DAV Class 8 Maths Chapter 1 Brain Teasers
Squares and Square Roots Brain Teasers
1. A. Tick (✓) the correct option.
a.
The difference between the squares of two consecutive numbers is equal to their
\(
\begin{aligned}
(i) \, & \, \text{difference} \\
(ii) \, & \, \text{sum} \\
(iii) \, & \, \text{product} \\
(iv) \, & \, \text{quotient} \\
\end{aligned}
\)
Answer \( \color{orange}(ii)\ \color{red} \ sum \)
b.
What will be the digit in the thousands place of \( (1111)^2 \)?
\(
\begin{aligned}
(i) \, & \, 3 \\
(ii) \, & \, 4 \\
(iii) \, & \, 2 \\
(iv) \, & \, 1 \\
\end{aligned}
\)
Solution\( (1111)^2 = 123{\color{red}4}321 \)
Answer \( \color{orange}(ii)\ \color{red} 4 \)
c.
Perfect squares cannot have 2, 3, ___ and ___ in its ones place.
\(
\begin{aligned}
(i) \, & \, 1, 7 \\
(ii) \, & \, 5, 6 \\
(iii) \, & \, 7, 8 \\
(iv) \, & \, 7, 9 \\
\end{aligned}
\)
Answer \( \color{orange}(iii)\ \color{red}7, 8 \)
d.
The smallest number by which 72 must be divided to make it a perfect square is
\(
\begin{aligned}
(a) \, & \, 4 \\
(b) \, & \, 5 \\
(c) \, & \, 3 \\
(d) \, & \, 2 \\
\end{aligned}
\)
Solution
\[ \begin{array}{c|c} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} 72 &= \boxed{2 \times 2} \times 2 \times \boxed{3 \times 3} \\ \\ & \text{Divide by } \color{red}\boxed{2} \\ \\ \frac{72}{2} &= 36 \\ \\ \sqrt{36} & = 6 \\ \end{align*} \]
Answer \( \color{orange}(iv)\ \color{red} 2 \)
e.
The square root of 3.052009 has ____ decimal places.
\(
\begin{aligned}
(a) \, & \, 3 \\
(b) \, & \, 4 \\
(c) \, & \, 5 \\
(d) \, & \, 1 \\
\end{aligned}
\)
Solution
\( \sqrt{3. {\boxed{05}} \ {\boxed{20}} \ {\boxed{09}}} \)
Answer \( \color{orange}(i)\ \color{red}3 \)
B. Answer the following questions.
(a) How many non-square numbers are there between \(13^2\) and \(14^2\)?
Solution
\[ \begin{align*} n &= 13 \\ 2n &= 2 \times 13 \\ & = 26 \end{align*} \]
Answer \( \color{red}26\ \text{non-square numbers} \)
(b) Write the first four triangular numbers.
Answer \( \color{red}1,\ 3,\ 6,\ 10 \)
(c) Is 5, 7, 9 a Pythagorean triplet? Why? Justify.
Solution
\[ \begin{align*} \text{Pythagorean} & \text{ triplet}\\ \implies \color{orange} a^2 + b^2 &= \color{orange} c^2 \\ a &= 5 \\ b &= 7 \\ c &= 9 \\ \\ \end{align*} \] \[ \begin{array}{c|c} \color{violet} LHS & \color{violet} RHS \\ = 5^2 + 7^2 & = 9^2 \\ = 25 + 49 & = 81 \\ = 74 & \\ \end{array} \] \[ \begin{align*} \implies 74 \ne 81 \end{align*} \]
Answer \( \color{red} \text{No, 5, 7, 9 is not a Pythagorean triplet.} \)
(d) Find \( \sqrt{9} \) by repeated subtraction method.
Solution
\[ \begin{align*} 9 - 1 &= 8 \\ 8 - 3 &= 5 \\ 5 - 5 &= 0 \\ \\ \text{Total steps} &= 3 \end{align*} \]
Answer \( \sqrt{9} = \color{red} 3 \)
(e) Find the measure of the side of a square handkerchief of area \( 324 \, \text{cm}^2 \).
Solution
\[ \begin{align*} \text{Area} &= 324 \, \text{cm}^2 \\ \text{(Side)}^2 &= 324 \\ \text{Side} &= \sqrt{324} \\ \text{Side} &= \sqrt{18 \times 18} \\ \text{Side} &= 18 \, cm \\ \end{align*} \]
Answer \( \color{red}18 \, \text{cm} \)
2. Find the square root of 10, correct to four places of decimal.
Solution
\[ \begin{array}{r|l} & \phantom{0} \ 3 \ . \ 1 \ 6 \ 2 \ 2 \ 7 \\ \hline 3 & \phantom{1}10 . \overline{00} \, \overline{00} \, \overline{00} \, \overline{00} \, \overline{00} \\ & -\ 9 \\ \hline 61 & \phantom{11}100 \\ & \phantom{ \ }-61 \\ \hline 626 & \phantom{--}3900 \\ & \phantom{1}-3756 \\ \hline 6322 & \phantom{00000}14400 \\ & \phantom{-1}-12644 \\ \hline 63242 & \phantom{000000}175600 \\ & \phantom{-00}-126484 \\ \hline 632447 & \phantom{0000000}4911600 \\ & \phantom{-000}-4427129 \\ \hline & \phantom{00000000}484471 \\ \end{array} \] \[ \begin{aligned} \sqrt{10} & = 3.16227 \\ \sqrt{10} & = 3.1623 \\ \end{aligned} \]
Answer \( \color{red}\sqrt{10} = 3.1623 \)
3. Find the values of: \(\sqrt{3.1428}\) and \(\sqrt{0.31428}\) correct to three decimal places.
Solution
\[ \begin{array}{r|l} & \phantom{0}1 \ . \ 7 \ 7 \ 2 \ 7 \\ \hline 1 & \phantom{-}3 . \overline{14} \,\overline{28} \,\overline{00} \,\overline{00} \\ & - 1 \\ \hline 27 & \phantom{00}214 \\ & -\ 189 \\ \hline 347 & \phantom{000}2528 \\ & \ -2429 \\ \hline 3542 & \phantom{00000}9900 \\ & \ \ \ \ \ -7084 \\ \hline 35447 & \phantom{00000}281600 \\ & \ \ \ \ \ -248129 \\ \hline & \phantom{000000}33471 \\ \end{array} \] \[ \begin{aligned} \sqrt{3.1428} &= 1.7727 \\ \color{green} \sqrt{3.1428} &= \color{green} 1.773 \\ \end{aligned} \]
\[ \begin{array}{r|l} & \ \ \ 0. \ 5 \ 6 \ 0 \ 6 \\ \hline 0 &\ \ \ 0 . \overline{31} \,\overline{42} \,\overline{80} \,\overline{00} \\ & - 0 \\ \hline 5 & \ \ \ \ 031 \\ & \ -25 \\ \hline 106 & \ \ \ \ \ \ \ \ 642 \\ & \ \ \ -636 \\ \hline 1120 & \ \ \ \ \ \ \ \ \ \ \ \ 680 \\ & \ \ \ \ \ \ \ \ \ \ - 0 \\ \hline 11206 & \ \ \ \ \ \ \ \ \ \ \ \ \ 68000 \\ & \ \ \ \ \ \ \ \ - 67236 \\ \hline & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 764 \\ \end{array} \] \[ \begin{aligned} \sqrt{0.31428} &= 0.5606 \\ \color{green} \sqrt{0.31428} &= \color{green} 0.561 \end{aligned} \]
Answer \( \sqrt{3.1428} = {\color{red} 1.773} \ and \ \sqrt{0.31428} = {\color{red} 0.561} \)
4. Simplify
(i) \( \displaystyle \frac{\sqrt{0.0441}}{\sqrt{0.000441}} \)
Solution
\[ \begin{aligned} & = \frac{\sqrt{0.0441}}{\sqrt{0.000441}} \\ \\ & = \sqrt{\frac{0.0441 \times 1000000}{0.000441 \times 1000000}} \\ \\ & = \sqrt{\frac{\cancel{441}^1 \times 100}{\cancel{441}_1}} \\ \\ & = \sqrt{100} \\ & = 10 \\ \end{aligned} \]
Answer \( \displaystyle \frac{\sqrt{0.0441}}{\sqrt{0.000441}} = {\color{red} 10} \)
(ii) \( \sqrt{49} + \sqrt{0.49} + \sqrt{0.0049} \)
Solution
\[ \begin{aligned} & = \sqrt{49} + \sqrt{0.49} + \sqrt{0.0049} \\ \\ & = \sqrt{49} + \sqrt{\frac{49}{100}} + \sqrt{\frac{49}{10000}} \\ \\ & = 7 + \frac{7}{10} + \frac{7}{100} \\ \\ & = 7 + 0.7 + 0.07 \\ & = 7.77 \\ \end{aligned} \]
Answer \( \sqrt{49} + \sqrt{0.49} + \sqrt{0.0049} = {\color{red} 7.77} \)
