ECZ 2024 · O-Level Paper 2 · Question 3 — Vectors In Two Dimensions
In the diagram, EFG is a triangle in which EH = 4a, EF = 4b, FJ : JG = 1 : 3 and H is the midpoint of EG. FH and EJ meet at P.
(a) The following program is written in the form of pseudocode. Start Enter r, s IF s < r THEN Print 'error, s is not valid' ELSE A = pi*r*s END IF Print A Stop Draw a flowchart corresponding to the pseudocode above.[5]
Verified working — step 1
Terminals (StartStop) are rounded, the input and output boxes are parallelograms, the t…(b(i)(a)) In the diagram, EFG is a triangle in which EH = 4a, EF = 4b, FJ : JG = 1 : 3 and H is the midpoint of EG. Express FH in terms of a andorb.[1]
Verified working — step 1
FH = EH - EF = …(b(i)(b)) Express FG in terms of a andorb.[1]
Verified working — step 1
H is the midpoint of EG, so EG = 2EH = …(b(i)(c)) Express EJ in terms of a andorb.[1]
Verified working — step 1
Since FJ : JG = 1 : 3, FJ = (14)FG = (14)(8a - 4b) = …The full step-by-step working and final answer are included with Premium — checked against ECZ marking standards.
Original examination question © Examinations Council of Zambia. Worked solution and commentary © G12 Titan — not to be reproduced without permission.
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