Triangles — scalene, isosceles, equilateral; acute, right, obtuse: Exercises

The vault topics you will lean on: the parent note, Pythagorean Theorem, Area of Triangles, Triangle Inequality, Interior Angles of Polygons, Trigonometric Ratios, Congruence Criteria (SSS, SAS, ASA), and Symmetry in Geometry.

Level 1 — Recognition

Goal: name the type from raw numbers. No formula needed, just careful comparison.

Recall Solution 1.1

What to compare: the three side lengths against each other. Two sides are equal ( $7 = 7$ ) and one differs ( $10$ ). "At least two equal" is the definition of isosceles. It is not equilateral, because not all three are equal. Answer: isosceles.

Recall Solution 1.2

Check the angles add to $180°$ first (else it is not a triangle): $35° + 55° + 90° = 180°$ . ✓ Exactly one angle equals $90°$ , the other two are below $90°$ . One right angle → right triangle. Answer: right triangle.

Level 2 — Application

Goal: plug into a formula (Pythagoras, area, angle sum) and compute.

The area formula rests on one hidden fact — the height $h = \frac{a 3}{2}$ . Before using it, let's earn that $3$ from a picture.

Figure — Triangles — scalene, isosceles, equilateral; acute, right, obtuse

Recall Solution 2.1

Which tool and why: we want area, so use base $\times$ height $\div 2$ . The base is $a$ , and from the box above the height is $h = \frac{a 3}{2}$ : $A = \frac{1}{2} \times a \times \frac{a 3}{2} = \frac{a ^{2} 3}{4} .$ Substitute $a = 6$ : $A = \frac{6 ^{2} 3}{4} = \frac{36 3}{4} = 93 cm^{2} .$ Numerically $93 \approx 9 \times 1.7320508 = 15.59 cm^{2}$ . Answer: $93 \approx 15.59 cm^{2}$ .

Recall Solution 2.2

Which tool and why: a right triangle links its three sides through Pythagorean Theorem: $c^{2} = a^{2} + b^{2} .$ Here $c$ is opposite the $90°$ corner, so it is the side we solve for. $c^{2} = 9^{2} + 1 2^{2} = 81 + 144 = 225 \Rightarrow c = 225 = 15.$ Answer: $c = 15$ . (This is the famous $9, 12, 15$ triangle — a scaled $3, 4, 5$ .)

Recall Solution 2.3

Read the figure first: the two equal legs are the blue slanted sides (marked with matching white ticks), and the dashed red line down the middle is the axis of symmetry — reflecting the triangle across it swaps the two base corners, so the two base angles must be equal (Symmetry in Geometry). The equal base angles are the green $70°$ marks at the bottom; the yellow $40°$ is the apex at the top. Which tool and why: the angles of any triangle sum to $180°$ (angle-sum, Interior Angles of Polygons). Let each base angle be $θ$ : $θ + θ + 40° = 180° \Rightarrow 2 θ = 140° \Rightarrow θ = 70°.$ Answer: each base angle is $70°$ .

Level 3 — Analysis

Goal: reason about why a type is forced, or extract a hidden fact.

Intuition Why the sign of

$c^{2} - (a^{2} + b^{2})$ decides the angle type Picture the two shorter sides $a, b$ hinged at their shared corner, opening like a pair of scissors; the third side $c$ is the "opening" between their far ends. There is a general law (the Law of Cosines, a big brother of Pythagoras) that says $c^{2} = a^{2} + b^{2} - 2 ab cos C,$ where $C$ is the angle between $a$ and $b$ — the very angle opposite $c$ .