5. The area of a square field is \( \displaystyle 101 \frac{1}{400} m^2 \). Find the length of one side of the field.
Solution
\[ \begin{aligned} \text{Area of the field} & = 101 \frac{1}{400} m^2 \\ \\ & = \frac{40401}{400} \\ \\ (Side)^2 & = \frac{40401}{400}\\ \\ Side & = \sqrt{\frac{40401}{400}}\\ \\ & = \frac{\sqrt{40401}}{\sqrt{400}}\\ \\ \end{aligned} \] \[ \begin{array}{r|l} & \phantom{0} 2 \ 0 \ 1 \\ \hline 2 & \phantom{-}4 \, \overline{04} \, \overline{01} \\ & - 4 \\ \hline 401 & \phantom{00}00401 \\ & \ \ \ -401 \\ \hline & \phantom{00000}0 \\ \end{array} \] \[ \begin{aligned} Side & = \frac{\sqrt{40401}}{\sqrt{400}}\\ \\ & = \frac{201}{20} \\ \\ \text{Side of the field}& = 10 \frac{1}{20} m \\ \\ \end{aligned} \]
Answer Side of the field \( = {\color{red}10 \displaystyle\frac{1}{20} m} \)
6. What is that number which when multiplied by itself gives 227.798649?
Solution
\( \text{Let the required number be } x \) \[ \begin{aligned} x \times x & = 227.798649 \\ x^2 & = 227.798649 \\ x & = \sqrt{227.798649} \\ \\ \end{aligned} \] \[ \begin{array}{r|l} & \ \ \ 1\ 5\ .\ 0\ 9\ 3 \\ \hline 2 & \ \ \ \overline 2 \ \overline{27}. \overline{79}\,\overline{86}\,\overline{49} \\ & - 1 \\ \hline 25 & \ \ \ \ 127 \\ & \, - 125 \\ \hline 3009 & \ \ \ \ \ \ \ \ 27986 \\ & \ \ \ - 27081 \\ \hline 30183 & \ \ \ \ \ \ \ \ \ \ \ \ 90549 \\ & \ \ \ \ \ \ \ - 90549 \\ \hline & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \\ \end{array} \] \[ \begin{aligned} \sqrt{227.798649} &= 15.093 \\ x &= 15.093 \end{aligned} \]
Answer Required number \( = {\color{red} 15.093 } \)
7. In a lecture hall, 8649 students are sitting in such a manner that there are as many students in a row as there are rows in the lecture hall. How many students are there in each row of the lecture hall?
Solution
Let the no. of rows be \(x\)
Let the no. of students in each row be \(x\)
\[
\begin{aligned}
x \times x &= 8649 \\
x^2 &= 8649 \\
x &= \sqrt{8649}
\end{aligned}
\]
\[
\begin{array}{r|l}
& \ \ \ 9\ 3 \\
\hline
9 & \ \ \ \overline{86}\,\overline{49} \\
& - 81 \\ \hline
183 & \ \ \ \ \ 549 \\
& \, \, \, - 549 \\ \hline
& \ \ \ \ \ \ 0
\end{array}
\]
\[
\begin{aligned}
\sqrt{8649} &= 93 \\
x &= 93
\end{aligned}
\]
Answer Each row has \( {\color{red}93} \) students.
8. A General wishing to draw up his 64019 men in the form of a square found that he had 10 men extra. Find the number of men in the front row.
Solution
\[ \begin{aligned} \text{No. of men} &= 64019 \\ \text{No. of extra men} &= 10 \\ \text{No. of remaining men} &= 64019 - 10 \\ &= 64009 \\ \text{No. of men in the front row} &= \sqrt{64009} \\ \end{aligned} \] \[ \begin{array}{r|l} & \ \ \ 2\ 5\ 3 \\ \hline 2 & \ \ \ \overline{6}\,\overline{40}\,\overline{09} \\ & - 4 \\ \hline 45 & \ \ \ \ 240 \\ & \, - 225 \\ \hline 503 & \ \ \ \ \ \ 1509 \\ & \ - 1509 \\ \hline & \ \ \ \ \ \ \ 0 \end{array} \] \[ \sqrt{64009} \;=\; 253 \]
Answer No. of men in the front row \( = \color{red}253 \, men\)