1991 - WAEC Mathematics Past Questions and Answers - page 5

(1/2)(1/4) = 1/8

Sum of ∠s in a pentagon = (n - 2)180 = 540°
x° + 2x° + x° + 60° + x° + 10° + x° - 10° = 540°
6x° + 60° = 540°; x = 80°

Since < URP = 25°, then < XPU = 4 x 25° = 100°

\(\therefore\) < TPQ = 180° - 100° = 80°

\(\therefore\) < PQT = 180° - (80° + 25°) = 75°

< SQR = 75° - 25° = 50° (exterior angle = 2 opp interior angles)

\(\therefore\) < USQ = 180° - 50° = 130°

Let x be the exterior angle; interior = 3x; but x + 3x = 180^o

4x = 180^o; x = 45

= (\frac{360}{45}) = 8

From the figure, < PQR = 68°

\(\therefore\) < QRS = < QSR = \(\frac{180 - 68}{2}\) (base angles of an isos. triangle)

= 56°

\(\therefore\) < PRS = 90° - 56° = 34° (angles in a semi-circle)

Area = ¹/₂ x 16 x 16sin60^o = 64√3cm²

Let the radius of the arc PQ = r and the radius of the arc XY = R.

Length of arc PQ = \(\frac{\theta}{360} \times 2\pi r = 1\)

Length of arc XY = \(\frac{\theta}{360} \times 2\pi R = 3\)

Ratio of the arc = \(\frac{r}{R} = \frac{360 \times 2\pi \theta}{2\pi \theta \times 360 \times 3}\)

= \(\frac{1}{3}\)

Ratio of their area = \((\frac{1}{3})^2 = \frac{1}{9}\)

= 1 : 9

Area of (\Delta)DCF = 24cm²

Area of Quad= 2 x 24 = 48cm²