MYP Inquiry Framework
Key Concept Space & Relationships
Related Concepts Measurement, Generalization, Models
Global Context Scientific and technical innovation (Structural architecture & navigation)
Statement of Inquiry Spatial relationships and geometrical measurements allow structural engineers to model and verify physical stability across two and three dimensions.
1.1

Triangle Congruence & Similarity Properties

  • Angle Sum & Exterior Angle: The interior angles of any planar triangle sum to $180^\circ$. An exterior angle equals the sum of the two opposite interior angles: $$\theta_{\text{ext}} = \alpha + \beta$$
  • Congruence Criteria ($\cong$): Two triangles are congruent (identical in shape and size) if any of the following tests are satisfied:
    • SSS: Three pairs of corresponding sides are equal.
    • SAS: Two sides and the included angle are equal.
    • ASA / AAS: Two angles and a corresponding side are equal.
    • RHS: Right-angle, hypotenuse, and one corresponding leg are equal.
  • Similarity Criteria ($\sim$): Two triangles are similar (same shape, proportional size) if:
    • AAA: All corresponding interior angles are equal.
    • SAS (Similarity): Two pairs of sides are in the same ratio and the included angle is equal.
    • SSS (Similarity): All three pairs of corresponding sides are in the same ratio ($k$).
  • Geometric Scale Factor Relationships: If the linear scale factor between two similar figures is $k$: $$\text{Linear Ratio} = k, \qquad \text{Area Ratio} = k^2, \qquad \text{Volume Ratio} = k^3$$
2D Pythagoras: a² + b² = c² b (base) a (height) c (hypotenuse) 3D Space Diagonal: d = √(l² + w² + h²) d
1.2

The Pythagorean Theorem in 2D & Converse

  • The Theorem: In any right-angled triangle with hypotenuse $c$ and legs $a$ and $b$: $$\mathbf{a^2 + b^2 = c^2} \implies c = \sqrt{a^2 + b^2}$$
  • Converse of Pythagoras: Given a triangle with side lengths $a, b, c$ (where $c$ is the longest side):
    • If $a^2 + b^2 = c^2$, the triangle is right-angled ($\angle C = 90^\circ$).
    • If $a^2 + b^2 > c^2$, the triangle is acute-angled ($\angle C < 90^\circ$).
    • If $a^2 + b^2 < c^2$, the triangle is obtuse-angled ($\angle C > 90^\circ$).
  • Common Integer Pythagorean Triples: $$(3, 4, 5), \quad (5, 12, 13), \quad (7, 24, 25), \quad (8, 15, 17), \quad (9, 40, 41)$$ Any integer multiple $k(a, b, c)$ is also a Pythagorean triple (e.g., $6-8-10$).
1.3

3D Pythagoras: Space Diagonals & Pyramids

  • Space Diagonal of a Rectangular Cuboid: Consider a cuboid with length $l$, width $w$, and height $h$.
    1. Base diagonal on the floor: $d_{\text{base}}^2 = l^2 + w^2$.
    2. Space diagonal right triangle with vertical height: $d^2 = d_{\text{base}}^2 + h^2 = (l^2 + w^2) + h^2$.
    $$\mathbf{d = \sqrt{l^2 + w^2 + h^2}}$$
  • Slant Height of a Regular Pyramid: For a right pyramid with square base of side $s$ and vertical height $H$: $$\text{Slant height } s_h = \sqrt{H^2 + \left(\frac{s}{2}\right)^2}$$ $$\text{Slant edge } e = \sqrt{s_h^2 + \left(\frac{s}{2}\right)^2} = \sqrt{H^2 + 2\left(\frac{s}{2}\right)^2}$$
  • Slant Height of a Cone: $$l = \sqrt{r^2 + h^2}$$

Worked Examples (Criterion A & C)

Example 1 3D Space Diagonal & Maximum Object Length

Problem: A closed delivery container has internal dimensions: length $60\text{ cm}$, width $40\text{ cm}$, and height $30\text{ cm}$.
(a) Calculate the length of the internal space diagonal.
(b) Can a rigid metal rod of length $80\text{ cm}$ fit completely inside the container?

