#Triangle : Experimental verification of properties of triangle
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The document outlines objectives and experimental verifications related to the properties of triangles, specifically focusing on angles in various types of triangles such as equilateral, isosceles, and right-angled triangles. Key findings include that the sum of interior angles of a triangle is always 180 degrees, base angles of isosceles triangles are equal, and other related angle properties are verified through experiments. It also includes multiple experiments with diagrams and conclusions drawn from measuring interior angles.
![• ∡C = ∡B [ Base angles of an isosceles triangle]
= 550
• ∡A +∡B + ∡C = 1800 [sum of interior angles of triangle]
∡A + 550 + 550 = 1800
∡A = 1800 – 1100
• = 700
6/27/2020 JANAK SINGH SAUD 15
1) Find unknown angles](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-15-320.jpg)
![• ∡N = 9 00 – 400[remaining angle of right
angled triangle]
• ∡N = 500
6/27/2020 JANAK SINGH SAUD 16
2)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-16-320.jpg)
![Find the unknown angles
X + 2x + x + 200 = 1800 [Sum of angles of triangle]
4x = 1800 -200
4x = 1600
X = 400
∡R = 400
∡Q = 2x = 2x400 = 800
∡P = x + 20 0 = 400 + 200 = 600
6/27/2020 JANAK SINGH SAUD 17
3)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-17-320.jpg)
![x0 + 600 + 300 = 1800 [sum of angles of triangle]
x0 + 900 = 1800
x0 = 1800 – 900
x = 900
6/27/2020 JANAK SINGH SAUD 18
4)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-18-320.jpg)
![6/27/2020 JANAK SINGH SAUD 20
• x = y [base angles of an isosceles triangle
x + y + 440 = 1800 [ sum of angles of triangle]
x + x + 440 = 1800 [ x = y]
2x = 1800 – 440
2x = 1360
x = 680
y = 680
Hence, x = y = 680
6)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-20-320.jpg)
![x + 200 = 900 [sum of remaining angles of right angled
triangle]
x = 900 - 200
x = 700
y = x [ base angles of isosceles triangle]
x = y = 700
6/27/2020 JANAK SINGH SAUD 21
7)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-21-320.jpg)
![x + 1500 = 1800 [Liner pairs]
x = 1800 – 1500
= 300
x + y = 900 [ Acute angles of right angled triangle]
300 + y = 900
y = 900 – 300
y = 600
Hence, x = 300 and y = 600
6/27/2020 JANAK SINGH SAUD 22
8)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-22-320.jpg)
![x + 2x0 + 6x0 = 1800 [Sum of angles of triangles]
Or, 9x0 = 1800
Or, x = 200
Now, x = 200
2x = 2 x 200 = 400
6x = 6 x 200 = 1200
6/27/2020 JANAK SINGH SAUD 23
9)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-23-320.jpg)
![i) x = y [Base angles of an isosceles triangle]
ii) x + y = 1200 [Exterior angles of triangle]
or, x + x = 1200 [x = y]
or, 2x = 1200
or, x = 600
ii) z + x = 1800 [Linear pair]
or, z + 600 = 1800
or z = 1800 – 600
or, z = 1200
Hence, x = y = 600 and z = 1200
6/27/2020 JANAK SINGH SAUD 24
10)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-24-320.jpg)
![• x + 600 = 1800 [Linear pair]
x = 1800 – 600
x = 1200
• y + 450 = x0 [Exterior angle of a triangle]
or, y + 450 = 1200
or, y = 1200 – 450
or, y = 750
Hence , x = 1200 and y = 750
6/27/2020 JANAK SINGH SAUD 25
11)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-25-320.jpg)
![• y0 = 380 [Alternate angles]
