In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm, \(\overline{...
Question 1 Report
In the diagram, \(\overline{PQ//RS}\) is a trapezium with QR//PS. U and T are points on \(\overline{PS}\) such that \(\overline{|PU|}\) = 5 cm,
\(\overline{|QU|}\) = 12 cm and ?PUQ= ?STR =90°. If the area of PQR = 20 cm\(^2\),
calculate, correct to the nearest whole number, the:
(a) perimeter; (b) area; of the trapezium
Answer Details
Reading the diagram. The trapezium has \(QR \parallel PS\). \(QU\) and \(RT\) are drawn perpendicular to \(PS\), so \(QU = RT = 12\text{ cm}\) is the vertical height of the trapezium. From the figure: \(|PU| = 5\text{ cm}\), \(|QU| = 12\text{ cm}\), \(\angle PUQ = \angle STR = 90^\circ\), the angle at \(S\) is \(50^\circ\), and the area of \(\triangle PQR = 20\text{ cm}^2\).
Step 1: The slant side \(PQ\). Triangle \(PUQ\) is right-angled at \(U\):
\[|PQ| = \sqrt{5^2 + 12^2} = \sqrt{25+144} = \sqrt{169} = 13\text{ cm}.\]Step 2: The top side \(QR\). Since \(QU \perp PS\), \(RT \perp PS\) and \(QR \parallel PS\), the figure \(QURT\) is a rectangle, so \(QR = UT\). Vertices \(Q\) and \(R\) both lie \(12\text{ cm}\) above line \(PS\), while \(P\) lies on \(PS\); hence the height of \(\triangle PQR\) on base \(QR\) is \(12\text{ cm}\):
\[\text{Area }\triangle PQR = \tfrac{1}{2}\times QR \times 12 = 6\,QR.\]\[6\,QR = 20 \;\Rightarrow\; QR = \tfrac{20}{6} = 3.33\text{ cm}.\]Step 3: The right-hand side using \(\angle S = 50^\circ\). Triangle \(STR\) is right-angled at \(T\) with \(RT = 12\):
\[|TS| = \frac{RT}{\tan 50^\circ} = \frac{12}{1.1918} = 10.07\text{ cm},\qquad |RS| = \frac{RT}{\sin 50^\circ} = \frac{12}{0.7660} = 15.67\text{ cm}.\]Step 4: The base \(PS\).
\[|PS| = |PU| + |UT| + |TS| = 5 + 3.33 + 10.07 = 18.40\text{ cm}.\](a) Perimeter of the trapezium.
\[P = |PQ| + |QR| + |RS| + |SP| = 13 + 3.33 + 15.67 + 18.40 = 50.40\text{ cm}.\]Perimeter \(\approx \mathbf{50\text{ cm}}\) (nearest whole number).
(b) Area of the trapezium. Parallel sides \(QR = 3.33\) and \(PS = 18.40\), height \(12\):
\[A = \tfrac{1}{2}(QR + PS)\times h = \tfrac{1}{2}(3.33 + 18.40)\times 12 = \tfrac{1}{2}(21.73)(12) = 130.4\text{ cm}^2.\]Area \(\approx \mathbf{130\text{ cm}^2}\) (nearest whole number).
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