Given Lengths 3 Sides
Can you form a triangle with sides of lengths 3, 4, and 6?
Yes
Explanation: For a triangle to exist, the sum of any two sides must be greater than the third side. Checking all combinations:
3 + 4 = 7 > 6 ✓
3 + 6 = 9 > 4 ✓
4 + 6 = 10 > 3 ✓ Therefore, these segments can form a triangle.
In a triangle, one side is 3 units and another side is 4 units long.
What is the possible range for the third side?
Given sides a = 3 and b = 4
|a - b| < c < (a + b)
|3 - 4| < c < (3 + 4)
1 < c < 7
The third side must be greater than 1 unit and less than 7 units.
In triangle ABC, angle A = 90°, angle B = 45°, and angle C = 45°.
Order the sides (a, b, c) of the triangle from shortest to longest.
The side opposite to angle A is called 'a', opposite to B is 'b', and opposite to C is 'c'. Since we know:
In any triangle, larger angles are opposite to longer sides
90° > 45° = 45°
Therefore: b = c < a
Triangle Angle Order Q: In triangle ABC, AB = 3 cm, BC = 4 cm, and AC = 5 cm.
Order the angles from smallest to largest.
Let's solve this step by step:
Angle A is opposite side BC (4 cm)
Angle B is opposite side AC (5 cm)
Angle C is opposite side AB (3 cm)
Since larger angles are opposite larger sides: 3 cm < 4 cm < 5 cm so ∠C < ∠A < ∠B
Answer: ∠C < ∠A < ∠B
In triangle ABC, angle A = 45° and angle B = 60°.
Find angle C.
Solution:
Sum of interior angles in a triangle = 180°
Angle C = 180° - (45° + 60°)
Angle C = 180° - 105° = 75°
Can you form a triangle with sides of lengths 5, 2, and 8?
No
Explanation: Using triangle inequality theorem:
5 + 2 = 7 < 8 ✗
5 + 8 = 13 > 2 ✓
2 + 8 = 10 > 5 ✓ Since 5 + 2 < 8, these segments cannot form a triangle because the sum of the two shorter sides is less than the longest side.
In a triangle, one side is 5 units and another is 12 units.
What is the possible range for the third side x, expressed as an interval?
Solution:
|a - b| < x < (a + b)
|5 - 12| < x < (5 + 12)
|7| < x < 17
7 < x < 17
The third side must be between 7 and 17 units: (7, 17)
In triangle PQR, angle P = 30°, angle Q = 60°, and angle R = 90°.
Order the sides (p, q, r) from shortest to longest.
30° is the smallest angle
60° is the medium angle
90° is the largest angle Therefore: p < q < r
Triangle with Equal Sides Q: In triangle DEF, DE = 6 cm, EF = 6 cm, and DF = 4 cm.
Order the angles from largest to smallest.
Angles opposite equal sides are equal
DE = EF, so ∠D = ∠F
DF is shortest side, so ∠E is smallest
Therefore: ∠D = ∠F > ∠E
Answer: ∠D = ∠F > ∠E
In triangle XYZ, angle X = 52° and angle Y = 73°.
Find:
a) Angle Z
b) The exterior angle at vertex Z
Solution:
a) Interior angle Z:
Z = 180° - (52° + 73°)
Z = 180° - 125° = 55°
b) Exterior angle at Z:
Exterior angle = 180° - interior angle
Exterior angle = 180° - 55° = 125°
Can you form a triangle with sides of lengths 12, 5, and 8?
No
Explanation: Let's check all combinations:
12 + 5 = 17 > 8 ✓
12 + 8 = 20 > 5 ✓
5 + 8 = 13 > 12 ✗ The sum of the two shorter sides (5 + 8 = 13) is barely larger than the longest side (12), but it still doesn't satisfy the triangle inequality theorem.
One side of a triangle is x units and another is (x+2) units, where x > 0.
If the third side is 10 units, find the possible values of x.
Solution:
Using triangle inequality three times:
From 2: x + 10 > x + 2 → 10 > 2 (always true)
From 3: x + 12 > x → 12 > 0 (always true)
Therefore, x > 4
Since x > 0 was given, x > 4
Answer: x must be greater than 4 units
In triangle DEF, angle D = 75°, angle E = 65°, and angle F = 40°.
Order the sides (d, e, f) from longest to shortest.
75° is the largest angle
65° is the medium angle
40° is the smallest angle Therefore: d > e > f
In triangle PQR, PQ = 5.5 cm, QR = 5.2 cm, and PR = 3.8 cm.
Order the angles from smallest to largest.
