(a) In the diagram, /PQ/ = 6 cm, /QR/ = 13 cm, /RS/ = 5 cm and < RSQ is a right- angled triangle. Calculate, correct to one decimal place, /PS/. (b) The dia...

(a) Calculating /PS/

From the diagram, \(P, Q, R\) lie on one straight line with \(PQ = 6\text{ cm}\) and \(QR = 13\text{ cm}\). The triangle \(QSR\) is right-angled at \(S\), with \(RS = 5\text{ cm}\) and hypotenuse \(QR = 13\text{ cm}\).

Step 1: Find QS using Pythagoras in triangle QSR.

\[ QS^2 = QR^2 - RS^2 = 13^2 - 5^2 = 169 - 25 = 144 \]

\[ QS = 12\text{ cm} \]

Step 2: Find angle RQS.

\[ \cos(\angle RQS) = \frac{QS}{QR} = \frac{12}{13} \]

Step 3: Use triangle PQS. Since \(P, Q, R\) are collinear, \(\angle PQS\) and \(\angle RQS\) are supplementary, so \(\cos(\angle PQS) = -\dfrac{12}{13}\).

Applying the cosine rule in triangle \(PQS\) with \(PQ = 6\), \(QS = 12\):

\[ PS^2 = PQ^2 + QS^2 - 2\,(PQ)(QS)\cos(\angle PQS) \]

\[ PS^2 = 6^2 + 12^2 - 2(6)(12)\left(-\tfrac{12}{13}\right) \]

\[ PS^2 = 36 + 144 + \frac{1728}{13} = 180 + 132.92 = 312.92 \]

\[ PS = \sqrt{312.92} \approx 17.7\text{ cm} \]

/PS/ \(\approx\) 17.7 cm.

(b) Surface area of the structure

The structure is a cone (height \(24\text{ cm}\), base radius \(7\text{ cm}\)) mounted on a hemisphere of the same radius \(7\text{ cm}\). The exposed surface is the curved surface of the cone plus the curved surface of the hemisphere (the joining circle is hidden).

Slant height of cone:

\[ l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25\text{ cm} \]

Curved surface area of cone:

\[ \pi r l = \frac{22}{7} \times 7 \times 25 = 550\text{ cm}^2 \]

Curved surface area of hemisphere:

\[ 2\pi r^2 = 2 \times \frac{22}{7} \times 7^2 = 2 \times 22 \times 7 = 308\text{ cm}^2 \]

Total surface area:

\[ 550 + 308 = 858\text{ cm}^2 \]

Surface area \(= 858\text{ cm}^2\) (to 3 significant figures).