Tag: squares and square roots

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Questions Related to squares and square roots

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

Estiamate the square root of $850$ 

  1. $29.15$
  2. $30.21$
  3. $98.23$
  4. $23.11$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The square root of $850$ is $\sqrt {850}=\sqrt {25 \times 34}=5\sqrt {34}$

Square root of $34$ lie between $5$ and $6$.
 square of $5.5$ is $30.25$ 
Now, we can say that square root of $34$ lie between $5.5$ and $6$.
Now, square of $5.75$ is $33.06$
So, square root of $34$ lie between $5.75$ and $6$.
Now, we have to choose the number $5.85$
$(5.85)^2=34.225$ which is greater than $34$ and close to $34$
So, assume a number $5.84$.
$(5.84)^2=34.1056$
$(5.83)^2=33.9889$
Hence, we can say that square root of 34 lie between $5.83$ and $5.84$.
So, $5\times 5.83=29.15$.

Multiple choice maths squares and square roots approximation of square roots estimating square roots square root of non perfect squares

The real number $(\sqrt [3]{\sqrt {75} - \sqrt {12}})^{-2}$ when expressed in the simplest form is equal to

  1. $\dfrac {1}{2}$
  2. $\dfrac {1}{3}$
  3. $\dfrac {1}{4}$
  4. $\dfrac {1}{5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Real number $(\sqrt[3]{\sqrt{75}-\sqrt{12}})^{-2}$

$\sqrt{75}=5\sqrt{3}$ and $\sqrt{12}=2\sqrt{3}$
$=(\sqrt[3]{5\sqrt{3}-2\sqrt{3}})^{-2}$
$=(\sqrt[3]{3\sqrt{3}})^{-2}$
$(3\sqrt{3})^{\dfrac{-2}{3}} ....... (1)$
$3\sqrt{3}=3^{\dfrac{1}{2}+1}=3^{\dfrac{3}{2}} ....... (ii)$
Substituting $(ii)$ in $(i)$
$\left[(3)^{\dfrac{3}{2}}\right]^{\dfrac{-2}{3}}$
$=\dfrac{1}{3}=(3)^{-1}$