(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 parallelogram in which |\(\overline{CD}\)| = 7 cm, I\(\overline{AD}\)I = 5 cm and < ADC= 125°.
(i) Illustrate the information in a diagram.
(ii) Find, correct to one decimal place, the area of the parallelogram.
(c) If x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\)). Evaluate (2x\(^2\) - 2x).
(a)
<ABE = <OBC (tangent and radius form a right angle)
<ABE = 40° (given)
Therefore, <OBC = 40°.
Since CD//AB, then <CDE = <ABE = 40° (alternate angles)
<ODE = <OBC + <CDE = 40° + 40° = 80°.
(b)
(i)
A ----------- B
\ /
\ /
\ /
\ /
\ /
\ /
D-------C
(125°)
(ii)
The area of the parallelogram is given by A = base x height. Since AD || BC, then the height of the parallelogram is given by the perpendicular distance between AD and BC, which is 7 cm. To find the length of the base, we use the cosine rule:
|\(\overline{AD}\)|² = |\(\overline{AB}\)|² + |\(\overline{BD}\)|² - 2|\(\overline{AB}\)||\(\overline{BD}\)| cos(ADC)
5² = x² + (x+7)² - 2x(x+7)cos(125°)
25 = 2x² + 14x + 49 + 2x(x+7)(-0.5736)
25 = 2x² + 14x + 49 - 1.1472x² - 10.0317x - 20.7032
0 = 0.8528x² + 4.0317x + 4.7032
Using the quadratic formula: x = (-b ± ?(b² - 4ac))/2a, we have:
x = (-4.0317 ± ?(4.0317² - 4(0.8528)(4.7032)))/(2(0.8528))
x = (-4.0317 ± 4.9988)/1.7056
x = 0.6559 or x = -3.6582
Since x represents a length, we discard the negative solution. Therefore, x = 0.6559 cm.
So, the area of the parallelogram is A = base x height = 0.6559 cm x 7 cm = 4.5913 cm² (to one decimal place).
(c)
x = \(\frac{1}{2}\)(1 - \(\sqrt{2}\))
2x² - 2x = 2\(\left(\frac{1}{2}\)(1 - \(\sqrt{2}\))²\) - 2(1 - \(\sqrt{2}\))
= (1 - 2\(\sqrt{2}\) + 2) - 2 + 2\(\sqrt{2}\)
= 1 + 2\(\sqrt{2}\).
Therefore, 2x² - 2x = 1 + 2\(\sqrt{2}\).
