WAEC 2012 · Paper 2 · Q3

(a)
In the diagram, $ABCD$ is part of a right-angled triangle $ODC$ (with $A$ on $OD$ and $B$ on $OC$ ). If $|AB| = 6\text{ cm}$ , $|CD| = 15\text{ cm}$ , $|BC| = 8\text{ cm}$ , $\angle BCD = 90^\circ$ and $AB \parallel DC$ , calculate, correct to 1 decimal place, the: (i) height; (ii) perimeter of the triangle $ODC$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)(i)

$AB \parallel DC$ , so triangles $OBA$ and $OCD$ are similar and their sides are in proportion: $\frac{OB}{OC} = \frac{AB}{DC}$ .
With $OC = OB + 8$ : $\frac{OB}{OB + 8} = \frac{6}{15}$ .
Cross-multiply: $15\,OB = 6\,OB + 48$ , so $9\,OB = 48$ and $OB = 5\frac13$ cm.
The height is $OC = 5\frac13 + 8$
$= 13\frac13$
$\approx 13.3$ cm.

(ii)

$\angle OCD = 90^\circ$ , so $OD = \sqrt{15^2 + (13\frac13)^2}$
$= \sqrt{225 + 177.78}$
$\approx 20.07$ cm.
Perimeter $= 13.33 + 15 + 20.07 \approx 48.4$ cm.

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