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A Tale of Three Intersecting Lines - Types of Triangles

Grade 7CBSE

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

🔑Concepts

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Classification by Sides: Triangles can be Scalene (no equal sides), Isosceles (two equal sides), or Equilateral (all three sides equal). In an equilateral triangle, all internal angles are exactly 60∘60^\circ.

An equilateral triangle with all angles marked as 60 degrees.
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Classification by Angles: Triangles are Acute-angled (all angles <90∘< 90^\circ), Right-angled (one angle =90∘= 90^\circ), or Obtuse-angled (one angle >90∘> 90^\circ).

A right-angled triangle with a 90 degree square symbol at the vertex.
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Angle Sum Property: The sum of the interior angles of any triangle is always 180∘180^\circ. This allows us to find a missing angle if two are known using x=180∘−(∠A+∠B)x = 180^\circ - (\angle A + \angle B).

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Exterior Angle Property: An exterior angle of a triangle is equal to the sum of its two interior opposite angles.

Triangle with one side extended showing the exterior angle equal to the sum of interior opposite angles.

📐Formulae

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

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

Exterior Angle=Sum of Interior Opposite Angles\text{Exterior Angle} = \text{Sum of Interior Opposite Angles}

a2+b2=c2 (Pythagoras Theorem for right-angled triangles)a^2 + b^2 = c^2 \text{ (Pythagoras Theorem for right-angled triangles)}

💡Examples

Problem 1:

In a triangle, two angles are 55∘55^\circ and 65∘65^\circ. Find the third angle and classify the triangle by its angles.

Solution:

Let the third angle be xx. By the Angle Sum Property: x+55∘+65∘=180∘x + 55^\circ + 65^\circ = 180^\circ First, calculate the sum of the known angles: 55+65120\begin{array}{r} 55 \\ + 65 \\ \hline 120 \end{array} So, x+120∘=180∘x + 120^\circ = 180^\circ. Now, subtract 120120 from 180180: 180−12060\begin{array}{r} 180 \\ - 120 \\ \hline 60 \end{array} The third angle is 60∘60^\circ. Since all angles (55∘55^\circ, 65∘65^\circ, 60∘60^\circ) are less than 90∘90^\circ, it is an acute-angled triangle.

Explanation:

We apply the Angle Sum Property which states that the sum of angles in a triangle is 180∘180^\circ. After finding the missing angle, we check if any angle exceeds 90∘90^\circ to classify it.

Problem 2:

An isosceles triangle has one vertex angle of 100∘100^\circ. Find the measure of the other two equal angles.

Solution:

Let each of the equal angles be xx. Using the Angle Sum Property: x+x+100∘=180∘x + x + 100^\circ = 180^\circ 2x+100∘=180∘2x + 100^\circ = 180^\circ 2x=180∘−100∘2x = 180^\circ - 100^\circ 2x=80∘2x = 80^\circ x=80∘2=40∘x = \frac{80^\circ}{2} = 40^\circ. Each equal angle is 40∘40^\circ.

Explanation:

In an isosceles triangle, the angles opposite the equal sides must be equal. We set up an equation where 2x2x plus the known vertex angle equals 180∘180^\circ.

Problem 3:

In the given triangle △PQR\triangle PQR, ∠P=40∘\angle P = 40^\circ and the exterior angle at RR is 110∘110^\circ. Find the measure of ∠Q\angle Q.

Triangle PQR with side PR extended to show exterior angle of 110 degrees and angle P as 40 degrees.

Solution:

  1. According to the Exterior Angle Property: Exterior Angle at R=∠P+∠Q\text{Exterior Angle at } R = \angle P + \angle Q
  2. Substitute the known values: 110∘=40∘+∠Q110^\circ = 40^\circ + \angle Q
  3. Solve for ∠Q\angle Q: ∠Q=110∘−40∘=70∘\angle Q = 110^\circ - 40^\circ = 70^\circ

Explanation:

The exterior angle is equal to the sum of the interior opposite angles. By subtracting the given interior angle from the exterior angle, we find the other interior opposite angle.

Problem 4:

In a right-angled triangle, one of the acute angles is 35∘35^\circ. Calculate the third angle.

A right-angled triangle with one angle marked 35 degrees and the top angle marked as x.

Solution:

  1. In a right-angled triangle, one angle is always 90∘90^\circ.
  2. Let the angles be ∠A=90∘\angle A = 90^\circ, ∠B=35∘\angle B = 35^\circ, and ∠C=x\angle C = x.
  3. Using the Angle Sum Property: ∠A+∠B+∠C=180∘\angle A + \angle B + \angle C = 180^\circ 90∘+35∘+x=180∘90^\circ + 35^\circ + x = 180^\circ 125∘+x=180∘125^\circ + x = 180^\circ
  4. Solve for xx: x=180∘−125∘=55∘x = 180^\circ - 125^\circ = 55^\circ

Explanation:

Since the sum of angles in a triangle is 180∘180^\circ and one angle is 90∘90^\circ, the two acute angles must sum to 90∘90^\circ. We subtract the given acute angle from 90∘90^\circ to find the third angle.

Types of Triangles Class 7 Notes & Examples | CBSE Maths