WAEC 2013 · Paper 2 · Q8

(a)
If $(3 - x)$ , $6$ , $(7 - 5x)$ are consecutive terms of a geometric progression (G.P.) with constant ratio $r > 0$ , find the: (i) value of $x$ ; (ii) constant ratio.
Separate values with commas, e.g. 3, −2
(b)
In the diagram, $ABCD$ is a quadrilateral with $\angle ABC = 90^\circ$ , $|AB| = 3\text{ cm}$ , $|BC| = 4\text{ cm}$ , $|CD| = 6\text{ cm}$ and $|DA| = 7\text{ cm}$ . Calculate $\angle ADC$ , correct to the nearest degree.

Worked solution (try it first)

(a)(i)

In a G.P. the ratio of consecutive terms is constant: $\frac{6}{3 - x} = \frac{7 - 5x}{6}$ .
Cross-multiply: $36 = (3 - x)(7 - 5x) = 21 - 22x + 5x^2$ .
So $5x^2 - 22x - 15 = 0$ , which factorises as $(5x + 3)(x - 5) = 0$ .
So $x = -\frac35$ or $x = 5$ .
With $x = 5$ the terms are $-2, 6, -18$ , whose ratio $-3$ is not positive.
So $x = -\frac35$ .

(ii)

With $x = -\frac35$ the terms are $3.6, 6, 10$ , and $r = \frac{6}{3.6} = \frac53$ .

(b)

Join $AC$ .
Triangle $ABC$ is right-angled at $B$ , so $|AC| = \sqrt{3^2 + 4^2} = 5\text{ cm}$ .
In triangle $ACD$ all three sides are known, so use the cosine rule for the angle at $D$ : $\cos\angle ADC = \frac{7^2 + 6^2 - 5^2}{2 \times 7 \times 6}$
$= \frac{60}{84}$
$\approx 0.7143$ .
So $\angle ADC \approx 44^\circ$ .

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