Step 1: Apply the 3D Pythagorean theorem formula $$d = \sqrt{l^2 + w^2 + h^2}$$
Step 2: Substitute dimensions $$d = \sqrt{60^2 + 40^2 + 30^2} = \sqrt{3600 + 1600 + 900} = \sqrt{6100}$$
Step 3: Calculate numerical value $$d = \sqrt{6100} \approx \mathbf{78.1\text{ cm}} \quad \text{(3 s.f.)}$$
Step 4: Draw a logical conclusion for part (b)

The maximum straight-line distance inside the container is its space diagonal ($78.1\text{ cm}$). Since the rod length is $80\text{ cm} > 78.1\text{ cm}$, the rod cannot fit completely inside the box.

Criterion A & D: Authentic spatial modeling with reasoned physical conclusion
Example 2 Similar Triangles Area Ratio Problem

Problem: In triangle $ABC$, a line parallel to $BC$ intersects side $AB$ at $D$ and side $AC$ at $E$. If $AD = 6\text{ cm}$, $DB = 4\text{ cm}$, and the area of $\triangle ADE$ is $45\text{ cm}^2$:
(a) Prove that $\triangle ADE \sim \triangle ABC$.
(b) Find the area of the quadrilateral $DBCE$.

Step 1: Prove similarity using angle properties

Since $DE \parallel BC$, $\angle ADE = \angle ABC$ (corresponding angles) and $\angle AED = \angle ACB$ (corresponding angles). $\angle A$ is common. By AAA, $\mathbf{\triangle ADE \sim \triangle ABC}$.

Step 2: Determine linear scale factor $k$

Total side $AB = AD + DB = 6 + 4 = 10\text{ cm}$.

$$k = \frac{AB}{AD} = \frac{10}{6} = \frac{5}{3}$$
Step 3: Apply the area ratio rule $$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle ADE)} = k^2 = \left(\frac{5}{3}\right)^2 = \frac{25}{9}$$ $$\text{Area}(\triangle ABC) = \frac{25}{9} \times 45 = 25 \times 5 = 125\text{ cm}^2$$
Step 4: Find the area of quadrilateral $DBCE$ $$\text{Area}(DBCE) = \text{Area}(\triangle ABC) - \text{Area}(\triangle ADE) = 125 - 45 = \mathbf{80\text{ cm}^2}$$ Criterion C: Explicit geometric deductive reasoning connecting scale factor to area

MYP 4–5 Practice Exercises: Triangle Properties & Pythagoras

Tier 1 (Standard Right-Angle & Congruence), Tier 2 (3D Diagonals & Similarity), and Tier 3 (Extended Proofs & Geometric Optimization).

Question 1 Tier 1: Standard 3 Marks

Find the length of the hypotenuse in a right-angled triangle with legs of $9\text{ cm}$ and $12\text{ cm}$.

View Full Method Markscheme
[M1] Apply Pythagoras: $c^2 = a^2 + b^2 = 9^2 + 12^2$.
[M1] Evaluate squares: $c^2 = 81 + 144 = 225$.
[A1] Take square root: $c = \sqrt{225} = \mathbf{15\text{ cm}}$.
Question 2 Tier 1: Standard 3 Marks

Use the converse of Pythagoras to determine whether a triangle with side lengths $10\text{ cm}$, $24\text{ cm}$, and $26\text{ cm}$ is right-angled.

View Full Method Markscheme
[M1] Test the sum of squares of the two shorter sides: $10^2 + 24^2 = 100 + 576 = 676$.
[M1] Square the longest side: $26^2 = 676$.
[A1] Since $10^2 + 24^2 = 26^2$ ($676 = 676$), by the converse of Pythagoras, $\mathbf{\text{the triangle is right-angled}}$.
Question 3 Tier 1: Standard 3 Marks

An isosceles triangle has equal sides of length $13\text{ cm}$ and a base of length $10\text{ cm}$. Find the perpendicular height and the area of the triangle.

View Full Method Markscheme
[M1] The altitude bisects the base into two segments of $5\text{ cm}$ each. Form right triangle: $h^2 + 5^2 = 13^2$.
[M1] Solve for $h$: $h^2 = 169 - 25 = 144 \implies h = \mathbf{12\text{ cm}}$.
[A1] Area: $A = \frac{1}{2} \times \text{base} \times h = \frac{1}{2} \times 10 \times 12 = \mathbf{60\text{ cm}^2}$.
Question 4 Tier 2: Intermediate 4 Marks

Find the exact length of the space diagonal of a cube with edge length $7\text{ cm}$.