• z0 = 420 [Alternate angles]
• x + z + 380 = 1800 [interior angles of a triangle]
or, x + 420 + 380 = 1800 [z = 420]
or, x + 800 = 1800
or, x = 1800 – 800
or, x = 1000
Hence, x = 1000, y = 380 and z = 420
6/27/2020 JANAK SINGH SAUD 26
12)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-26-320.jpg)
![• x = 600 [Corresponding angles with // lines]
• (y + 600) + 650 = 1800 [Co-interior angles with // lines]
or, y + 1250 = 1800
or, y = 1800 – 1250
or, y = 550
Hence, x = 600 and y = 550
6/27/2020 JANAK SINGH SAUD 27
13)](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-27-320.jpg)
![14) Find the value of x and y
• x + 2x + 450 = 1800 [Interior angles of a triangle]
or, 3x = 1800 – 450
or, 3x = 1350
or, x = 450
• y = 2x [Alternate angles with // lines]
or, y = 2 x 450
or, y = 900
Hence x = 450 and y = 900
6/27/2020 JANAK SINGH SAUD 28](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-28-320.jpg)
![6/27/2020 JANAK SINGH SAUD 29
15) In the figure ABCD is an equilateral triangle and BCD is an isosceles triangle
in which BC = CD. Find ∡CBA
i. ∡ABD = 600 [∆ABD is an equilateral
triangle]
ii. ∡CBD = ∡CDB [In ∆BCD, BC = CD]
iii. ∡CBD + ∡CDB + 1000 = 1800 [Sun of angles of a triangle]
∡CBD + ∡CBD + 1000 = 1800
2 ∡CBD = 1800 – 1000
2 ∡CBD = 800
∴ ∡CBD = 400
Lastly, ∡CBA = ∡ABD + ∡CBD [Whole part axiom]
= 400 + 600
= 1000](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-29-320.jpg)
![6/27/2020 JANAK SINGH SAUD 30
15) If a base angle is double of the vertical angle of an isosceles triangle, then find
all the angles.
Let ABC is an isosceles triangle in which AB = AC and the vertical angle = ∡A = x
Then by the question ∡B = ∡C = 2x
Now, ∡A + ∡B + ∡C = 1800 [Sum of angles of triangle]
or, x + 2x + 2x = 1800
or, 5x = 1800
or,
5𝑥
5
=
1800
5
or, x = 360
Hence, the vertical angle = ∡A = x = 360
Each base angles = ∡B = ∡ C = 2x = 2(360) = 720
A
B C
x
2x 2x](https://image.slidesharecdn.com/triangleview-200705133744/85/Triangle-Experimental-verification-of-properties-of-triangle-30-320.jpg)
#Triangle : Experimental verification of properties of triangle
- 1. 6/27/2020 JANAK SINGH SAUD 1
- 2. Objectives: Students will be able to • Verify experimentally that the sum of sum of interior angles of a triangle is 1800 • Verify experimentally that Base angles of an isosceles triangle are always equal • Verify that the Interior angles of an equilateral triangle are always 600 • Verify experimentally that the Base angles of an isosceles right angled triangle are 450 • Verify experimentally that The line joining from vertex of an isosceles triangle to join the mid-point of the base is perpendicular to the base 6/27/2020 JANAK SINGH SAUD 2
- 3. Review : Types of a triangle • A triangle having three equal sides is called an equilateral triangle • A triangle having any two sides equal is called an isosceles triangle 6/27/2020 JANAK SINGH SAUD 3
- 4. Verification of properties of triangles 6/27/2020 JANAK SINGH SAUD 4 A B C A B C Sum of interior angles of a triangle ∡BAC + ∡ABC + ∡BCA = ? Relation of base angles of an isosceles triangle Relation between ∡ABC and ∡BCA Relation of interior angles of equilateral triangle Given AB = BC = CA, then Relation between ∡BAC, ∡ABC and ∡BCA A B C
- 5. Experimental verification of sum of interior angles of a triangle 6/27/2020 JANAK SINGH SAUD 5 https://www.geogebra.org/m/FAhtKpR5 https://www.geogebra.org/m/YqaqaUt3 https://www.geogebra.org/m/qyrXkXRZ
- 6. EXPERIMENT - 1 • Three triangle of different shapes and sizer are drawn • Measuring all the interior angles and tabulated : • Conclusion: the sum of interior angles of a triangle is always 1800 6/27/2020 JANAK SINGH SAUD 6 A B C A B C A B C Fig. (i) Fig. (ii) Fig. (iii) Figure ∡𝑩𝑨𝑪 ∡ABC ∡BCA Result (i) ∡BAC + ∡ABC + ∡BCA = (ii) ∡BAC + ∡ABC + ∡BCA = (iii) ∡BAC + ∡ABC + ∡BCA =
- 7. Base angles of an isosceles triangle 6/27/2020 JANAK SINGH SAUD 7 https://www.geogebra.org/m/zQyvhJZv https://www.geogebra.org/m/MfzdgSYw
- 8. Relation of Base angles of an isosceles triangle • Three isosceles triangles ABC with different base BC are drawn • Measuring the angles of each triangles and tabulated • Conclusion: Base angles of an isosceles triangle are always equal. 6/27/2020 JANAK SINGH SAUD 8 EXPERIMENT - 2 A B C A B C A B C Figure ∡𝑩𝑨𝑪 ∡ABC ∡BCA Result (i) (ii) (iii)
- 9. Interior angles of an equilateral triangle 6/27/2020 JANAK SINGH SAUD 9 https://www.geogebra.org/m/VqjH4BPZ
- 10. Relation of angles of an equilateral triangle triangle • Three isosceles triangles ABC with different base BC are drawn • Measuring the angles of each triangles and tabulated • Conclusion: Base angles of an isosceles triangle are always equal. 6/27/2020 JANAK SINGH SAUD 10 EXPERIMENT - 3 A B C A B C A B C Figure ∡𝑩𝑨𝑪 ∡ABC ∡BCA Result (i) (ii) (iii)
- 11. The line joining from vertex of an isosceles triangle to join the mid-point of the base is perpendicular to the base 6/27/2020 JANAK SINGH SAUD 11 https://www.geogebra.org/m/Au4rzFcJ https://www.geogebra.org/m/ErD5dhGt
- 12. The line joining from vertex of an isosceles triangle to join the mid-point of the base is perpendicular to the base • Three isosceles triangles ABC with different base BC are drawn • Measuring the angles BMC and AMC in each triangles and tabulated 6/27/2020 JANAK SINGH SAUD 12 EXPERIMENT - 4 Figure ∡𝑩𝑴𝑪 ∡AMC Result (i) (ii) (iii) M M M Conclusion: The line joining from the vertex of an isosceles triangle to join the mid-point of base is perpendicular to the base.
- 13. Base angles of an isosceles right angled triangle 6/27/2020 JANAK SINGH SAUD 13 https://www.geogebra.org/m/ksRTgq9N
- 14. Base angles of an isosceles right angled triangle • Three isosceles right angled triangles ABC at C of different sizes are drawn • Measuring the ∡ABC and ∡CBA in each triangles and tabulated 6/27/2020 JANAK SINGH SAUD 14 EXPERIMENT - 5 Figure ∡𝑨𝑩𝑪 ∡𝐂𝐁𝐀 Result (i) ∡𝑨𝑩𝑪 = ∡𝐂𝐁𝐀 = 450 (ii) (iii) Conclusion: The base angles of an isosceles right angled triangle are always 450 Fig. (i) Fig. (ii) Fig. (iii)
- 15. • ∡C = ∡B [ Base angles of an isosceles triangle] = 550 • ∡A +∡B + ∡C = 1800 [sum of interior angles of triangle] ∡A + 550 + 550 = 1800 ∡A = 1800 – 1100 • = 700 6/27/2020 JANAK SINGH SAUD 15 1) Find unknown angles
- 16. • ∡N = 9 00 – 400[remaining angle of right angled triangle] • ∡N = 500 6/27/2020 JANAK SINGH SAUD 16 2)
- 17. Find the unknown angles X + 2x + x + 200 = 1800 [Sum of angles of triangle] 4x = 1800 -200 4x = 1600 X = 400 ∡R = 400 ∡Q = 2x = 2x400 = 800 ∡P = x + 20 0 = 400 + 200 = 600 6/27/2020 JANAK SINGH SAUD 17 3)
