त्रिभुज नोट्स (Triangles: Notes and Questions)_00.1
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त्रिभुज नोट्स (Triangles: Notes and Questions)

एक त्रिभुज एक बहुभुज है, 2-आयामी वस्तु है जिसमें 3 भुजाएँ और 3 शीर्ष होते हैं। त्रिकोणीय आकृतियों का क्षेत्रफल समस्याओं या प्रश्नों को हल करते समय उपयोग किए जाने वाले एक सरल फार्मूला का उपयोग करके निर्धारित किया जाता है। इसके अलावा, कई केंद्रीय और राज्य परीक्षा पास हैं इच्छुक उम्मीदवारों के लिए पदों का एक समूह जिसमें गणित एक प्रमुख हिस्सा है। हमने इन प्रतिष्ठित परीक्षाओं पर ध्यान केंद्रित करने वाले महत्वपूर्ण नोट्स और प्रश्नों को कवर किया है।

त्रिभुज

1). केन्द्रक: माध्य का अंतर्बिंदु
त्रिभुज नोट्स (Triangles: Notes and Questions)_50.1
त्रिभुज नोट्स (Triangles: Notes and Questions)_60.1
त्रिभुज नोट्स (Triangles: Notes and Questions)_70.1
2). अन्तः केन्द्र → [आंतरिक कोण द्विभाजक का प्रतिछेदन बिंदु]
त्रिभुज नोट्स (Triangles: Notes and Questions)_80.1
3).परिकेंद्र :  [लम्बवत द्विभाजक का प्रतिछेदन बिंदु]
त्रिभुज नोट्स (Triangles: Notes and Questions)_90.1
4). लंब केंद्र: → [ऊंचाई का प्रतिछेदन बिंदु] 
त्रिभुज नोट्स (Triangles: Notes and Questions)_100.1
Important Points:
(a)  समकोण त्रिभुज का लांब केंद्र ⇒   समकोण के शीर्ष पर
(b)  समकोण का परिकेंद्र ⇒ मध्य बिंदु का कर्ण
(c)  एक त्रिकोण के आतंरिक और परित्रिज्या के मध्य की दूरी
त्रिभुज नोट्स (Triangles: Notes and Questions)_110.1
(d)  In Equilateral triangle / समबाहु त्रिभुज में,
त्रिभुज नोट्स (Triangles: Notes and Questions)_120.1

Questions

Q. शीर्ष के निर्देशांक A(x1, y1), B(x2, y2), C(x3, y3) वाले त्रिभुज का अंत:केंद्र ज्ञात कीजिये

त्रिभुज नोट्स (Triangles: Notes and Questions)_130.1

Solution:

By geometry, we know that BD/DC = AB/AC (since AD bisects ÐA).
The lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC = c/b.
Coordinates of D are (bx2+cx3/b+c, by2+cy3/b+c)
IB bisects DB. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.
Let the coordinates of I be (x, y).
Then x = ax1+bx2+cx3/a+b+c, y = ay1+by2+cy3/a+b+c.

Q. यदि (0, 1), (1, 1) और (1, 0) एक त्रिभुज की भुजाओं के मध्य बिंदु हैं, तो इसका अंत: केंद्र ज्ञात कीजिये

Solution:

Let A(x1, y1), B(x2, y2) and C(x3, y3)be teh vertices of a triangle.
x1 + x2 = 0, x2 + x= 0, x3 + x1 = 0
y1 + y2 = 0, y2 + y3 = 0, y3 + y1 = 0.
Solving these equations, we get A(0, 0), B(0, 2) and C(2, 0).
Now, a = BC = 2√ 2, b = CA = 2 and c = AB = 2.
Thus, incentre of the triangle ABC is (2-√ 2, 2-√ 2).

Q. यदि किसी त्रिभुज की भुजाओं के मध्यबिंदु (0, 4), (6, 4) और (6, 0) हैं, तो त्रिभुज के शीर्ष, केन्द्रक और परिधि ज्ञात करें।
त्रिभुज नोट्स (Triangles: Notes and Questions)_140.1Solution:
Let points A (x1, y1), B (x2, y2) and C (x3, y3) be vertices of ΔABC.
x1 + x3 = 0 , y1 + y3 = 8
x2 + x3 = 12 , y2 + y3 = 8
x1 + x2 = 12, y1 + y2 = 0
Solving we get A (0, 0), B (12, 0) and C (0, 8)
Hence ΔABC is right angled triangle. ∠A = π/2
Circumcentre is midpoint of hypotenuse which is (6, 4) itself and centroid
(x1+x2+x3)/3 , (y1+y2+y3)/3 = (4 , 8/3)

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