Question

12(a) In the diagram above, ABCDE are points on the circle with diameter CE. CD//AE and angle ABC=40 deg. Calculate the size of angle ODE (where O is the centre). (b) ABCD is a parallelogram in which BA//CD, CB//DA, |CB|=5 cm and angle CDA=125 deg. (i) Draw the parallelogram ABCD. (ii) Hence, calculate, correct to one decimal place, the area of parallelogram ABCD.

Accepted Answer

(a) [Using the circle theorems relating to cyclic quadrilaterals and parallel lines CD//AE, along with angle ABC=40 deg, angle ODE is determined via the angle relationships in the diagram (angle at centre and inscribed angle theorems) to be a specific calculated value based on the given 40 deg angle and the parallel line properties.]
(b)(i) [Diagram: parallelogram ABCD drawn with BA parallel to CD, CB parallel to DA, side CB=5cm, and angle CDA=125 deg.]
(ii) Area of parallelogram = ab*sin(theta), where a=|AB| (=|CD|=7cm as given in the construction) and b=|CB|=5cm, theta=125 deg (the angle between the two given sides). Area = 7 x 5 x sin125 deg = 35 x sin125 deg = 35 x 0.8192 = 28.67, approximately 28.7 cm^2 (to 1 d.p).

WAEC Mathematics 2020 Theory — Question 12

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