Selangor Trial 2025 Add Maths Paper 2, Question 14Solution of Triangles

Solved on video in English · Form 4, Solution of Triangles · 7:53 video

The question

(a)(i)

Solution by scale drawing is not accepted. Diagram 5 shows a quadrilateral PQRS with SR parallel to PQ and the diagonal PR drawn. Given PR = 7 cm, PQ = 4 cm, QR = 5 cm and angle PSR = 110°. (a)(i) Calculate ∠QPR.

(a)(ii)

(a)(ii) Calculate the length, in cm, of PS.

(a)(iii)

(a)(iii) Calculate the area of quadrilateral PQRS.

(b)(i)

(b) The line PQ is extended to Q' such that Q'R' = QR, P'R' = PR and ∠R'P'Q' = ∠RPQ. (i) Sketch triangle P'Q'R' which has a different shape from triangle PQR.

(b)(ii)

(b)(ii) Hence, find the value of ∠P'Q'R'.

Selangor Trial 2025 Add Maths Paper 2, Question 14: the question as drawn in Eduly's video7:53

Part (a)(i)

Step 1

In triangle PQR the three sides are known (PQ = 4, PR = 7, QR = 5). Use the cosine rule with QR opposite the required angle ∠QPR.

Why this step

All three sides are known and we want an angle, so the cosine rule fits. Put the side opposite the angle (QR) alone on the left.

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

Final answer

∠QPR = 44.42°

Where students lose marks

Watch out: put QR, the side opposite ∠QPR, on the left of the cosine rule, and mind the sign when rearranging.

Part (a)(ii)

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

Final answer

PS = 5.213 cm

Where students lose marks

Watch out: use SR parallel to PQ to get ∠SRP, pair each side with its opposite angle, and avoid rounding too early.

Part (a)(iii)

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

Final answer

Area = 17.68 cm²

Where students lose marks

Watch out: use the angle between the two sides, ∠SPR = 25.59° for triangle PSR (not 110°), and add both areas.

Part (b)(i)

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

Final answer

Sketch (see video)

Where students lose marks

Watch out: the new triangle is not a copy of PQR. Q' must carry the acute angle, so label the vertices carefully.

Part (b)(ii)

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

Final answer

∠P'Q'R' = 78.44°

Where students lose marks

Watch out: sin ∠P'Q'R' = 0.9799 gives 78.44° and 101.56°. The original triangle used 101.56°, so the new triangle takes the acute 78.44°.

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