Subject: Compulsory Maths
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.
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.
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°
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.
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°
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 midpoint of the base of an isosceles triangle is perpendicular to the base.
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.
Find the value of x and y from the following figure.
Solution:
or, x=y(base angles of an isoscles triangle.)
or, x+y+70^{o}=180^{o}(sum of angles of a triangle is 180^{o})
or, x+x=180^{o}70^{o }(\(\therefore\)x=y)
or, 2x=110^{o}
or, \(\frac{110^o}{2}\)=55^{o}
Hence, x=y=55^{o}
Find the value of x and y from the following figure.
Solution:
\(\angle\)P + \(\angle\)Q + \(\angle\)R = 180^{o} (sum of angles of a triangle is 180^{o})
or, 3x+ 3x+ 3x=180^{o}
or, 9x=180^{o}
\(\therefore\) x=20^{o}
Also,
3x+y=180^{o }(straight angle)
or, 3\(\times\)20^{o}+y=180^{o}
or, 60^{o}+y=180^{o}
or, y=180^{o} 60^{o}
\(\therefore\) y=120^{o}
Find the value of x, y and z from the following figure.
Solution:
Here, \(\angle\)FEG=20^{o}+20^{o}=40^{o}
\(\angle\)EFG= \(\angle\)EGF=x=y [ \(\therefore\)
Now, \(\angle\)FEG+ \(\angle\)EFG+ \(\angle\)EFG=180^{o} [\(\therefore\)sum of angle FEG is 180^{o}]
or, 40^{o} + x + x=180^{o}
or, 2x=180^{o}40^{o}=140^{o}
or, x= \(\frac{140^o}{2}\)
\(\therefore\)x =y=70^{o}
Find the value of x, y and z from the following figure.
Solution,
Here, \(\angle\)P= \(\angle\)Q= \(\angle\)R [\(\therefore\) ]
or, x^{o}=y^{o}=z^{o}
or, x=y=z
Now, \(\angle\)P+ \(\angle\)Q+ \(\angle\)R=180^{o} [\(\therefore\)sum of angle FEG is 180^{o}]
or, x+x+x=180^{o}
or, 3x=180^{o}
or, x=\(\frac{180^o}{3}\)=60^{o}
or, x=60^{o}
\(\therefore\) x=y=z=60^{o}
Find the value of x and y from the following figure.
Solution:
Here,
\(\angle\)P+\(\angle\)Q+\(\angle\)R=180^{o }[ \(\therefore\) sum of angle ABC]
and x=y
now,
\(\angle\)P+\(\angle\)Q+\(\angle\)R=180^{o} [ \(\therefore\) sum of angle ABC is 180^{o}]
or, x+90^{o}+y=180^{o}
or, x+90^{o}+x=180^{o}
or, 2x=180^{o}90^{o}
or, x=\(\frac{90^o}{2}\)
or, x=45^{o}
\(\therefore\) x=y=45^{o}
Hence, \(\angle\)P= \(\angle\)R = 45^{o}
Find the value of x, y and z from the following figure.
Solution:
Here, \(\angle\) ABC= \(\angle\)ACB=x^{o}=y^{o} [ \(\therefore\) ]
or, \(\angle\)BAC+ \(\angle\)DAC=180^{o} [ \(\therefore\) Sum of BAC and DAC is 180^{o}]
or, \(\angle\)BAC+120^{o}=180^{o}
or, \(\angle\)BAC=180^{o}120^{o}=60^{o}
Again, \(\angle\)BAC+ \(\angle\)ABC+ \(\angle\)ACB=180^{o} [\(\therefore\) ]
or, 60^{o}+x+x=180^{o}
or, 2x=180^{o}60^{o}=120^{o}
or, x=\(\frac{120^o}{2}\)=60^{o}
\(\therefore\)x=y=60^{o}
Again, x^{o}+z^{o}=180^{o}[ \(\therefore\)having adjacent angles]
or, 60^{o}+z^{o}=180^{o}
or, z^{o}=180^{o}60^{o}=120^{o}
Hence, x=y=60^{o}, z=120^{o}
Construct three different triangle ABC where AC > AB
Figure  AC  AB  \(\angle\)ACB  \(\angle\)  Result 
A  
B  
C 
Solution:
Conclusion:
Figure  AC  AB  \(\angle\)ACB  \(\angle\)ABC  Results  
A  2.0 cm  1.6 cm  42^{o}  75^{o}  AC>AB,\(\angle\)ABc>\(\angle\)ACB  
B  3.2 cm  1.5 cm  37^{o}  91^{o} 
 
C  2.8 cm  1.5 cm  35^{o}  90^{o} 

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