The problem consists of two parts. Part (a) requires simplifying the algebraic expression $\frac{x^2 - y^2}{3x + 3y}$. Part (b) involves a rectangle $PQRS$ with $|PK| = 15$ cm, $|SK| = |KR|$, and $\angle PKS = 37^{\circ}$. We need to find (i) $|PS|$, (ii) $|SK|$, and (iii) the area of the shaded portion.

1. Problem Description

The problem consists of two parts. Part (a) requires simplifying the algebraic expression $\frac{x^2 - y^2}{3x + 3y}$. Part (b) involves a rectangle $PQRS$ with $|PK| = 15$ cm, $|SK| = |KR|$, and $\angle PKS = 37^{\circ}$. We need to find (i) $|PS|$, (ii) $|SK|$, and (iii) the area of the shaded portion.

2. Solution Steps

(a) Simplifying the expression:

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

Since $x+y$ is a common factor in the numerator and the denominator, we can cancel it out, assuming $x+y \ne 0$:

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

(b) (i) Finding $|PS|$:

In the right triangle $PKS$, we have $\angle PKS = 37^{\circ}$ and $|PK| = 15$ cm. We want to find $|PS|$, which is the side adjacent to the angle $37^{\circ}$. We can use the cosine function:

$\cos(\angle PKS) = \frac{|PS|}{|PK|}$

$|PS| = |PK| \cos(37^{\circ})$

$|PS| = 15 \cos(37^{\circ})$

Using a calculator, $\cos(37^{\circ}) \approx 0.7986355$.

$|PS| = 15 \times 0.7986355 \approx 11.9795325 \approx 12.0$ cm (to three significant figures).

(ii) Finding $|SK|$:

In the right triangle $PKS$, we have $\angle PKS = 37^{\circ}$ and $|PK| = 15$ cm. We want to find $|SK|$, which is the side opposite to the angle at $P$. We can use the sine function (or tangent since we know $|PS|$):

$\sin(\angle PKS) = \frac{|PK|}{|SK|}$ is INCORRECT. $|PK|$ is the hypotenuse.

Instead use: $\tan(37^\circ) = \frac{opposite}{adjacent} = \frac{PK}{PS}$ where we want to find $SK$ such that $|SK|=|KR|$, and $|SR| = |SK| + |KR| = 2|SK|$. Thus, we can use the fact that PQRS is a rectangle to deduce $|SR|=|PQ|$ and use $\tan$. We can also find $|SK|$ using the tangent:

$\tan(\angle PKS) = \frac{|PK|}{|PS|}$ which doesn't help.

Correct approach:

$\tan (37^\circ) = \frac{PK}{PS}$

$PS = 15 \cos(37^\circ)$

$PK = \frac{15}{\cos(37^\circ)} = \sqrt{PS^2 + SK^2} = 15$

$\tan(37^{\circ}) = \frac{|PK|}{|SK|}$

$|SK| = |PS| \tan(37^\circ)$

$|SK| = \frac{|PK|}{\tan(37^{\circ})} = \frac{15}{\tan(37^{\circ})}$

$\tan(37^{\circ}) \approx 0.753554$

$|SK| = \frac{15}{0.753554} \approx 19.9054$ cm

Since $|SK| = |KR|$, $|SR| = 2|SK| = 2 \times 19.9054 = 39.8108$ cm.

So $|SK| \approx 19.9$ cm (to three significant figures).

(iii) Finding the area of the shaded portion:

The area of the rectangle $PQRS$ is $|PS| \times |SR| = |PS| \times 2|SK|$.

Area of rectangle $= 11.98 \times (2 \times 19.91) = 11.98 \times 39.82 = 477.0436$ cm$^2$.

The area of the unshaded triangle $PKS = \frac{1}{2} \times |PS| \times |SK| = \frac{1}{2} \times 11.98 \times 19.91 = \frac{1}{2} \times 238.5218 = 119.2609$ cm$^2$.

Area of shaded portion = Area of rectangle - Area of triangle $PKS$.

Area of shaded portion $= 477.0436 - 119.2609 = 357.7827 \approx 358$ cm$^2$ (to three significant figures).

Alternate Calculation using more precise values

Area of rectangle $= |PS| \cdot |SR|$

$|PS| = 15 \cos(37) \approx 15(0.7986) \approx 11.979 \approx 12.0$

$|SK| = \frac{15}{\tan(37)} \approx \frac{15}{0.7536} \approx 19.905 \approx 19.91$

$|SR| = 2|SK| \approx 39.81$

Area of rectangle $\approx 11.979(39.81) \approx 477.084 \approx 477$

Area of triangle PKS $= \frac{1}{2} |PS||SK| \approx \frac{1}{2}(11.979)(19.905) \approx \frac{1}{2}(238.44) \approx 119.22 \approx 119$

Area of shaded portion = $477 - 119 = 358$ (to 3 significant figures)

3. Final Answer

(a) $\frac{x-y}{3}$

(b) (i) $|PS| = 12.0$ cm

(ii) $|SK| = 19.9$ cm

(iii) Area of shaded portion $= 358$ cm$^2$