(a) In the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm, /PS/ = 14.4 cm and /TQ/ = 6 cm. Calculate the area of the quadrilateral QRST. (b) Two opposite sides ...

Answer Details

(a) Area of quadrilateral \(QRST\). From the diagram, \(P, T, S\) lie on a straight top line with \(PS = 14.4\text{ cm}\); \(TQ = 6\text{ cm}\) is perpendicular to \(PS\) at \(T\) (\(\angle PTQ = 90^\circ\)); \(SR\) is perpendicular to \(PS\) at \(S\) (\(\angle PSR = 90^\circ\)); and \(P, Q, R\) are collinear along the slant \(PR\), with \(PQ = 10\text{ cm}\).

Step 1 - the small triangle \(PTQ\). It is right-angled at \(T\), so by Pythagoras

\[PT = \sqrt{PQ^2 - TQ^2} = \sqrt{10^2 - 6^2} = \sqrt{100-36} = \sqrt{64} = 8\text{ cm}.\]

Step 2 - similar triangles. Triangles \(PTQ\) and \(PSR\) share \(\angle P\) and each has a right angle, so they are similar with scale factor

\[k = \frac{PS}{PT} = \frac{14.4}{8} = 1.8.\]

Hence

\[SR = k\cdot TQ = 1.8\times 6 = 10.8\text{ cm}.\]

Step 3 - subtract the areas. The quadrilateral \(QRST\) is the large triangle \(PSR\) with the small triangle \(PTQ\) removed:

\[[PSR] = \tfrac12\cdot PS\cdot SR = \tfrac12\times 14.4\times 10.8 = 77.76\text{ cm}^2,\]\[[PTQ] = \tfrac12\cdot PT\cdot TQ = \tfrac12\times 8\times 6 = 24\text{ cm}^2.\]\[\text{Area of }QRST = 77.76 - 24 = \boxed{53.76\text{ cm}^2}.\]

(b) Ratio of the rectangle's area to the square's. Let the square have side \(s\), so its area is \(s^2\).

Two opposite sides are decreased by \(10\%\) to \(0.9s\); the other two are increased by \(15\%\) to \(1.15s\). The rectangle's area is

\[(0.9s)(1.15s) = 1.035\,s^2.\]

Therefore

\[\frac{\text{Rectangle}}{\text{Square}} = \frac{1.035\,s^2}{s^2} = 1.035 = \frac{1035}{1000} = \boxed{207 : 200}.\]