Tag: vedic mathematics

Quizzes (3) Questions (151)

Questions Related to vedic mathematics

Multiple choice addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

If $A, {A} _{1}, {A} _{2}, {A} _{3}$ be the area of the in-circle and ex-circles, then $\dfrac {1}{\sqrt {{A} _{1}}}+\dfrac {1}{\sqrt {{A} _{2}}}+\dfrac {1}{\sqrt {{A} _{3}}}$ is equal to

  1. $\dfrac {1}{\sqrt {{A}}}$
  2. $\dfrac {2}{\sqrt {{A}}}$
  3. $\dfrac {3}{\sqrt {{A}}}$
  4. $None$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A _1={\pi}{r _1}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-a)^{2}}$

$A _2={\pi}{r _2}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-b)^{2}}$
$A _3={\pi}{r _3}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s-c)^{2}}$
$A={\pi}{r}^{2}=\dfrac{{\pi}{\Delta^{2}}}{(s)^{2}}$
$\dfrac{1}{\sqrt{A _1}}+\dfrac{1}{\sqrt{A _2}}+\dfrac{1}{\sqrt{A _3}}=\dfrac{1}{\sqrt{\pi}}\bigg[\dfrac{s-a}{\Delta}+\dfrac{s-b}{\Delta}+\dfrac{s-c}{\Delta}\bigg]=\dfrac{1}{\sqrt{\pi}\Delta}[3{s}-(a+b+c)]=\dfrac{s}{\sqrt{\pi}\Delta}=\dfrac{1}{\sqrt{A}}$

Multiple choice addition and subtraction vedic methods of multiplication vedic mathematics history of mathematics maths

Subtract $347657$ by $238294$ using vinculum numbers.

  1. $109363$
  2. $100363$
  3. $109373$
  4. $109563$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

1) The first thing we do is write these numbers one on top of the other.

347657
238294

2) Now we just start subtracting vertically, and whenever our number is negative we represent it as a vinculum number.

347657
238294

3) This next subtraction is 7 – 8 = -1…so we just write this as 1 and continue on.

347657
238294
1 1 1 443

4) The ‘1‘ and ‘4‘ here are considered as two separate groups (since theres a non-vinculum number in-between them), so each is subtracted from 10. This has the effect of reducing the “previous one” by one.

109363