Triangle Theorems

Subject: Compulsory Maths

Find Your Query

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.

Scholarships after +2 Abroad Studies Opportunities
.
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:

 

1
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.
Questions and Answers

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:

.

Conclusion:

Figure AC AB \(\angle\)ACB \(\angle\)ABC Results
A 2.0 cm 1.6 cm 42o 75o AC>AB,\(\angle\)ABc>\(\angle\)ACB
B 3.2 cm 1.5 cm 37o 91o
AC>AB,\(\angle\)ABc>\(\angle\)ACB
C 2.8 cm 1.5 cm 35o 90o
AC>AB,\(\angle\)ABc>\(\angle\)ACB

© 2019-20 Kullabs. All Rights Reserved.