View Full Method Markscheme
[M1] In a cube, $l = w = h = 7$. Formula: $d = \sqrt{s^2 + s^2 + s^2} = \sqrt{3s^2}$.
[M1] Substitute $s = 7$: $d = \sqrt{3(7^2)} = \sqrt{3 \times 49}$.
[A1] Exact length: $\mathbf{7\sqrt{3}\text{ cm}}$ ($\approx 12.1\text{ cm}$).
Question 5 Tier 2: Intermediate 4 Marks

Two similar triangles have areas of $32\text{ cm}^2$ and $72\text{ cm}^2$. If the perimeter of the smaller triangle is $20\text{ cm}$, find the perimeter of the larger triangle.

View Full Method Markscheme
[M1] Find area ratio: $\frac{A_2}{A_1} = \frac{72}{32} = \frac{9}{4}$.
[M1] Linear scale factor is square root of area ratio: $k = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1.5$.
[M1] Multiply perimeter by linear scale factor: $P_2 = k \times P_1 = 1.5 \times 20$.
[A1] Perimeter of larger triangle: $\mathbf{30\text{ cm}}$.
Question 6 Tier 2: Intermediate 4 Marks

A right square-based pyramid has a base side of $16\text{ cm}$ and a vertical height of $15\text{ cm}$. Calculate:
(a) The slant height of each triangular face.
(b) The length of each sloping corner edge.

View Full Method Markscheme
[M1] Distance from center to side midpoint is $\frac{16}{2} = 8\text{ cm}$. Slant height: $s_h = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289}$.
[A1] Slant height is $\mathbf{17\text{ cm}}$.
[M1] Corner edge right triangle: $e = \sqrt{s_h^2 + 8^2} = \sqrt{17^2 + 8^2} = \sqrt{289 + 64} = \sqrt{353}$.
[A1] Corner edge length: $\mathbf{\sqrt{353} \approx 18.8\text{ cm}}$.
Question 7 Tier 2: Intermediate 4 Marks

A ladder of length $6.5\text{ m}$ rests against a vertical wall with its base $2.5\text{ m}$ away from the wall. If the top of the ladder slips down by $0.8\text{ m}$, by how much does the bottom slide outward?

View Full Method Markscheme
[M1] Initial height on wall: $h_1 = \sqrt{6.5^2 - 2.5^2} = \sqrt{42.25 - 6.25} = \sqrt{36} = 6.0\text{ m}$.
[M1] New height after slipping: $h_2 = 6.0 - 0.8 = 5.2\text{ m}$.
[M1] New base distance: $b_2 = \sqrt{6.5^2 - 5.2^2} = \sqrt{42.25 - 27.04} = \sqrt{15.21} = 3.9\text{ m}$.
[A1] Outward slide distance: $\Delta b = 3.9 - 2.5 = \mathbf{1.4\text{ m}}$.
Question 8 Tier 3: Extended 5 Marks

Prove that for any right-angled triangle with inradius $r$, hypotenuse $c$, and legs $a, b$: $$r = \frac{a + b - c}{2}$$

View Full Method Markscheme
[M1] By tangent properties from an external point to an incircle, the segments of the legs meeting at the right angle each equal the inradius $r$.
[M1] The remaining portions of legs $a$ and $b$ are $(a - r)$ and $(b - r)$ respectively.
[M1] The tangents to the hypotenuse from vertices $A$ and $B$ have lengths equal to these remaining portions: $c = (a - r) + (b - r)$.
[M1] Simplify: $c = a + b - 2r \implies 2r = a + b - c$.
[A1] Divide by $2$: $\mathbf{r = \frac{a + b - c}{2}}$. [Proved]
Question 9 Tier 3: Extended 5 Marks

In a triangle, the sides are $x$, $x+1$, and $x+2$.
(a) Find the value of $x$ for which this triangle is right-angled.
(b) State the values of $x$ for which this triangle is acute-angled.

