(a) In the diagram, < PTQ = < PSR = 90°, /PQ/ = 10 cm,
(a).jpg)
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 of a square are each decreased by 10% while the other two are each increased by 15% to form a rectangle. Find the ratio of the area of the rectangle to that of the square.
Explanation
(a) \((PT)^{2} + (TQ)^{2} = (PQ)^{2}\)
\((PT)^{2} + 6^{2} = 10^{2}\)
\((PT)^{2} = 100 - 36 = 64\)
\(\therefore PT = 8 cm\)
\(\therefore \frac{8}{6} = \frac{14.4}{SR}\)
\(SR = \frac{6 \times 14.4}{8}\)
\(SR = 10.8 cm\)
Area of the quadrilateral QRST = \(\frac{1}{2} (6 + 10.8) \times 6.4\)
= \(53.76 cm^{2}\)
(b) Opp sides of a square = \((\frac{(100 - 10)x}{100} = 0.9x\)
other two sides of the square = \(\frac{(100 + 15)x}{100} = 1.15x\)
Area of the square : \(x \times x = x^{2}\)
Area of the rectangle : \(0.9x \times 1.15x = 1.035x^{2}\)
\(\therefore \text{The ratio of the area of the rectangle to the square} = 1.035x^{2} : x^{2}\)
= \(1.035 : 1\)