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3 years ago

Explain how Ashoka expanded the Mauryan Empire. What effect did war have on Ashoka's religious beliefs?

1 answer:

5 0

Ashoka expanded the Mauryan Empire through waging war, which in turn influenced his religious beliefs after the conquests were over.

Explanation:

Ashoka was a great leader in the Indian subcontinent. When he came to power, he was only ruling over a small kingdom. He focused on strengthening the military and the kingdom, and once this was done, he started to attack the other small kingdoms one by one. Ashoka was becoming more and more powerful as he was conquering more and more kingdoms, and eventually he managed to create a large unified empire.

All the wars in which Asoka engaged actually had big influence on him. All of the horrors of war he saw made him change his mind about wagging wars. That resulted in a relative peace in the time to come, and Ashoka converted to Buddhism. Ashoka even made the Buddhism the official religion in his empire, as he thought that the people will live better and in peace.

Ashoka managed to conquer:

Nanda Empire
Seleucid Empire
the remaining satraps in India by the Macedonian Empire

Learn more about the Buddhism brainly.com/question/12315545

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wariber [46]

Answer : The value of $x_6$ is $\sqrt{7}$ .

Explanation :

As we are given 6 right angled triangle in the given figure.

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$x_2=\sqrt{(1)^2+(\sqrt{2})^2}$

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Now we have to calculate the value of $x_3$ .

Using Pythagoras theorem in triangle 3 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_3)^2=(1)^2+(X_2)^2$

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$x_3=\sqrt{(1)^2+(\sqrt{3})^2}$

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Now we have to calculate the value of $x_4$ .

Using Pythagoras theorem in triangle 4 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_4)^2=(1)^2+(X_3)^2$

$(x_4)^2=(1)^2+(\sqrt{4})^2$

$x_4=\sqrt{(1)^2+(\sqrt{4})^2}$

$x_4=\sqrt{5}$

Now we have to calculate the value of $x_5$ .

Using Pythagoras theorem in triangle 5 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_5)^2=(1)^2+(X_4)^2$

$(x_5)^2=(1)^2+(\sqrt{5})^2$

$x_5=\sqrt{(1)^2+(\sqrt{5})^2}$

$x_5=\sqrt{6}$

Now we have to calculate the value of $x_6$ .

Using Pythagoras theorem in triangle 6 :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(x_6)^2=(1)^2+(X_5)^2$

$(x_6)^2=(1)^2+(\sqrt{6})^2$

$x_6=\sqrt{(1)^2+(\sqrt{6})^2}$

$x_6=\sqrt{7}$

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