When $C = 90°$ , $cos C = 0$ , the middle term vanishes and $c^{2} = a^{2} + b^{2}$ (plain Pythagoras) → right.
When $C < 90°$ (a sharp opening), $cos C > 0$ , so we subtract something positive: $c^{2} < a^{2} + b^{2}$ → acute.
When $C > 90°$ (a wide opening), $cos C < 0$ , so $- 2 ab cos C$ is positive and gets added: $c^{2} > a^{2} + b^{2}$ → obtuse. So comparing $c^{2}$ against $a^{2} + b^{2}$ is really just reading off the sign of $cos C$ , i.e. whether the biggest angle is under, exactly, or over $90°$ .

Recall Solution 3.1

Which tool and why: by the reasoning above, the sign of $c^{2} - (a^{2} + b^{2})$ for the longest side $c$ (opposite the largest angle) tells the angle type:

$c^{2} = a^{2} + b^{2}$ → right,
$c^{2} < a^{2} + b^{2}$ → largest angle acute → acute triangle,
$c^{2} > a^{2} + b^{2}$ → largest angle obtuse → obtuse triangle.

Longest side $c = 10$ : compare $1 0^{2} = 100$ with $5^{2} + 6^{2} = 25 + 36 = 61$ . Since $100 > 61$ , the angle opposite $10$ is obtuse. Answer: obtuse triangle.

Recall Solution 3.2

Which tool and why: Triangle Inequality — any two sides must together exceed the third, or the short sides can't reach across to close the shape. Check the two shortest against the longest: $2 + 3 = 5$ , which is less than $6$ . $5 < 6$ , so sides $2$ and $3$ can never meet if the third is $6$ . Answer: no such triangle exists.

Recall Solution 3.3

Which tool and why: angle-sum $= 180°$ , and one angle is fixed at $90°$ . Let the smaller acute angle be $x$ ; the other is $2 x$ . $x + 2 x + 90° = 180° \Rightarrow 3 x = 90° \Rightarrow x = 30°.$ So the acute angles are $30°$ and $60°$ . Answer: $30°, 60°, 90°$ — the standard 30-60-90 triangle.

Level 4 — Synthesis

Goal: combine side-and-angle classification, symmetry and a formula in one problem.

Recall Solution 4.1

Read the figure: the two blue legs meeting at corner $P$ each have length $5$ ; the little yellow square at $P$ marks the $90°$ right angle; the yellow slanted top edge is the hypotenuse; the dashed red line is the axis of symmetry from the right angle to the midpoint of the hypotenuse — reflecting across it swaps the two $45°$ corners. (a) Hypotenuse — Pythagoras. The legs $5 cm, 5 cm$ meet at the $90°$ corner: $c^{2} = 5^{2} + 5^{2} = 50 \Rightarrow c = 50 = 52 \approx 7.07 cm .$ (b) Angles — symmetry + angle-sum. Equal legs force equal base angles $θ$ (reflect across the dashed axis in the figure): $θ + θ + 90° = 180° \Rightarrow θ = 45°.$ Angles: $45°, 45°, 90°$ . (c) Area — the two legs are base and height because they meet at a right angle: $A = \frac{1}{2} \times 5 \times 5 = 12.5 cm^{2} .$ Answers: $c = 52 \approx 7.07 cm$ ; angles $45°, 45°, 90°$ ; area $= 12.5 cm^{2}$ .

Recall Solution 4.2

Set up each shape from its perimeter. Equilateral side $= 24/3 = 8$ . Which tool and why (triangle area): for an equilateral triangle we use $A_{△} = \frac{a ^{2} 3}{4}$ which comes from dropping an altitude $h = \frac{a 3}{2}$ that splits the base in half, then $A = \frac{1}{2} a h$ (the same $3$ derivation as Exercise 2.1; see Area of Triangles). With $a = 8$ : $A_{△} = \frac{8 ^{2} 3}{4} = 163 \approx 27.71.$ Square side $= 24/4 = 6$ , so $A_{□} = 6^{2} = 36$ . Compare: $36 > 27.71$ , difference $= 36 - 163 \approx 8.29$ . Answer: the square is larger by $\approx 8.29$ square units. (More sides packed into the same perimeter → more area — a taste of the isoperimetric idea.)

