WAEC 2010 · Paper 2 · Q9

$OABCD$ is a right pyramid with a rectangular base $ABCD$ . Its vertical height is $OG$ . If $|AB| = 6\text{ cm}$ , $|BC| = 8\text{ cm}$ and each slant edge is $13\text{ cm}$ ,

(a)
Calculate, correct to one decimal place, the: (i) vertical height $|OG|$ ; (ii) angle between a slant edge and the base $ABCD$ ; (iii) angle between the triangle $OAB$ and the base $ABCD$ .
Separate values with commas, e.g. 3, −2
(b)
Find the volume of the pyramid.

Worked solution (try it first)

(a)(i)

$G$ is the centre of the base, so $|AG|$ is half the diagonal: $|AC| = \sqrt{6^2 + 8^2} = 10$ , so $|AG| = 5\text{ cm}$ .
Pythagoras in triangle $OGA$ : $|OG|^2 = 13^2 - 5^2 = 144$ , so $|OG| = 12.0\text{ cm}$ .

(ii)

The angle between the edge $OA$ and the base is $\angle OAG$ : $\cos \angle OAG = \dfrac{5}{13}$ .
So $\angle OAG = 67.4^\circ$ .

(iii)

Let $X$ be the mid-point of $AB$ .
$GX$ is perpendicular to $AB$ and $|GX| = \frac12|BC| = 4\text{ cm}$ .
The angle between face $OAB$ and the base is $\angle OXG$ : $\tan \angle OXG = \dfrac{12}{4} = 3$ , so the angle is $71.6^\circ$ .

(b)

Volume $= \frac13 \times \text{base area} \times \text{height}$
$= \frac13 \times 6 \times 8 \times 12$
$= 192\text{ cm}^3$ .

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