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Lesson 1, Topic 1
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# Triangles and Polygons

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The study of Geometry brings out important theorems that serve as guiding principles in solving problems on triangles and polygons. Some of these theorems include:

### 1. The sum of the Angles of a Triangle is 180º:

Given: any triangle XYZ

Required to prove: $$\scriptsize \hat{X} \: + \: \hat{Y} \: + \: \hat{Z} = 180^o$$

Construction: extend $$\scriptsize \bar{YZ} \: to \: A$$

Draw $$\scriptsize \bar{ZB}$$ parallel to $$\scriptsize \bar{YX}$$

### Proof:

From Fig 1.1 above

x1 = x2 (alternate angles)

y1 = y2 (corresponding angles)

z + x1 + y1 = 180º   (Sum of angles on a straight line)

z + x2 + y2 = 180º (Sum of angles in a triangle)

∴ Z + X + Y = 180º

∴ X + Y + Z = 180º

### 2. The exterior angle of a triangle is equal to the sum of the opposite interior angles:

Consider $$\scriptsize \Delta XYZ \: i.e \: \hat{Z} \: = \hat{X} \: + \: \hat{Y}$$

Let’s prove this theorem!

Given: XYZ, YZ extended to point A

To prove: $$\scriptsize \angle XZA = \angle X \: + \: \angle Y$$

### Proof:

⇒ $$\scriptsize \angle XZA \: + \: \angle Z = 180^o \:……….(1)$$

(angles on a straight line = 180°)

⇒ $$\scriptsize \angle X \: + \: \angle Y \: + \: \angle Z = 180^o \:……….(2)$$

(sum of angles in a triangle = 180°)

∴ $$\scriptsize \angle XZA \: + \: \angle Z = \angle X \: + \: \angle Y \: + \: \angle Z$$

⇒ $$\scriptsize \angle XZA = \angle X \: + \: \angle Y \: + \: \angle Z \: – \: \angle Z$$

⇒ $$\scriptsize \angle XZA = \angle X \: + \: \angle Y$$

### 3. The sum of the interior angles of any n-sided convex polygon is (2n-4) right angles:

In fig. 1.3 above

To prove: ∠A + ∠B +∠C +…= (2n-4) right angles

### Proof:

ABCDEFGH is an “n” sided polygon (The polygon has 8 sides so n = 8). Take any point O inside the polygon. Join OA, OB, OC…etc.

For “n” sided polygon, the polygon forms “n” triangles.

Recall that the sum of the angles of a triangle is equal to 180 degrees

Therefore, the sum of the angles of n triangles = n × 180°

From the above statement, we can say that

Sum of interior angles + Sum of the angles at O = 2n × 90° …………..(1)

But, the sum of the angles at O = 360°

Substitute the above value in (1), we get

Sum of interior angles + 360° = 2n × 90°

So, the sum of the interior angles = (2n × 90°) – 360°

Take 90 as common, then it becomes

2n(90°) – 4(90°)

The sum of the interior angles = (2n – 4) × 90°

Therefore, the sum of “n” interior angles is (2n – 4) × 90°

(Recall a convex polygon does not have any reflex angles, however, a re-entrant polygon has reflex angles.)

### 4. The sum of the exterior angles of any convex polygon is 4 right angles:

To prove: Sum of the exterior angles = 360°

### Proof:

Consider fig. 1.4

ABCDEFGH is a complex polygon with angles A1, B2, C3, D4, E5, F6, G7, H8

// = parallel to

$$\scriptsize \bar{AB} \: // \: \bar{L} \\ \scriptsize \bar{BC} \: // \: \bar{M} \\ \scriptsize \bar{CD} \: // \: \bar{N} \\ \scriptsize \bar{DE}\: // \: \bar{O} \\ \scriptsize \bar{EF}\: // \: \bar{P} \\ \scriptsize \bar{FG}\: // \: \bar{Q} \\ \scriptsize \bar{GH} \: // \: \bar{R} \\ \scriptsize \bar{HA} \: // \: \bar{S}$$

We know that parallel lines have equal angles,

Therefore,

A1 = a
B2 = b
C3 = c
D4 = d
E5 = e
F6 = f
G7 = g
H8 = h

Adding: A1 + B2 + C3 + D4 + E5 + F6 + G7 + H8 = a + b + c + d + e + f + g + h = 360° (sum of angles at point)

= 4 right angles.

Evaluation:

1. The angles of a triangle are x, 2x, and 3x. Find the value of x in degrees.
2. An Isosceles triangle is such that each base angle is twice the vertical angle. Find the angles of the triangle.
3. In a right-angled triangle, one of the acute angles is 20º greater than the other. Find the angles of the triangle.

#### Responses

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Thank you

1. Thank you for your comment, and yes are working on future updates. 🙂

2. x+2x+3x=180(sum of <s in triangle)
6x=180
x=30
2x=2(30)=60
3x=3(30)=90 error: