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Geometry - The Triangle and its Properties: Angle Sum Property, Exterior Angle Property

Grade 7ICSE

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

🔑Concepts

A triangle is a closed polygon made of three line segments. It has three vertices (points), three sides (line segments), and three interior angles. Visually, any three non-collinear points connected by straight lines form this basic geometric shape.

The Angle Sum Property states that the sum of all three interior angles of a triangle is always 180180^\circ. For any ABC\triangle ABC, the relationship is represented as A+B+C=180\angle A + \angle B + \angle C = 180^\circ. If you imagine tearing off the three corners of a paper triangle and placing them together, they would perfectly form a straight line.

An exterior angle is formed when one side of a triangle is extended outwards. For instance, if side BCBC of ABC\triangle ABC is produced to a point DD, the angle ACD\angle ACD is the exterior angle at vertex CC. This angle sits outside the triangle but shares a vertex with an interior angle.

The Exterior Angle Property states that the measure of an exterior angle of a triangle is equal to the sum of the measures of its two interior opposite angles. Visually, if you look at the exterior angle at one corner, its size is exactly the 'sum' of the openings of the two angles located at the other two corners.

The interior angle and its adjacent exterior angle at any vertex always form a linear pair. This means they lie on the same straight line extension and their sum is always 180180^\circ. For example, Interior C+Exterior C=180\text{Interior } \angle C + \text{Exterior } \angle C = 180^\circ.

In a right-angled triangle, where one angle is exactly 9090^\circ, the sum of the other two acute angles must be 9090^\circ to satisfy the Angle Sum Property. Visually, these two angles 'complement' each other to form a perfect corner if added together.

A triangle can have only one right angle or only one obtuse angle (greater than 9090^\circ). If a triangle had two angles of 9090^\circ or more, the sum would already meet or exceed 180180^\circ, leaving no room for the third angle or for the sides to close and form a vertex.

📐Formulae

1+2+3=180\angle 1 + \angle 2 + \angle 3 = 180^\circ

Exterior =Sum of interior opposite \angles\text{Exterior } \angle = \text{Sum of interior opposite } \angles

ext=int1+int2\angle \text{ext} = \angle \text{int1} + \angle \text{int2}

Interior +Adjacent Exterior =180\text{Interior } \angle + \text{Adjacent Exterior } \angle = 180^\circ

💡Examples

Problem 1:

In PQR\triangle PQR, the measures of P\angle P and Q\angle Q are 4545^\circ and 7575^\circ respectively. Find the measure of R\angle R.

Solution:

  1. According to the Angle Sum Property: P+Q+R=180\angle P + \angle Q + \angle R = 180^\circ
  2. Substitute the given values: 45+75+R=18045^\circ + 75^\circ + \angle R = 180^\circ
  3. Add the known angles: 120+R=180120^\circ + \angle R = 180^\circ
  4. Subtract 120120^\circ from both sides: R=180120\angle R = 180^\circ - 120^\circ
  5. R=60\angle R = 60^\circ

Explanation:

To find a missing interior angle when two are known, we use the property that all three must add up to 180180^\circ.

Problem 2:

In ABC\triangle ABC, the side BCBC is produced to DD. If the exterior angle ACD=115\angle ACD = 115^\circ and the interior opposite angle A=50\angle A = 50^\circ, find the measure of B\angle B.

Solution:

  1. By the Exterior Angle Property: Exterior ACD=A+B\text{Exterior } \angle ACD = \angle A + \angle B
  2. Substitute the known values: 115=50+B115^\circ = 50^\circ + \angle B
  3. Rearrange the equation to solve for B\angle B: B=11550\angle B = 115^\circ - 50^\circ
  4. B=65\angle B = 65^\circ

Explanation:

The exterior angle is always equal to the sum of the two interior angles that are not adjacent to it. By subtracting the given interior opposite angle from the exterior angle, we find the second interior opposite angle.