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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 mid-point 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.
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}
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}
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}
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how to measure traingle
how to measure traingle using protactor
Mar 26, 2017
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Adarsha
If the triangle is equilateral triangle and the angles are only given how to find the sides? Please I hope a reply. Thank you
Mar 18, 2017
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