(a) Simplify : \(\frac{x^{2} - y^{2}}{3x + 3y}\) (b) In the diagram, PQRS is a rectangle. /PK/ = 15 cm, /SK/ = /KR/ and <PKS = 30°. Calculate, correct to th...

(a) Simplify \(\dfrac{x^{2}-y^{2}}{3x+3y}\).

Factorise the numerator as a difference of two squares and take out the common factor in the denominator:

\[\frac{x^{2}-y^{2}}{3x+3y} = \frac{(x-y)(x+y)}{3(x+y)} = \frac{x-y}{3}.\]

(b) Reading the diagram. \(PQRS\) is a rectangle with \(P\) top-left, \(Q\) top-right, \(R\) bottom-right and \(S\) bottom-left. \(K\) lies on the base \(SR\) with \(|SK| = |KR|\) (so \(K\) is the midpoint of \(SR\)). The line \(PK = 15\text{ cm}\) is drawn, and \(\angle PKS = 30^\circ\). Triangle \(PSK\) has its right angle at the rectangle corner \(S\).

(i) Find \(|PS|\). In right-angled triangle \(PSK\), \(PS\) is opposite the \(30^\circ\) angle at \(K\) and \(PK\) is the hypotenuse:

\[|PS| = PK\sin 30^\circ = 15\times 0.5 = 7.50\text{ cm}.\]

(ii) Find \(|SK|\). \(SK\) is adjacent to the \(30^\circ\) angle:

\[|SK| = PK\cos 30^\circ = 15\times 0.8660 = 12.99 \approx 13.0\text{ cm}.\]

(iii) Area of the shaded portion. The unshaded region is triangle \(PSK\); the shaded portion is the rest of the rectangle. Since \(K\) is the midpoint of \(SR\), the full base is \(SR = 2|SK| = 2\times 12.99 = 25.98\text{ cm}\), and the height is \(|PS| = 7.50\text{ cm}\).

\[\text{Area of rectangle} = 25.98\times 7.50 = 194.85\text{ cm}^2,\]

\[\text{Area of triangle } PSK = \tfrac{1}{2}\times |SK|\times |PS| = \tfrac{1}{2}\times 12.99\times 7.50 = 48.71\text{ cm}^2.\]

\[\text{Shaded area} = 194.85 - 48.71 = 146.14 \approx 146\text{ cm}^2.\]

Answers: (a) \(\dfrac{x-y}{3}\); (b)(i) \(|PS| = 7.50\text{ cm}\); (ii) \(|SK| = 13.0\text{ cm}\); (iii) shaded area \(= 146\text{ cm}^2\) (3 s.f.).