Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 13Solution of Triangles

Solved on video in English · Form 4, Solution of Triangles · 6:19 video

The question

(a)

Diagram 9 shows a cuboid ABCDEFGH. FH, FA and AH are straight lines that form a shaded triangle FAH. It is given that AC = 17 cm, CH = 8 cm and GH = 5 cm.

(a) Find ∠AFH.

(b)

Diagram 9 shows a cuboid ABCDEFGH with shaded triangle FAH, where AC = 17 cm, CH = 8 cm and GH = 5 cm.

(b) Find the area of the inclined plane FAH.

(c)

(c) Find the shortest distance from F to the line AH.

(d)

(d) Sketch a triangle F'A'H' which has a different shape from triangle FAH such that F'A' = FA, F'H' = FH and ∠F'H'A' = ∠FHA. Hence, state the size of ∠F'A'H'.

Kuala Lumpur Trial 2025 Add Maths Paper 2, Question 13: the question as drawn in Eduly's video6:19

Part (a)

Step 1

Find the three sides of triangle FAH from the cuboid edges (5, 8 and 17): AF = √(8² + 5²) = √89, AH = √(8² + 17²) = √353, FH = √(5² + 17²) = √314.

Why this step

Each side of the triangle is a diagonal of one face of the box, so Pythagoras on that face's two edges gives it. Pair the edges carefully.

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

Final answer

∠AFH = 81.40°

Where students lose marks

Watch out: √353 (AH) is opposite ∠AFH, so it is the side on its own in the cosine rule.

Part (b)

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

Final answer

Area = 82.65 cm²

Where students lose marks

Watch out: use the two sides around the angle (AF and FH with ∠AFH), and do not round s early in Heron's formula.

Part (c)

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

Final answer

8.798 cm

Where students lose marks

Watch out: take AH (√353) as the base, and remember ½ × base × height, so double the area before dividing.

Part (d)

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

Final answer

∠F'A'H' = 111.17°

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