(a) In the diagram, AB is a tangent to the circle with centre O, and COB is a straight line. If CD//AB and < ABE = 40°, find: < ODE. (b) ABCD is a parallelo...

(a) Finding \( \angle ODE \) from the diagram

Reading the diagram: \(AB\) is a tangent touching the circle at \(A\); \(C\), \(O\) and \(B\) lie on one straight line (so \(CB\) passes through the centre \(O\)); \(E\) is the point where this line \(CB\) meets the circle on the right, so \(CE\) is a diameter. \(CD \parallel AB\) and \( \angle ABE = 40^\circ \).

Step 1: Use the tangent. A radius is perpendicular to a tangent at the point of contact, so \( \angle OAB = 90^\circ \).

In \( \triangle OAB \), \( \angle ABO = 40^\circ \), hence

\[ \angle AOB = 180^\circ - 90^\circ - 40^\circ = 50^\circ. \]

Step 2: Use the parallel chord. The line \(CB\) is a transversal cutting the parallel lines \(AB\) and \(CD\). By alternate angles,

\[ \angle DCB = \angle ABE = 40^\circ, \] so \( \angle DCO = 40^\circ \).

Step 3: Base angles of an isosceles triangle. In \( \triangle OCD \), \(OC = OD\) (both radii), so it is isosceles with

\[ \angle ODC = \angle OCD = 40^\circ. \]

Step 4: Angle in a semicircle. Since \(CE\) is a diameter and \(D\) lies on the circle, the angle it subtends is a right angle:

\[ \angle CDE = 90^\circ. \]

Step 5: Combine. The radius \(OD\) lies inside \( \angle CDE \), so

\[ \angle ODE = \angle CDE - \angle ODC = 90^\circ - 40^\circ = 50^\circ. \]

\( \angle ODE = 50^\circ \).

(b) Parallelogram \(ABCD\)

(i) Illustration. Draw parallelogram \(ABCD\) with vertices labelled in order. Mark side \(DC = 7\ \text{cm}\) along the base and side \(AD = 5\ \text{cm}\) meeting it at \(D\), with the interior angle \( \angle ADC = 125^\circ \) between them. The opposite sides are equal and parallel: \(AB = DC = 7\ \text{cm}\), \(BC = AD = 5\ \text{cm}\), and \( \angle ABC = 125^\circ \), while \( \angle DAB = \angle BCD = 55^\circ \).

(ii) Area. For a parallelogram, area equals the product of two adjacent sides and the sine of the included angle:

\[ \text{Area} = |DC| \times |AD| \times \sin(\angle ADC). \]

\[ \text{Area} = 7 \times 5 \times \sin 125^\circ = 35 \times 0.8192 = 28.67\ \text{cm}^2. \]

Area \( \approx 28.7\ \text{cm}^2 \) (to one decimal place).

(c) Evaluate \( 2x^2 - 2x \) when \( x = \tfrac{1}{2}(1 - \sqrt{2}) \)

First compute \( x^2 \):

\[ x^2 = \left(\frac{1-\sqrt{2}}{2}\right)^2 = \frac{(1-\sqrt{2})^2}{4} = \frac{1 - 2\sqrt{2} + 2}{4} = \frac{3 - 2\sqrt{2}}{4}. \]

Then

\[ 2x^2 = \frac{3 - 2\sqrt{2}}{2}, \qquad 2x = 1 - \sqrt{2} = \frac{2 - 2\sqrt{2}}{2}. \]

Therefore

\[ 2x^2 - 2x = \frac{3 - 2\sqrt{2}}{2} - \frac{2 - 2\sqrt{2}}{2} = \frac{3 - 2\sqrt{2} - 2 + 2\sqrt{2}}{2} = \frac{1}{2}. \]

\( 2x^2 - 2x = \dfrac{1}{2} \).