Melaka Trial 2025 Add Maths Paper 1, Question 14Circular Measure

Solved on video in English · Form 5, Circular Measure · 5:56 video

The question

(a)

Diagram 6 shows sector JOK with centre O and a radius of 24 cm. Point L lies on the straight line OJ such that OL:OJ = 3:4. [Use π = 3.142]

(a) Calculate the value of θ, in radians.

(b)

Point L lies on OJ such that OL:OJ = 3:4, and LK is perpendicular to OJ. 142]

(b) Calculate the perimeter of the shaded region.

(c)

142]

Step 1

OL:OJ = 3:4 with OJ = radius = 24 cm, so OL = (3/4)(24) = 18 cm.

Why this step

OJ is the whole length (4 parts) and OL is 3 of those parts, so OL = 3/4 of 24. A wrong OL ruins every later step.

3 more steps, each with the reason for it, in the full solution

Final answer

θ = 0.7228 rad

Where students lose marks

Watch out: OL is adjacent to θ and OK is the hypotenuse, so use cos θ = OL/OK. Give θ in radians, not degrees.

5 more steps, each with the reason for it, in the full solution

Final answer

39.22 cm

Where students lose marks

Watch out: perimeter = LK + arc JK + LJ only. Leave out OL and OK, and don't forget LJ = 6 cm.

4 more steps, each with the reason for it, in the full solution

Final answer

65.2959 cm²

Where students lose marks

Watch out: subtract triangle OLK from sector JOK, use r = 24 (not 18) for the sector, and don't round θ early.

Watch every step of this question solved on video (5:56), with the reason behind each step and a box that checks your own answer.

No card. Every past-year question on video, in English and BM.