In the diagram, ABCD is a trapezium in which \(AD... | WAEC SSCE - General Mathematics - 1999

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In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate :

(a) |AB| ;

(b) the area of the triangle BCD ;

(d) perimeter of the trapezium.

Answer

Answer Details

Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).

(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):

\[|AB|^2 = |BD|^2 - |AD|^2 = 17^2 - 15^2 = 289 - 225 = 64,\]\[|AB| = \sqrt{64} = \mathbf{8\text{ cm}}.\]

(b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):

\[\text{Area} = \tfrac{1}{2}\times |BC| \times |AB| = \tfrac{1}{2}\times 9 \times 8 = \mathbf{36\text{ cm}^2}.\]

(c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then

\[|CD| = \sqrt{(15-9)^2 + (8-0)^2} = \sqrt{6^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = \mathbf{10\text{ cm}}.\]

(d) Perimeter of the trapezium.

\[P = |AB| + |BC| + |CD| + |DA| = 8 + 9 + 10 + 15 = \mathbf{42\text{ cm}}.\]

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In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate : ...

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In the diagram, ABCD is a trapezium in which \(AD \parallel BC\) and \(< ABC\) is a right angle. If |AD| = 15 cm, |BD| = 17 cm and |BC| = 9 cm, calculate : ...

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