(ii) In A ABC, seg AD perpendicular seg BC. Prove : AB square+CDsquare=BDsquare+ ACsquare​

Answer:

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Explanation:

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क) संस्कृत भाषा प्राचीनतमा।

क) संस्कृत भाषा प्राचीनतमा।(ख) शून्यस्य प्रतिपादनं आर्यभटः अकरोत् ।

(ग) कौटिल्येन रचितं शास्त्रं अर्थशास्त्रं अस्ति।

(ग) कौटिल्येन रचितं शास्त्रं अर्थशास्त्रं अस्ति।(घ) संस्कृत भाषायाः काव्यसौन्दर्यम् अनुपमम्।

(ग) कौटिल्येन रचितं शास्त्रं अर्थशास्त्रं अस्ति।(घ) संस्कृत भाषायाः काव्यसौन्दर्यम् अनुपमम्।(ङ) संस्कृते विद्यमानाः सूक्तयः अभ्युदयाय प्रेरयन्ति

Answer:

Kalaburagi division, formerly known as Gulbarga division, is one of the four divisions of the Indian state of Karnataka and has seven districts, namely: Bellary district, Bidar district, Kalaburagi district, Koppal district, Raichur district, Yadgir district and Vijayanagar District. Kalaburagi city is the headquarters of Kalaburagi division.[1] The total area of the division is 44,138 sq.km.[2] The total population as of 2011 census is 11,215,224.

Explanation:

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Answer:

1. DIAMETER
2. 90°
3. ARC
4. AO AND OC
5. 8^2+6^2=64+36=√100=10^2

• SO THE AREA OF ABC IS
• √12(4)(6)(2)
• √2*3*2*2*2*2*3*2
• 2*2*2*3
• 24cm^2

Step-by-step explanation:

Answer:

• The line AOC is diameter of the circle with centre O.
• Angle ABC is right angle i.e. 90⁰ .
• Area AMB is arc of circle with centre O.
• The radius of circle with centre O is 5 units .
• The area of ∆ ABC is 24 sq. units.

Step-by-step explanation:

PART 1 :

• Line AOC is a straight line which passes through centre and intersect the circle at two points A and C.
• Hence, it is diameter of the circle with centre O.

PART 2 :

• The inscribed angle theorem states that: " An angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. "
• Angle AOC subtends 180⁰ at the centre of the circle and angle ABC is half of the angle subtended at the centre.
• Angle ABC is :

[tex] \frac{ {180}^{0} }{2} = {90}^{0} [/tex]

• Hence, Angle ABC is a right angle.

PART 3 :

• Area AMB is the arc of the circle with centre o.

PART 4:

• From geometry of the figure ABC, it is evident that ∆ABC is a right angled triangle with right angle at B.
• By Pythagoras theorem:

(AB)² + (BC)² = (AC)²

• Given that AB = 8 cm and BC = 6 cm.
• This gives AC =

[tex] \: \: \: \: \: \: \: \: = \sqrt{ {8}^{2} + {6}^{2} } \\ \: \: \: \: \: \: \: = \sqrt{ 64 + 36 } \\ = \sqrt{100} \\ = 10 \: cm[/tex]

• AC is the diameter of the circle with centre O. Radius of the circle is

[tex] \frac{10}{2} = 5 \: cm[/tex]

PART 5 :

• ∆ ABC is a right angled triangle with right angle at B.
• Base of triangle = 6 cm
• Height of triangle = 8 cm
• Area of triangle is given by :

[tex] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times base \times height \\ = \frac{1}{2} \times 8 \times 6 \\ = 24 \: sq \: cm[/tex]

Hence, area of ∆ ABC is 24 sq. cm .