# How did aryabhata calculate pi?

## How did aryabhata calculate pi?

What did Aryabhata discover? Aryabhata discovered an approximation of pi, 62832/20000 = 3.1416. He also correctly believed that the planets and the Moon shine by reflected sunlight and that the motion of the stars is due to Earth’s rotation.

What was Brahmagupta’s biggest contribution to mathematics?

Brahmagupta (ad 628) was the first mathematician to provide the formula for the area of a cyclic quadrilateral. His contributions to geometry are significant. He is the first person to discuss the method of finding a cyclic quadrilateral with rational sides.

Who did Brahmagupta mention the theory described by him?

Brahmagupta was the first to give rules to compute with zero….

Brahmagupta
Born c. 598 CE
Died c. 668 CE
Known for Zero Modern number system Brahmagupta’s theorem Brahmagupta’s identity Brahmagupta’s problem Brahmagupta–Fibonacci identity Brahmagupta’s interpolation formula Brahmagupta’s formula
Scientific career

### What is the theorem of Brahmagupta in geometry?

In geometry, Brahmagupta’s theorem states that if a cyclic quadrilateral is orthodiagonal (that is, has perpendicular diagonals ), then the perpendicular to a side from the point of intersection of the diagonals always bisects the opposite side.

What kind of criticism did Brahmagupta receive?

Brahmagupta had a plethora of criticism directed towards the work of rival astronomers, and in his Brahmasphutasiddhanta is found one of the earliest attested schisms among Indian mathematicians. The division was primarily about the application of mathematics to the physical world, rather than about the mathematics itself.

How is the algebra of Brahmagupta syncopated?

Like the algebra of Diophantus, the algebra of Brahmagupta was syncopated. Addition was indicated by placing the numbers side by side, subtraction by placing a dot over the subtrahend, and division by placing the divisor below the dividend, similar to our notation but without the bar.

## How are the Rupas subtracted in Brahmagupta’s equation?

The rupas are [subtracted on the side] below that from which the square and the unknown are to be subtracted. which is a solution for the equation bx + c = dx + e where rupas refers to the constants c and e. The solution given is equivalent to x = e − c b − d. He further gave two equivalent solutions to the general quadratic equation 18.44.
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