Let's solve this step by step:
Angle P is opposite QR (5.2 cm)
Angle Q is opposite PR (3.8 cm)
Angle R is opposite PQ (5.5 cm)
Ordering sides: 3.8 cm < 5.2 cm < 5.5 cm
Therefore: ∠Q < ∠P < ∠R
Answer: ∠Q < ∠P < ∠R
In triangle PQR, angle P = 35° and angle Q = 82°.
Find:
a) Angle R
b) All exterior angles
Solution:
a) Angle R:
R = 180° - (35° + 82°)
R = 180° - 117° = 63°
b) Exterior angles:
At P: 180° - 35° = 145°
At Q: 180° - 82° = 98°
At R: 180° - 63° = 117°
Can you form a triangle with sides of lengths x, x+2, and 2x, where x = 4?
Yes
Explanation: Substitute x = 4: The sides would be 4, 6, and 8 Check all combinations:
4 + 6 = 10 > 8 ✓
4 + 8 = 12 > 6 ✓
6 + 8 = 14 > 4 ✓ These segments can form a triangle.
In a triangle with sides of length 8 and y units, the third side must be at least 3 units long.
Find the range of possible values for y.
Solution:
Let the third side be z
Given: z ≥ 3
Using triangle inequality:
Combining inequalities: y > 5 and y < 11
Answer: 5 < y < 11
In triangle XYZ, angle X = 55°, and angle Y = 55°.
Without knowing angle Z, order as many sides (x, y, z) as possible relative to each other.
Since angles X and Y are equal (55°), their opposite sides must be equal
Therefore x = y
Since angle Z must be 70° (180° - 55° - 55°), and 70° > 55° Therefore: x = y < z
In right triangle XYZ, XY = 10 cm, YZ = 8 cm, and XZ = 6 cm.
Order the angles from largest to smallest.
Let's solve this step by step:
It's a right triangle, so one angle is 90°
By Pythagorean theorem, the right angle must be at Y
∠Y = 90° (opposite longest side)
Angle X is opposite YZ (8 cm)
Angle Z is opposite XY (6 cm)
Therefore: ∠Y > ∠X > ∠Z
Answer: ∠Y > ∠X > ∠Z
In triangle DEF, angle D = 42° and exterior angle at E = 128°.
Find:
a) Interior angle E
b) Angle F
Solution:
a) Interior angle E:
Interior angle = 180° - exterior angle
E = 180° - 128° = 52°
b) Angle F:
F = 180° - (42° + 52°)
F = 180° - 94° = 86°
Can you form a triangle with sides of lengths x, x+3, and 2x+1, where x = 5?
No
Explanation: Substitute x = 5: The sides would be 5, 8, and 11 Check all combinations:
5 + 8 = 13 > 11 ✓
5 + 11 = 16 > 8 ✓
8 + 11 = 19 > 5 ✓
Although all combinations satisfy the triangle inequality theorem numerically, there's an additional consideration: when the sum of two sides is barely larger than the third side, the triangle would be degenerate (the three points would lie on a straight line). In this case, 5 + 8 = 13 is too close to 11, making it impossible to form a proper triangle.
Triangle ABC has AB = 6 units and BC = 9 units.
If AC must be an integer and the perimeter of the triangle must be less than 24 units, find all possible values for AC.
Solution:
Let AC = x where x is an integer
From triangle inequality: |6 - 9| < x < (6 + 9)
3 < x < 15
Given perimeter < 24: 6 + 9 + x < 24
15 + x < 24
x < 9
Combining: 3 < x < 9
Since x must be an integer: x ∈ {4, 5, 6, 7, 8}
The possible integer values for AC are 4, 5, 6, 7, and 8 units.
n triangle ABC, we know that angle A is obtuse (> 90°).
Without knowing the other angles, can you determine which side is definitely the longest?
Since angle A is obtuse, it must be the largest angle in the triangle (a triangle can only have one obtuse angle)
The side opposite to the largest angle is always the longest side
Therefore, side 'a' must be the longest side
We cannot determine the order of the other two sides without more information Therefore: b, c < a
In triangle LMN, LM = x+2, MN = x+4, and LN = x+1, where x > 0.
Order the angles from smallest to largest.
Let's solve this step by step:
Angle L is opposite MN (x+4)
Angle M is opposite LN (x+1)
Angle N is opposite LM (x+2)
For any positive x: x+1 < x+2 < x+4
Therefore: ∠M < ∠N < ∠L
Answer: ∠M < ∠N < ∠L
In triangle MNO, angle M = 38° and the exterior angle at N is 135°.
Find:
a) Interior angle N
b) Angle O
c) The exterior angle at O
Solution:
a) Interior angle N:
N = 180° - 135° = 45°
b) Angle O:
O = 180° - (38° + 45°)
O = 180° - 83° = 97°
c) Exterior angle at O:
Exterior angle = 180° - interior angle
Exterior angle = 180° - 97° = 83°