2.प्रश्नानाम् एकपदेन उत्तराणि लिखत- (क) का भाषा प्राचीनतमा? संस्कृत ) (ख) शून्यस्य प्रतिपादनं कः अकरोत्? यत्र (ग) कौटिल्येन रचितं शास्त्रं किम्? अर्थ (घ) कस्याः भाषायाः काव्यसौन्दर्यम् अनुपमम्? (ङ) काः अभ्युदयाय प्रेरयन्ति? सूत (​

Answer:

it shows the presence of jointed appendages. it is triploblastic, coelomate, bilaterally symmetrical, and segmented. it has a chitinous exoskeleton around its body. it respires with the help of tracheal system.

Explanation:

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Answer:

In cockroaches we see that they have head, thorax and abdomen, so due to this they have been classified in the class insecta and phylum Arthropoda.

Explanation:

Not only cockroaches have the above said features, but also have locomotive functions and sensory organs due to the structure called as appendages which help them with this feature.

Answer:

5834 is divisible by 2

It is not divisible by 3 so it is not divisible by 6

35000 is divisible by 4

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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Question :- In triangle ABC, segment AD is perpendicular to BC. Prove that

[tex]\rm \: {AB}^{2} + {CD}^{2} = {BD}^{2} + {AC}^{2} \\ [/tex]

[tex]\large\underline{\sf{Solution-}}[/tex]

As it is given that

[tex]\rm \: AD \: \perp \: BC \\ [/tex]

[tex]\rm\implies \angle\:ADB \: = \: \angle ADC \: = \: 90 \degree \\ [/tex]

Now, in right angle triangle ADB

By using Pythagoras Theorem, we have

[tex]\rm \: {AB}^{2} = {AD}^{2} + {BD}^{2} \\ [/tex]

[tex]\rm\implies \: {AD}^{2} = {AB}^{2} - {BD}^{2} - - - (1) \\ [/tex]

Now, in right angle triangle ADC

By using Pythagoras Theorem, we have

[tex]\rm \: {AC}^{2} = {AD}^{2} + {CD}^{2} \\ [/tex]

[tex]\rm\implies \: {AD}^{2} = {AC}^{2} - {CD}^{2} - - - (2) \\ [/tex]

From equation (1) and (2), we get

[tex]\rm \: {AB}^{2} - {BD}^{2} = {AC}^{2} - {CD}^{2} \\ [/tex]

[tex]\rm\implies \:\boxed{ \rm{ \: {AB}^{2} + {CD}^{2} = {BD}^{2} + {AC}^{2} \: \: }} \\ [/tex]

Hence, Proved

[tex]\rule{190pt}{2pt}[/tex]

Additional Information :-

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

This theorem states that : - If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

by cancelling AD from right hand side we get BD square -CD square

and interchanging the position of negative segments we can prace this proof