Level 5 — Mastery

Goal: multi-step proof-like reasoning, unusual configuration, full case coverage.

Recall Solution 5.1

Turn the ratio into real angles using angle-sum. Let the angles be $2 k, 3 k, 4 k$ : $2 k + 3 k + 4 k = 180° \Rightarrow 9 k = 180° \Rightarrow k = 20°.$ Angles: $40°, 60°, 80°$ . Classify: the largest angle is $80° < 90°$ , so all angles are acute. Answer: acute triangle. (And scalene by angles, since all three differ — the sides opposite them differ too.)

Recall Solution 5.2

Which tool and why: the longest side $x + 2$ must be the hypotenuse (opposite the $90°$ corner, always the longest — Pythagorean Theorem). So $(x + 2)^{2} = x^{2} + (x + 1)^{2} .$ Expand each square: $x^{2} + 4 x + 4 = x^{2} + x^{2} + 2 x + 1.$ Simplify: $x^{2} + 4 x + 4 = 2 x^{2} + 2 x + 1 \Rightarrow 0 = x^{2} - 2 x - 3.$ Factor: $x^{2} - 2 x - 3 = (x - 3) (x + 1) = 0 \Rightarrow x = 3$ or $x = - 1$ . Case check (never skip): a length must be positive, so $x = - 1$ is rejected. Take $x = 3$ . Sides: $3, 4, 5$ . Verify: $3^{2} + 4^{2} = 25 = 5^{2}$ . ✓ Answer: $x = 3$ ; sides $3, 4, 5$ .

Recall Solution 5.3

(a) Existence — Triangle Inequality on all three pairings. Every pair of sides must together exceed the third: $5 + 12 > k, 5 + k > 12, 12 + k > 5.$ The third, $12 + k > 5$ , is automatic for any positive $k$ , so it never bites. The other two give $k < 17 and k > 7.$ So $7 < k < 17$ . The boundary values $k = 7$ and $k = 17$ are degenerate: at $k = 7$ we'd have $5 + 7 = 12$ (the sides collapse onto a straight line, zero area), and at $k = 17$ we'd have $5 + 12 = 17$ (same collapse). These are excluded because the inequality is strict. Integer values: $k \in {8, 9, 10, 11, 12, 13, 14, 15, 16}$ .

(b) Right-angled — two cases, because we don't know which side is the hypotenuse. Case A: $12$ is the hypotenuse (so $12 > k$ ): $5^{2} + k^{2} = 1 2^{2} \Rightarrow k^{2} = 144 - 25 = 119$ . Then $k = 119 \approx 10.9$ — not an integer. Reject. Case B: $k$ is the hypotenuse (so $k > 12$ ): $5^{2} + 1 2^{2} = k^{2} \Rightarrow k^{2} = 25 + 144 = 169 \Rightarrow k = 13$ . Integer, and $13$ lies in $(7, 17)$ . ✓ Answers: (a) $8 \leq k \leq 16$ ; (b) $k = 13$ (the $5, 12, 13$ right triangle).

Recall Quick self-quiz (

question ::: answer — cover the right side and test yourself) Sides $7, 7, 10$ classify by sides ::: isosceles Sides $5, 6, 10$ — acute/right/obtuse ::: obtuse (since $100 > 61$ ) Do sides $2, 3, 6$ form a triangle ::: no ( $2 + 3 < 6$ ) Right triangle legs $9, 12$ , hypotenuse ::: $15$ Equilateral area, side $6$ ::: $93 \approx 15.59$ Isosceles right, legs $5$ : hypotenuse ::: $52 \approx 7.07$ Angles ratio $2 : 3 : 4$ — type ::: acute ( $40°, 60°, 80°$ ) Sides $x, x + 1, x + 2$ right triangle ::: $3, 4, 5$ Fifth side for $5, 12, k$ right triangle, integer ::: $k = 13$