(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\) (b) The diagram shows a rectangle PQRS from which a square of side x cm has been...
Question 1 Report
(a) Solve the inequality : \(4 + \frac{3}{4}(x + 2) \leq \frac{3}{8}x + 1\)
(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.
Answer Details
(a) Solving the inequality
\[4 + \frac{3}{4}(x+2) \le \frac{3}{8}x + 1\]
Multiply every term by 8 (the LCM of the denominators) to clear fractions:
\[8(4) + 8\cdot\frac{3}{4}(x+2) \le 8\cdot\frac{3}{8}x + 8(1)\]
\[32 + 6(x+2) \le 3x + 8\]
\[32 + 6x + 12 \le 3x + 8\]
\[6x + 44 \le 3x + 8\]
Collect like terms:
\[6x - 3x \le 8 - 44\]
\[3x \le -36\]
\[x \le -12\]
(b) Finding x from the rectangle
From the diagram, rectangle PQRS has height \(|PS| = 20\ \text{cm}\). Along the base SR a square of side x cm is cut out, leaving 10 cm on the left and 10 cm on the right. So the full length of the rectangle is
\[|SR| = 10 + x + 10 = (20 + x)\ \text{cm}\]
Area of the whole rectangle:
\[(20 + x)\times 20 = 400 + 20x\]
Area of the square removed:
\[x \times x = x^2\]
The shaded (remaining) area is 484 cm\(^2\):
\[(400 + 20x) - x^2 = 484\]
\[-x^2 + 20x + 400 - 484 = 0\]
\[-x^2 + 20x - 84 = 0\]
\[x^2 - 20x + 84 = 0\]
Solve using factorisation (\(-6\) and \(-14\) multiply to 84 and add to \(-20\)):
\[(x - 6)(x - 14) = 0\]
\[x = 6 \quad\text{or}\quad x = 14\]
Both values are geometrically valid (the cut height stays within the 20 cm rectangle), so
\[x = 6\ \text{cm} \quad\text{or}\quad x = 14\ \text{cm}\]
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