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Basic Geometrical Ideas - Triangles

Grade 6CBSE

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

🔑Concepts

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A triangle is a three-sided polygon, the simplest closed figure made of line segments. It has three vertices, three sides, and three angles.

A triangle ABC showing vertices A, B, and C and the three sides.
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The interior of a triangle consists of all points enclosed by the three sides. Points can lie in the interior, in the exterior, or on the boundary (the sides) of the triangle.

Triangle showing points in the interior, exterior, and on the boundary.
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Triangles can be classified by their sides: Scalene (all sides different), Isosceles (at least two sides equal), or Equilateral (all three sides equal).

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Triangles can be classified by their angles: Acute-angled (all angles <90∘< 90^{\circ}), Right-angled (one angle =90∘= 90^{\circ}), or Obtuse-angled (one angle >90∘> 90^{\circ}).

📐Formulae

Sum of angles: ∠A+∠B+∠C=180∘\text{Sum of angles: } \angle A + \angle B + \angle C = 180^{\circ}

Perimeter of ΔABC=AB+BC+CA\text{Perimeter of } \Delta ABC = AB + BC + CA

Triangle Inequality: a+b>c\text{Triangle Inequality: } a + b > c

💡Examples

Problem 1:

In ΔPQR\Delta PQR, the measures of two angles are ∠P=55∘\angle P = 55^{\circ} and ∠Q=65∘\angle Q = 65^{\circ}. Find the measure of the third angle ∠R\angle R.

Solution:

Step 1: Use the Angle Sum Property of a triangle, which states that ∠P+∠Q+∠R=180∘\angle P + \angle Q + \angle R = 180^{\circ}. Step 2: Substitute the known values: 55∘+65∘+∠R=180∘55^{\circ} + 65^{\circ} + \angle R = 180^{\circ}. Step 3: Add the known angles: 120∘+∠R=180∘120^{\circ} + \angle R = 180^{\circ}. Step 4: Subtract 120∘120^{\circ} from both sides: ∠R=180∘−120∘=60∘\angle R = 180^{\circ} - 120^{\circ} = 60^{\circ}.

Explanation:

We apply the property that all interior angles of a triangle must add up to 180∘180^{\circ} to find the unknown value.

Problem 2:

Check if it is possible to form a triangle with side lengths 33 cm, 44 cm, and 88 cm.

Solution:

Step 1: Identify the three side lengths: a=3a = 3, b=4b = 4, and c=8c = 8. Step 2: Apply the Triangle Inequality Property, which states that the sum of any two sides must be greater than the third side. Step 3: Check a+b>ca + b > c: 3+4=73 + 4 = 7. Since 77 is not greater than 88 (7<87 < 8), the condition fails. Step 4: Conclusion: A triangle cannot be formed with these lengths.

Explanation:

According to the Triangle Inequality Property, the sum of the two shorter sides (3+4=73 + 4 = 7) must be greater than the longest side (88). Since 7<87 < 8, the sides cannot meet to form a closed triangle.

Problem 3:

Identify the sides, vertices, and angles of the given triangle XYZXYZ.

A triangle with vertices labeled X, Y, and Z.

Solution:

Vertices: X,Y,Z\text{Vertices: } X, Y, Z Sides: XY,YZ,ZX\text{Sides: } XY, YZ, ZX Angles: ∠YXZ,∠XYZ,∠XZY\text{Angles: } \angle YXZ, \angle XYZ, \angle XZY

Explanation:

A triangle is named by its vertices. The line segments forming the triangle are its sides, and the space between the meeting segments at the vertices forms the angles.

Problem 4:

In the right-angled triangle LMNLMN shown below, if ∠M=90∘\angle M = 90^{\circ} and ∠L=40∘\angle L = 40^{\circ}, calculate the measure of ∠N\angle N.

A right-angled triangle LMN with right angle at M and 40 degrees at L.

Solution:

In ΔLMN,∠L+∠M+∠N=180∘\text{In } \Delta LMN, \angle L + \angle M + \angle N = 180^{\circ} 40∘+90∘+∠N=180∘40^{\circ} + 90^{\circ} + \angle N = 180^{\circ} 130∘+∠N=180∘130^{\circ} + \angle N = 180^{\circ} ∠N=180∘−130∘\angle N = 180^{\circ} - 130^{\circ} ∠N=50∘\angle N = 50^{\circ}

Explanation:

Using the Angle Sum Property of a triangle, the sum of all interior angles is always 180∘180^{\circ}. We subtract the sum of the two known angles from 180∘180^{\circ} to find the third.