Secondary 4 Mathematics-Cultural, Social and Technical Option Ministerial Examination - Québec study materials
Secondary 4 Mathematics-Cultural, Social and Technical Option Ministerial Examination - Québec study resources

Secondary 4 Mathematics-Cultural, Social and Technical Option Ministerial Examination - Québec: Geometric Reasoning

Secondary 4 Mathematics-Cultural, Social and Technical Option Ministerial Examination - Québec · Work through your exam, one topic at a time.

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Geometric Reasoning

What this domain tests

Geometric Reasoning assesses plane figures and solids, metric and trigonometric relations, similarity, and transformations. The exam expects you to choose the right relation, justify it, and compute with correct units. Many items are embedded in physical contexts such as land, packaging, or construction.

Core rules and concepts

  • Plane figures: perimeter and area formulas for triangles, quadrilaterals, and circles. The area of a trapezoid is $\frac{(b_1+b_2)h}{2}$; the area of a circle is $\pi r^2$.
  • Solids: volume of a prism or cylinder is base area times height; a pyramid or cone is one-third of that. Surface area sums the areas of all faces, including bases.
  • Metric relations in right triangles: the Pythagorean theorem $a^2+b^2=c^2$; the altitude-to-hypotenuse relations $h^2=mn$, $a^2=mc$, $b^2=nc$.
  • Trigonometry: $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Use the inverse functions to recover an angle.
  • Similarity: corresponding angles are equal and corresponding sides are proportional; the ratio of areas is the square of the scale factor, and the ratio of volumes is its cube.
  • Transformations: translations, rotations, reflections, and dilations preserve shape; only dilations change size, and they change area by the square of the factor.

Worked example

A right triangle has legs 6 cm and 8 cm. Find the hypotenuse, then the altitude to the hypotenuse.

Hypotenuse: $c^2=6^2+8^2=36+64=100$, so $c=10$ cm.

The altitude $h$ splits the hypotenuse into segments $m$ and $n$. Using $a^2=mc$ with $a=6$: $36=10m$, so $m=3.6$ and $n=10-3.6=6.4$. Then $h^2=mn=3.6\times6.4=23.04$, giving $h=4.8$ cm. Check with the area: $\frac{6\times8}{2}=24$ and $\frac{10\times4.8}{2}=24$, which agrees.

Traps that cost marks

  • Mixing units (cm with m) before computing, or reporting area in cm instead of cm².
  • Applying the Pythagorean theorem to a triangle that is not right-angled.
  • Using the sine ratio when the cosine is required because the wrong side was labelled adjacent.
  • Forgetting that doubling the dimensions of a solid multiplies volume by 8, not 2.
  • Leaving the calculator in the wrong angle mode (degrees versus radians).