Triangle Properties

Cheatsheet Content

### Introduction to Triangles - A **triangle** is a closed shape with three sides and three angles. - The symbol for a triangle is $\triangle$. - **Vertices:** The corners of the triangle (e.g., A, B, C). - **Sides:** The line segments forming the triangle (e.g., AB, BC, CA). - **Angles:** The angles formed at each vertex (e.g., $\angle A, \angle B, \angle C$). ### Types of Triangles (Based on Sides) 1. **Scalene Triangle:** All three sides are different lengths. All three angles are different. *Example:* A triangle with sides 3 cm, 4 cm, 5 cm. 2. **Isosceles Triangle:** Two sides are equal in length. The angles opposite to the equal sides are also equal. *Example:* A triangle with sides 5 cm, 5 cm, 7 cm. The angles opposite the 5 cm sides will be equal. 3. **Equilateral Triangle:** All three sides are equal in length. All three angles are also equal, each measuring $60^\circ$. *Example:* A triangle with all sides 6 cm long. All angles are $60^\circ$. #### **Solved Example 1: Identifying Triangle Type** **Question:** A triangle has sides of length 8 cm, 8 cm, and 10 cm. What type of triangle is it? **Solution:** 1. **Look at the side lengths:** Two sides are 8 cm, and one side is 10 cm. 2. **Compare lengths:** Since two sides are equal (8 cm = 8 cm), it is an Isosceles Triangle. ### Types of Triangles (Based on Angles) 1. **Acute-Angled Triangle:** All three angles are acute (less than $90^\circ$). *Example:* A triangle with angles $60^\circ, 70^\circ, 50^\circ$. 2. **Right-Angled Triangle:** One angle is exactly $90^\circ$. *Example:* A triangle with angles $30^\circ, 60^\circ, 90^\circ$. The side opposite the $90^\circ$ angle is called the **hypotenuse** (it's always the longest side). * **Pythagorean Theorem:** In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. * $a^2 + b^2 = c^2$, where $c$ is the hypotenuse and $a, b$ are the other two sides. 3. **Obtuse-Angled Triangle:** One angle is obtuse (greater than $90^\circ$). *Example:* A triangle with angles $20^\circ, 30^\circ, 130^\circ$. #### **Solved Example 2: Identifying Triangle Type by Angles** **Question:** A triangle has angles $45^\circ, 45^\circ, 90^\circ$. What type of triangle is it? **Solution:** 1. **Look at the angles:** One angle is exactly $90^\circ$. 2. **Identify type:** Since it has a $90^\circ$ angle, it is a Right-Angled Triangle. (Bonus: Since two angles are equal, it's also an Isosceles Right-Angled Triangle!) #### **Solved Example 12: Using Pythagorean Theorem** **Question:** A right-angled triangle has sides of length 6 cm and 8 cm. Find the length of the hypotenuse. **Solution:** 1. **Identify type:** It's a right-angled triangle. 2. **Recall Pythagorean Theorem:** $a^2 + b^2 = c^2$, where $c$ is the hypotenuse. 3. **Substitute known values:** Let $a = 6$ cm and $b = 8$ cm. * $6^2 + 8^2 = c^2$ * $36 + 64 = c^2$ * $100 = c^2$ 4. **Solve for c:** $c = \sqrt{100} = 10$ cm. 5. **Result:** The length of the hypotenuse is 10 cm. #### **Solved Example 13: Finding a Missing Side (Pythagorean Theorem)** **Question:** A right-angled triangle has a hypotenuse of 13 cm and one side of 5 cm. Find the length of the other side. **Solution:** 1. **Recall Pythagorean Theorem:** $a^2 + b^2 = c^2$. 2. **Substitute known values:** Let $c = 13$ cm and $a = 5$ cm. We need to find $b$. * $5^2 + b^2 = 13^2$ * $25 + b^2 = 169$ 3. **Solve for $b^2$:** $b^2 = 169 - 25 = 144$. 4. **Solve for b:** $b = \sqrt{144} = 12$ cm. 5. **Result:** The length of the other side is 12 cm. ### Angle Sum Property of a Triangle - The sum of all interior angles of any triangle is always $180^\circ$. - For $\triangle ABC$, $\angle A + \angle B + \angle C = 180^\circ$. #### **Solved Example 3: Finding a Missing Angle** **Question:** In $\triangle PQR$, $\angle P = 50^\circ$ and $\angle Q = 70^\circ$. Find $\angle R$. **Solution:** 1. **Recall Angle Sum Property:** $\angle P + \angle Q + \angle R = 180^\circ$. 2. **Substitute known values:** $50^\circ + 70^\circ + \angle R = 180^\circ$. 3. **Add known angles:** $120^\circ + \angle R = 180^\circ$. 4. **Solve for $\angle R$:** $\angle R = 180^\circ - 120^\circ$. 5. **Result:** $\angle R = 60^\circ$. #### **Solved Example 4: Angles in an Isosceles Triangle** **Question:** An isosceles triangle has one angle equal to $100^\circ$. Find the other two angles. **Solution:** 1. **Identify Property:** In an isosceles triangle, angles opposite to equal sides are equal. Let the two equal angles be $x$. 2. **Apply Angle Sum Property:** $100^\circ + x + x = 180^\circ$. 3. **Simplify:** $100^\circ + 2x = 180^\circ$. 4. **Solve for $2x$:** $2x = 180^\circ - 100^\circ = 80^\circ$. 5. **Solve for $x$:** $x = \frac{80^\circ}{2} = 40^\circ$. 6. **Result:** The other two angles are $40^\circ$ each. #### **Solved Example 5: Angles in an Equilateral Triangle** **Question:** What are the measures of the angles in an equilateral triangle? **Solution:** 1. **Identify Property:** In an equilateral triangle, all three sides are equal, and all three angles are equal. Let each angle be $x$. 2. **Apply Angle Sum Property:** $x + x + x = 180^\circ$. 3. **Simplify:** $3x = 180^\circ$. 4. **Solve for $x$:** $x = \frac{180^\circ}{3} = 60^\circ$. 5. **Result:** Each angle in an equilateral triangle is $60^\circ$. ### Exterior Angle Property - When a side of a triangle is extended, the angle formed outside the triangle is called an **exterior angle**. - The exterior angle is equal to the sum of the two interior opposite angles. - For $\triangle ABC$ with side BC extended to D, $\angle ACD$ is an exterior angle. * $\angle ACD = \angle A + \angle B$. #### **Solved Example 6: Using Exterior Angle Property** **Question:** In $\triangle ABC$, side BC is extended to D. If $\angle A = 60^\circ$ and $\angle B = 70^\circ$, find $\angle ACD$. **Solution:** 1. **Identify Property:** The exterior angle $\angle ACD$ is equal to the sum of the interior opposite angles $\angle A$ and $\angle B$. 2. **Apply Property:** $\angle ACD = \angle A + \angle B$. 3. **Substitute values:** $\angle ACD = 60^\circ + 70^\circ$. 4. **Result:** $\angle ACD = 130^\circ$. #### **Solved Example 7: Finding an Interior Angle using Exterior Angle** **Question:** An exterior angle of a triangle is $110^\circ$. One of the interior opposite angles is $50^\circ$. Find the other interior opposite angle. **Solution:** 1. **Let the angles be:** Exterior angle $= 110^\circ$. One interior opposite angle $= 50^\circ$. Let the other interior opposite angle $= x$. 2. **Apply Exterior Angle Property:** Exterior angle = Sum of two interior opposite angles. 3. **Substitute values:** $110^\circ = 50^\circ + x$. 4. **Solve for $x$:** $x = 110^\circ - 50^\circ$. 5. **Result:** $x = 60^\circ$. The other interior opposite angle is $60^\circ$. #### **Solved Example 8: Combining Properties (HOTS)** **Question:** In $\triangle ABC$, $\angle B = 30^\circ$, $\angle C = 40^\circ$. Side BC is extended to D. Also, AE is drawn such that AE || CD. Find $\angle CAE$. **Solution:** 1. **Find Exterior Angle:** First, find $\angle ACD$ using Exterior Angle Property. * $\angle ACD = \angle B + \angle A$. Oh wait, we don't have $\angle A$. * Let's find $\angle A$ first using Angle Sum Property: $\angle A + \angle B + \angle C = 180^\circ$. * $\angle A + 30^\circ + 40^\circ = 180^\circ \implies \angle A + 70^\circ = 180^\circ \implies \angle A = 110^\circ$. * Now, $\angle ACD = \angle A + \angle B = 110^\circ + 30^\circ = 140^\circ$. 2. **Use Parallel Lines Property:** We are given AE || CD. AC is a transversal. * When two parallel lines are cut by a transversal, alternate interior angles are equal. * $\angle EAC$ and $\angle ACD$ are alternate interior angles if AE || CD. No, they are not. * $\angle EAC$ and $\angle ACB$ are not related directly. * $\angle CAE$ and $\angle ACE$ are not related. * Let's look at transversal AC. Since AE || CD, the angles $\angle EAC$ and $\angle ACD$ are NOT alternate interior angles. $\angle EAC$ is an angle inside the triangle. * The angle alternate to $\angle ACD$ would be an angle formed by AE and a line parallel to AC. * Let's re-evaluate. Since AE || CD, and AC is a transversal, then $\angle EAC$ and $\angle ACD$ are NOT alternate interior angles. * Consider transversal AC. $\angle CAE$ and $\angle ECA$ are not related. * Consider transversal AC. We have AE || BD (since CD is extension of BC). * So, $\angle EAC$ and $\angle ACD$ are **consecutive interior angles** (if we think of AC as a transversal cutting parallel lines AE and CD). No, this is incorrect. * Let's use **alternate interior angles** or **corresponding angles**. * If AE || CD, then $\angle EAC$ is an angle. The angle *alternate interior* to $\angle ACD$ would be if we extended CA past A, and then the angle formed with AE. This is getting complicated. * Let's use **Corresponding Angles**: $\angle EAB$ and $\angle CBD$ (exterior angle). No. * **Alternate Interior Angles**: If we consider AC as a transversal cutting parallel lines AE and CD, then $\angle EAC$ and $\angle BCA$ are not alternate interior. * The angle $\angle EAC$ and $\angle ACD$ are not directly related as alternate interior or corresponding. * Let's rethink. We have AE || BD. Consider AC as transversal. * $\angle EAC$ is the angle we want. * We know $\angle C = 40^\circ$. * Since AE || BD, and AC is a transversal, then $\angle EAC$ and $\angle ACB$ are **alternate interior angles** if we consider AE and BC as parallel lines cut by AC. This is incorrect. AE is parallel to CD, which means AE || BC. * If AE || BC, then: * $\angle EAC = \angle ACB$ (Alternate Interior Angles, with AC as transversal). * So, $\angle CAE = 40^\circ$. * Let's check this. If AE || BC, and AC is a transversal, then $\angle EAC = \angle ACB$. This seems correct. * **Result:** $\angle CAE = 40^\circ$. * This problem relies on understanding parallel lines and transversals, combined with triangle properties. **Step-by-step breakdown for Solved Example 8:** 1. **Identify parallel lines and transversal:** We are given AE || CD. Since CD is an extension of BC, it means AE || BC. AC is a transversal cutting these parallel lines. 2. **Identify angle relationship:** When two parallel lines (AE and BC) are cut by a transversal (AC), the alternate interior angles are equal. 3. **Apply alternate interior angles:** Therefore, $\angle CAE = \angle ACB$. 4. **Substitute known value:** We are given $\angle C = 40^\circ$, which is $\angle ACB$. 5. **Conclusion:** $\angle CAE = 40^\circ$. ### Median and Altitude - **Median:** A line segment that joins a vertex of a triangle to the midpoint of the opposite side. * A triangle has three medians. * The point where all three medians meet is called the **centroid**. - **Altitude:** A line segment from a vertex that is perpendicular to the opposite side (or its extension). It represents the height of the triangle. * A triangle has three altitudes. * The point where all three altitudes meet is called the **orthocenter**. #### **Solved Example 9: Understanding Medians/Altitudes** **Question:** In $\triangle PQR$, PS is a line segment from vertex P to side QR. If S is the midpoint of QR, what is PS called? If PS is perpendicular to QR, what is PS called? **Solution:** 1. **S is midpoint:** If S is the midpoint of QR, then PS is a **median**. 2. **PS is perpendicular:** If PS is perpendicular to QR, then PS is an **altitude**. ### Properties of Isosceles and Equilateral Triangles (HOTS) - **Isosceles Triangle:** * If two sides are equal, then the angles opposite to these sides are equal. * If two angles are equal, then the sides opposite to these angles are equal. * The altitude from the vertex connecting the two equal sides bisects the base and the vertex angle. - **Equilateral Triangle:** * All sides are equal. * All angles are $60^\circ$. * Each median is also an altitude and an angle bisector. #### **Solved Example 10: Applying Isosceles Triangle Properties** **Question:** In $\triangle ABC$, AB = AC. If $\angle B = 50^\circ$, find $\angle A$ and $\angle C$. **Solution:** 1. **Identify Property:** Since AB = AC, $\triangle ABC$ is an isosceles triangle. The angles opposite the equal sides are equal. 2. **Apply Property:** The angle opposite AB is $\angle C$, and the angle opposite AC is $\angle B$. So, $\angle C = \angle B$. 3. **Substitute known value:** Given $\angle B = 50^\circ$, so $\angle C = 50^\circ$. 4. **Use Angle Sum Property:** $\angle A + \angle B + \angle C = 180^\circ$. 5. **Substitute values:** $\angle A + 50^\circ + 50^\circ = 180^\circ$. 6. **Solve for $\angle A$:** $\angle A + 100^\circ = 180^\circ \implies \angle A = 180^\circ - 100^\circ = 80^\circ$. 7. **Result:** $\angle A = 80^\circ$ and $\angle C = 50^\circ$. #### **Solved Example 11: Finding Unknowns (HOTS)** **Question:** In the figure, $\triangle PQR$ is an isosceles triangle with PQ = PR. The exterior angle at R is $120^\circ$. Find $\angle P$. **Solution:** 1. **Find $\angle PRQ$:** $\angle PRQ$ and the exterior angle $120^\circ$ form a linear pair. * $\angle PRQ + 120^\circ = 180^\circ$ (Linear Pair) * $\angle PRQ = 180^\circ - 120^\circ = 60^\circ$. 2. **Apply Isosceles Property:** Since PQ = PR, the angles opposite these sides are equal. * $\angle PQR = \angle PRQ$. * So, $\angle PQR = 60^\circ$. 3. **Use Angle Sum Property:** In $\triangle PQR$, $\angle P + \angle Q + \angle R = 180^\circ$. * $\angle P + 60^\circ + 60^\circ = 180^\circ$. * $\angle P + 120^\circ = 180^\circ$. * $\angle P = 180^\circ - 120^\circ = 60^\circ$. 4. **Result:** $\angle P = 60^\circ$. (Bonus: Since all angles are $60^\circ$, $\triangle PQR$ is also an equilateral triangle!)