w \ge 20$.

If $w = 30$, then $600 - 2x^2 = 484$, so $2x^2 = 116$, and $x^2 = 58$, so $x = \sqrt{58}$. This seems like a reasonable value of x.

Let us assume the total area is 400, and the area removed is $2x^2 = 400-484 = -84$. That is impossible, so perhaps the area of shaded portion given is wrong.

Let's look for an area such that x is an integer. We need $400 - \text{area} = 2x^2$ or that $(400 - \text{area})$ is an even number.

Let's assume $x$ is the width of two rectangles $10 \times h$.

Then $400 - (10h+10h) = 484$, or $20h = 400-484 = -84$, so $h = -4.2$, which doesn't make sense.

3. Final Answer

(a) $x \le -12$

(b) Assuming the area of the rectangle PQRS is 400 $cm^2$ and the area of the shaded region is 484 $cm^2$, which is not possible as the shaded area cannot be greater than the original area.

If the area of shaded portion is $384$ $cm^2$, then $x = 2\sqrt{2}$ cm.