View Full Method Markscheme
[M1] Part (a): Set up Pythagoras with longest side $(x+2)$: $x^2 + (x+1)^2 = (x+2)^2$.
[M1] Expand: $x^2 + x^2 + 2x + 1 = x^2 + 4x + 4 \implies x^2 - 2x - 3 = 0$.
[M1] Factorise: $(x - 3)(x + 1) = 0$. Since side length $x > 0$, $\mathbf{x = 3}$ (giving the $3-4-5$ triangle).
[M1] Part (b): For triangle to be acute, $a^2 + b^2 > c^2 \implies x^2 - 2x - 3 > 0 \implies (x - 3)(x + 1) > 0$. For positive $x$, this requires $x > 3$.
[A1] Furthermore, by the triangle inequality $x + (x+1) > x+2 \implies 2x+1 > x+2 \implies x > 1$. Thus, the triangle is acute for all $\mathbf{x > 3}$.
Question 10 Tier 3: Extended 5 Marks

A spider sits at the bottom corner $A(0, 0, 0)$ of a rectangular room measuring $8\text{ m} \times 6\text{ m} \times 3\text{ m}$. A fly rests at the opposite top corner $B(8, 6, 3)$.
(a) Find the straight-line distance through the air between spider and fly.
(b) Find the shortest distance the spider can crawl along the walls/floor/ceiling to reach the fly.

View Full Method Markscheme
[M1] Part (a): Space diagonal $d = \sqrt{8^2 + 6^2 + 3^2} = \sqrt{64 + 36 + 9} = \sqrt{109} \approx \mathbf{10.44\text{ m}}$.
[M1] Part (b): Flatten the net of the cuboid. The crawl path is the hypotenuse of a 2D unfolded net with legs $(L+H)$ and $W$, or $(W+H)$ and $L$, or $(L+W)$ and $H$.
[M1] Route 1: $\sqrt{(8 + 3)^2 + 6^2} = \sqrt{11^2 + 36} = \sqrt{121 + 36} = \sqrt{157} \approx 12.53\text{ m}$.
[M1] Route 2: $\sqrt{(6 + 3)^2 + 8^2} = \sqrt{9^2 + 64} = \sqrt{81 + 64} = \sqrt{145} \approx 12.04\text{ m}$.
[A1] Shortest crawl path is $\mathbf{\sqrt{145}\text{ m} \approx 12.0\text{ m}}$.

Triangle & Pythagoras Formulas

  • 2D Pythagoras: $a^2 + b^2 = c^2 \quad (c = \text{hypotenuse})$
  • 3D Cuboid Space Diagonal: $d = \sqrt{l^2 + w^2 + h^2}$
  • Cube Space Diagonal: $d = s\sqrt{3}$
  • Scale Factor Relationships: Linear $= k \implies$ Area $= k^2 \implies$ Volume $= k^3$
  • Pythagorean Converse:
    • $a^2 + b^2 = c^2 \implies$ Right-angled
    • $a^2 + b^2 > c^2 \implies$ Acute-angled
    • $a^2 + b^2 < c^2 \implies$ Obtuse-angled

Top Common MYP Pitfalls

1. Adding When Finding a Leg

Using $a^2 = c^2 + b^2$. To find a leg, you must subtract: $a^2 = c^2 - b^2$!

2. Linear vs Area Scale Factor

Multiplying area by linear scale factor $k$. When lengths scale by $k$, area scales by $k^2$!

3. Confusing Slant Height with Vertical Height

In pyramids, using the vertical height $H$ in face area calculations. Face area uses the slant height $s_h$!

4. Non-Included Angle in Congruence

Assuming ASS (two sides and non-included angle) guarantees congruence. ASS is ambiguous and invalid in Euclidean proofs.

Interactive Mini Test: Triangle Properties & Pythagoras

Answer the questions to see your score and feedback.

Question 1 of 5 Pythagorean Triple Calculation

In a right-angled triangle, the hypotenuse is $25\text{ cm}$ and one leg is $7\text{ cm}$. What is the length of the other leg?

Question 2 of 5 3D Cuboid Diagonal

What is the space diagonal of a rectangular prism with dimensions $4\text{ cm} \times 12\text{ cm} \times 3\text{ cm}$?

Question 3 of 5 Converse of Pythagoras

A triangle has sides $6\text{ cm}$, $8\text{ cm}$, and $11\text{ cm}$. What type of triangle is it?

Question 4 of 5 Similar Triangles Area Ratio

If the side lengths of triangle $B$ are $3$ times those of triangle $A$, how many times larger is the area of triangle $B$?

Question 5 of 5 Invalid Congruence Criterion

Which of the following is NOT a valid congruence criterion for triangles?