ARCH-M1-10: 3D Mensuration (Cylinder/Sphere/Cone) & Right-Triangle Trigonometry in Degrees

Full 15-Q Timed Drill

Step 1 of 6 • The Core Idea

ARCH-M1-10: Why the 155s Classroom Method Fails Under the 90s Clock

Standard Classroom Method (Too Slow)~155s (Blows 90s Clock)

First find the diagonal of the

3\text{ cm}

4\text{ cm}

base:

d_{\text{base}} = \sqrt{3^2 + 4^2} = 5\text{ cm}

. Then use Pythagoras with the vertical edge

12\text{ cm}

d = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ cm}

(~90s).

The Fast ShortcutTarget: 45s (Saves 110s)

**Line 1 (Chained Triples

3\text{–}4\text{–}5 \to 5\text{–}12\text{–}13

):**

d = \sqrt{3^2 + 4^2 + 12^2}

.
Line 2 (Evaluate):

\sqrt{25 + 144} = 13\text{ cm}

(15s).

Where Students Lose Time: In 20 past UAT-UK questions (8.89% of Mathematics 1), candidates lose 110s by computing curved surface area 2*pi*r*h or pi*r*l only and forgetting the closed circular end(s). Instead, apply M5 with Maths 1 skill M5.11.

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