 ## Triangle Theorems

Subject: Compulsory Maths

#### Overview

Triangles are governed by two important inequalities. The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides. This note contains information about triangle theorem.

##### Triangle Theorems

Triangles are governed by two important inequalities. The first is often referred to as the triangle inequality. It states that the length of a side of a triangle is always less than the sum of the lengths of the other two sides.

The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides.

### Theorem 1

The sum of interior angles of a triangle is 180°

Draw three different triangles in your notebook. Measure ∠X, ∠Y and ∠Z using a protector and fill in the table.

Verification:

 Figure ∠X ∠Y ∠Z ∠X +∠Y+∠Z (i) (ii) (iii)

Look at the figure and complete the table given below.  Statements Reasons a+b+c = 180° Sum of adjacent angles on a straight line a = m, c = n Corresponding angles m+b+n = 180° ?

Conclusion: The sum of interior angles of a triangle is 180°

### Theorem 2

Base angles of an isosceles triangle are equal.

Draw three different triangles making AB = AC, ∠B and ∠C opposite to AC and AB respectively are the base angles. Measure ∠ABC and ∠ACB using a protector and fill in the table.

Verification:

 Figure ∠ABC ∠ACB Result (i) ∠ABC =∠ACB (ii) (iii)

Conclusion: Base angles of an isosceles triangle are equal.

### Theorem 3

Each of the base angles of an isosceles right triangle is 45°.

Draw three triangles making ∠B = 90° and AC = BC. Measure ∠BAC and ∠ACB and fill in the table.

Verification:

 Figure ∠BAC ∠ACB Result (i) ∠BAC =∠ACB = 45° (ii) (iii)

Conclusion: Each of the base angles of an isosceles right triangle is 45°

### Theorem 4

The line joining the vertex and midpoint of the base of an isosceles triangle is perpendicular to the base.

Draw three triangles making AB = AC. Join the midpoint P of BC and A, in each figure. Measure the angles APB and APC and fill in the blanks.

Verification:

 Figure ∠APB ∠APC Result (i) ∠APB =∠APC =90° (ii) (iii)

Conclusion: The line joining the vertex and mid-point of the base of an isosceles triangle is perpendicular to the base.

### Theorem 5

All the angles of an equilateral triangle are equal.

Draw three triangles making AB = BC =CA in each figure. Measure∠ABC, ∠BCA and ∠CAB and fill in the table given below.

Verification:

 Figure ∠ABC ∠BCA ∠CAB Result (i) ∠ABC =∠BCA =∠CAB (ii) (iii)

Conclusion: All angles of an equilateral triangle are equal.

##### Things to remember
• Triangles are governed by two important inequalities.
• A triangle cannot be constructed from three line segments if any of them is longer than the sum of the other two.
• The triangle inequality theorem states that any side of a triangle is always shorter than the sum of the other two sides.
• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.

Solution:

or, x=y(base angles of an isoscles triangle.)

or, x+y+70o=180o(sum of angles of a triangle is 180o)

or, x+x=180o-70o ($\therefore$x=y)

or, 2x=110o

or, $\frac{110^o}{2}$=55o

Hence, x=y=55o

Solution:

$\angle$P + $\angle$Q + $\angle$R = 180o (sum of angles of a triangle is 180o)

or, 3x+ 3x+ 3x=180o

or, 9x=180o

$\therefore$ x=20o

Also,

3x+y=180o (straight angle)

or, 3$\times$20o+y=180o

or, 60o+y=180o

or, y=180o- 60o

$\therefore$ y=120o

Solution:

Here, $\angle$FEG=20o+20o=40o

$\angle$EFG= $\angle$EGF=x=y [ $\therefore$

Now, $\angle$FEG+ $\angle$EFG+ $\angle$EFG=180o [$\therefore$sum of angle FEG is 180o]

or, 40o + x + x=180o

or, 2x=180o-40o=140o

or, x= $\frac{140^o}{2}$

$\therefore$x =y=70o

Solution,

Here, $\angle$P= $\angle$Q= $\angle$R [$\therefore$ ]

or, xo=yo=zo

or, x=y=z

Now, $\angle$P+ $\angle$Q+ $\angle$R=180o [$\therefore$sum of angle FEG is 180o]

or, x+x+x=180o

or, 3x=180o

or, x=$\frac{180^o}{3}$=60o

or, x=60o

$\therefore$ x=y=z=60o

Solution:

Here,

$\angle$P+$\angle$Q+$\angle$R=180o [ $\therefore$ sum of angle ABC]

and x=y

now,

$\angle$P+$\angle$Q+$\angle$R=180o [ $\therefore$ sum of angle ABC is 180o]

or, x+90o+y=180o

or, x+90o+x=180o

or, 2x=180o-90o

or, x=$\frac{90^o}{2}$

or, x=45o

$\therefore$ x=y=45o

Hence, $\angle$P= $\angle$R = 45o

Solution:

Here, $\angle$ ABC= $\angle$ACB=xo=yo [ $\therefore$ ]

or, $\angle$BAC+ $\angle$DAC=180o [ $\therefore$ Sum of BAC and DAC is 180o]

or, $\angle$BAC+120o=180o

or, $\angle$BAC=180o-120o=60o

Again, $\angle$BAC+ $\angle$ABC+ $\angle$ACB=180o [$\therefore$ ]

or, 60o+x+x=180o

or, 2x=180o-60o=120o

or, x=$\frac{120^o}{2}$=60o

$\therefore$x=y=60o

Again, xo+zo=180o[ $\therefore$having adjacent angles]

or, 60o+zo=180o

or, zo=180o-60o=120o

Hence, x=y=60o, z=120o

Solution:

Figure AC AB $\angle$ACB $\angle$ABC Results
A 2.0 cm 1.6 cm 42o 75o AC>AB,$\angle$ABc>$\angle$ACB
 AC>AB,$\angle$ABc>$\angle$ACB
 AC>AB,$\angle$ABc>$\angle$ACB