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Triangles - Properties of a Triangle

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Angle Sum Property: The sum of the interior angles of a triangle is always 180∘180^\circ. If you imagine a triangle △ABC\triangle ABC, the three interior corners ∠A\angle A, ∠B\angle B, and ∠C\angle C can be torn off and placed together to form a straight line, representing a total of 180∘180^\circ.

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Exterior Angle Property: If a side of a triangle is produced (extended), the exterior angle so formed is equal to the sum of the two interior opposite angles. Visually, if side BCBC of △ABC\triangle ABC is extended to point DD, the angle ∠ACD\angle ACD (the outside corner) is equal to ∠BAC+∠ABC\angle BAC + \angle ABC.

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Triangle Inequality Property: The sum of any two sides of a triangle is always greater than the third side. For a triangle with sides aa, bb, and cc, this means a+b>ca + b > c, b+c>ab + c > a, and c+a>bc + a > b. If you try to draw a triangle where the two shorter sides don't add up to more than the longest, the lines will never meet to form a vertex.

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Isosceles Triangle Theorem: Angles opposite to equal sides of an isosceles triangle are equal. Conversely, sides opposite to equal angles of a triangle are equal. If you see a triangle where two sides are marked with identical tick marks, the angles at the base of those sides will be identical in measure.

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Side-Angle Relationship: In any triangle, the side opposite to the larger angle is longer, and the angle opposite to the longer side is larger. For example, in a right-angled triangle, the hypotenuse is always the longest side because it is opposite the 90∘90^\circ angle, which is the largest angle.

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Congruence Criteria: Two triangles are congruent if they are exact copies of each other. The main criteria used to prove this are SAS (Side-Angle-Side), ASA (Angle-Side-Angle), AAS (Angle-Angle-Side), SSS (Side-Side-Side), and RHS (Right angle-Hypotenuse-Side). When triangles are congruent, their corresponding parts (CPCT) are equal.

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Medians and Altitudes: A median is a line segment connecting a vertex to the midpoint of the opposite side, bisecting that side into two equal parts. An altitude is a perpendicular segment dropped from a vertex to the opposite side, forming a 90∘90^\circ angle with the base.

📐Formulae

∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ

Exterior ∠ACD=∠A+∠B\text{Exterior } \angle ACD = \angle A + \angle B

AB+BC>ACAB + BC > AC (Triangle Inequality)

In △ABC, if AB=AC⇒∠C=∠B\text{In } \triangle ABC, \text{ if } AB = AC \Rightarrow \angle C = \angle B

Area of a triangle=12×base×height\text{Area of a triangle} = \frac{1}{2} \times \text{base} \times \text{height}

Perimeter of △ABC=AB+BC+CA\text{Perimeter of } \triangle ABC = AB + BC + CA

💡Examples

Problem 1:

In △XYZ\triangle XYZ, the measure of ∠X=45∘\angle X = 45^\circ and ∠Y=75∘\angle Y = 75^\circ. Find the measure of ∠Z\angle Z.

Solution:

Step 1: Identify the Angle Sum Property, which states that ∠X+∠Y+∠Z=180∘\angle X + \angle Y + \angle Z = 180^\circ. Step 2: Substitute the known values: 45∘+75∘+∠Z=180∘45^\circ + 75^\circ + \angle Z = 180^\circ. Step 3: Simplify the equation: 120∘+∠Z=180∘120^\circ + \angle Z = 180^\circ. Step 4: Subtract 120∘120^\circ from both sides: ∠Z=180∘−120∘\angle Z = 180^\circ - 120^\circ. Step 5: Therefore, ∠Z=60∘\angle Z = 60^\circ.

Explanation:

This problem uses the fundamental property that all internal angles of any triangle must add up to exactly 180180 degrees.

Problem 2:

In an isosceles triangle ABCABC, AB=ACAB = AC and the vertex angle ∠A=50∘\angle A = 50^\circ. Find the measures of the base angles ∠B\angle B and ∠C\angle C.

Solution:

Step 1: Given AB=ACAB = AC, we know from the Isosceles Triangle Theorem that ∠B=∠C\angle B = \angle C. Let ∠B=∠C=x\angle B = \angle C = x. Step 2: Use the Angle Sum Property: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ. Step 3: Substitute the values: 50∘+x+x=180∘50^\circ + x + x = 180^\circ. Step 4: Combine like terms: 50∘+2x=180∘50^\circ + 2x = 180^\circ. Step 5: Isolate xx: 2x=180∘−50∘⇒2x=130∘2x = 180^\circ - 50^\circ \Rightarrow 2x = 130^\circ. Step 6: Solve for xx: x=130∘2=65∘x = \frac{130^\circ}{2} = 65^\circ. Step 7: So, ∠B=65∘\angle B = 65^\circ and ∠C=65∘\angle C = 65^\circ.

Explanation:

The solution relies on two properties: first, that equal sides imply equal opposite angles in a triangle, and second, that all angles must sum to 180180 degrees.