sqrt{33124} times sqrt{2601} - (83)^{2} = (?) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) First,calculate the square roots: $\sqrt{33124} = 182$ and $\sqrt{2601} = 51$.Next,calculate the squares: $(83)^{2} = 6889$ and $(37)^{2} = 1369$.

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) First,calculate the square roots: $\sqrt{33124} = 182$ and $\sqrt{2601} = 51$.Next,calculate the squares: $(83)^{2} = 6889$ and $(37)^{2} = 1369$.Substitute these values into the equation: $(182 	imes 51) - 6889 = (?)^{2} + 1369$.Calculate the product: $182 	imes 51 = 9282$.Now,the equation becomes: $9282 - 6889 = (?)^{2} + 1369$.Simplify the left side: $2393 = (?)^{2} + 1369$.Subtract $1369$ from both sides: $(?)^{2} = 2393 - 1369 = 1024$.Finally,find the square root: $? = \sqrt{1024} = 32$.

$\sqrt{33124} \times \sqrt{2601} - (83)^{2} = (?)^{2} + (37)^{2}$

Similar Questions

Given that $\sqrt{4096} = 64$,the value of $\sqrt{4096} + \sqrt{40.96} + \sqrt{0.004096}$ is

$\frac{\sqrt{4375}}{\sqrt{7}} = ?$

If $\sqrt{5} = 2.236$,then the value of $\frac{\sqrt{5}}{2} - \frac{10}{\sqrt{5}} + \sqrt{125}$ is:

The least number by which $14175$ should be divided to make it a perfect square is

$\sqrt{\frac{0.324 \times 0.081 \times 4.624}{1.5625 \times 0.0289 \times 72.9 \times 64}} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module