- 18. x0 + 600 + 300 = 1800 [sum of angles of triangle] x0 + 900 = 1800 x0 = 1800 – 900 x = 900 6/27/2020 JANAK SINGH SAUD 18 4)
- 19. • x = y = 450[ Base angles of an isosceles right angled triangle 6/27/2020 JANAK SINGH SAUD 19 5)
- 20. 6/27/2020 JANAK SINGH SAUD 20 • x = y [base angles of an isosceles triangle x + y + 440 = 1800 [ sum of angles of triangle] x + x + 440 = 1800 [ x = y] 2x = 1800 – 440 2x = 1360 x = 680 y = 680 Hence, x = y = 680 6)
- 21. x + 200 = 900 [sum of remaining angles of right angled triangle] x = 900 - 200 x = 700 y = x [ base angles of isosceles triangle] x = y = 700 6/27/2020 JANAK SINGH SAUD 21 7)
- 22. x + 1500 = 1800 [Liner pairs] x = 1800 – 1500 = 300 x + y = 900 [ Acute angles of right angled triangle] 300 + y = 900 y = 900 – 300 y = 600 Hence, x = 300 and y = 600 6/27/2020 JANAK SINGH SAUD 22 8)
- 23. x + 2x0 + 6x0 = 1800 [Sum of angles of triangles] Or, 9x0 = 1800 Or, x = 200 Now, x = 200 2x = 2 x 200 = 400 6x = 6 x 200 = 1200 6/27/2020 JANAK SINGH SAUD 23 9)
- 24. i) x = y [Base angles of an isosceles triangle] ii) x + y = 1200 [Exterior angles of triangle] or, x + x = 1200 [x = y] or, 2x = 1200 or, x = 600 ii) z + x = 1800 [Linear pair] or, z + 600 = 1800 or z = 1800 – 600 or, z = 1200 Hence, x = y = 600 and z = 1200 6/27/2020 JANAK SINGH SAUD 24 10)
- 25. • x + 600 = 1800 [Linear pair] x = 1800 – 600 x = 1200 • y + 450 = x0 [Exterior angle of a triangle] or, y + 450 = 1200 or, y = 1200 – 450 or, y = 750 Hence , x = 1200 and y = 750 6/27/2020 JANAK SINGH SAUD 25 11)
- 26. • y0 = 380 [Alternate angles] • z0 = 420 [Alternate angles] • x + z + 380 = 1800 [interior angles of a triangle] or, x + 420 + 380 = 1800 [z = 420] or, x + 800 = 1800 or, x = 1800 – 800 or, x = 1000 Hence, x = 1000, y = 380 and z = 420 6/27/2020 JANAK SINGH SAUD 26 12)
- 27. • x = 600 [Corresponding angles with // lines] • (y + 600) + 650 = 1800 [Co-interior angles with // lines] or, y + 1250 = 1800 or, y = 1800 – 1250 or, y = 550 Hence, x = 600 and y = 550 6/27/2020 JANAK SINGH SAUD 27 13)
- 28. 14) Find the value of x and y • x + 2x + 450 = 1800 [Interior angles of a triangle] or, 3x = 1800 – 450 or, 3x = 1350 or, x = 450 • y = 2x [Alternate angles with // lines] or, y = 2 x 450 or, y = 900 Hence x = 450 and y = 900 6/27/2020 JANAK SINGH SAUD 28
- 29. 6/27/2020 JANAK SINGH SAUD 29 15) In the figure ABCD is an equilateral triangle and BCD is an isosceles triangle in which BC = CD. Find ∡CBA i. ∡ABD = 600 [∆ABD is an equilateral triangle] ii. ∡CBD = ∡CDB [In ∆BCD, BC = CD] iii. ∡CBD + ∡CDB + 1000 = 1800 [Sun of angles of a triangle] ∡CBD + ∡CBD + 1000 = 1800 2 ∡CBD = 1800 – 1000 2 ∡CBD = 800 ∴ ∡CBD = 400 Lastly, ∡CBA = ∡ABD + ∡CBD [Whole part axiom] = 400 + 600 = 1000
- 30. 6/27/2020 JANAK SINGH SAUD 30 15) If a base angle is double of the vertical angle of an isosceles triangle, then find all the angles. Let ABC is an isosceles triangle in which AB = AC and the vertical angle = ∡A = x Then by the question ∡B = ∡C = 2x Now, ∡A + ∡B + ∡C = 1800 [Sum of angles of triangle] or, x + 2x + 2x = 1800 or, 5x = 1800 or, 5𝑥 5 = 1800 5 or, x = 360 Hence, the vertical angle = ∡A = x = 360 Each base angles = ∡B = ∡ C = 2x = 2(360) = 720 A B C x 2x 2x