DAV Class 8 Maths Chapter 1 Practice Worksheet
Squares and Square Roots Practice Worksheet
A. Tick (✓) the correct option.
1.
The natural numbers between \(10^2\) and \(11^2\) is –
\(
\begin{aligned}
(a) \, & \, 22 \\
(b) \, & \, 21 \\
(c) \, & \, 20 \\
(d) \, & \, 221 \\
\end{aligned}
\)
Solution
\[ \begin{align*} n &= 10 \\ \text{Natural numbers} &= 2n \\ &= 2 \times 10 \\ &= 20 \end{align*} \]
Answer \( \color{red} 20 \ \color{orange} (c) \)
2.
The difference between \(98^2\) and \(97^2\) is –
\(
\begin{aligned}
(a) \, & \, 195 \\
(b) \, & \, 196 \\
(c) \, & \, 1 \\
(d) \, & \, 185 \\
\end{aligned}
\)
Solution
\[ \begin{align*} &= (98)^2 - (97)^2 \\ &= 98 + 97 \\ &= 195 \end{align*} \]
Answer \( \color{red} 195 \ \color{orange} (a) \)
3.
Square of 0.7 is –
\(
\begin{aligned}
(a) \, & \, 4.9 \\
(b) \, & \, 0.049 \\
(c) \, & \, 0.49 \\
(d) \, & \, 1.4 \\
\end{aligned}
\)
Solution
\[ \begin{align*} &= (0.7)^2 \\ &= 0.49 \end{align*} \]
Answer \( \color{red} 0.49 \ \color{orange} (c) \)
4.
The square of the number which has 9 in its unit place ends with –
\(
\begin{aligned}
(a) \, & \, 9 \\
(b) \, & \, 1 \\
(c) \, & \, 3 \\
(d) \, & \, 8 \\
\end{aligned}
\)
Solution
\[ \begin{align*} 9^2 &= 81 \\ (19)^2 &= 361 \end{align*} \]
Answer \( \color{red} 1 \ \color{orange} (b) \)
5.
If \( \sqrt{2} = 1.414 \), then the value of \( 32 \) is –
\(
\begin{aligned}
(a) \, & \, 5.656 \\
(b) \, & \, 56.56 \\
(c) \, & \, 4.656 \\
(d) \, & \, 46.56 \\
\end{aligned}
\)
Solution
\[ \begin{array}{c|c} 2 & 32 \\ \hline 2 & 16 \\ \hline 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} \sqrt{32} &= \sqrt{2^2 \times 2^2 \times 2} \\ &= 2 \times 2 \times \sqrt{2} \\ &= 4 \times \sqrt{2} \\ &= 4 \times 1.414 \\ &= 5.656 \end{align*} \]
Answer \( \color{red} 5.656 \ \color{orange} (a) \)
6.
The number of zeroes in the square of 10000 is –
\(
\begin{aligned}
(a) \, & \, 5 \\
(b) \, & \, 6 \\
(c) \, & \, 3 \\
(d) \, & \, 8 \\
\end{aligned}
\)
Solution
\[ \begin{align*} (10000)^2 &= 100000000 \end{align*} \]
Answer \( \color{red} 8 \ \color{orange} (d) \)
7.
The unit digit of square of 34564 is –
\(
\begin{aligned}
(a) \, & \, 8 \\
(b) \, & \, 6 \\
(c) \, & \, 4 \\
(d) \, & \, 16 \\
\end{aligned}
\)
Solution
\[ \begin{align*} 4^2 &= 16 \\ &= \text{6 is in the units place} \end{align*} \]
Answer \( \color{red} 6 \ \color{orange} (b) \)
8.
The third triangular number is –
\(
\begin{aligned}
(a) \, & \, 6 \\
(b) \, & \, 1 \\
(c) \, & \, 3 \\
(d) \, & \, 10 \\
\end{aligned}
\)
Solution
\[ \begin{align*} 1,3,6,10 \end{align*} \]
Answer \( \color{red} 6 \ \color{orange} (a) \)
9.
The value of \( \sqrt{0.2} \times \sqrt{9.8} \) is –
\(
\begin{aligned}
(a) \, & \, 1.8 \\
(b) \, & \, 1.2 \\
(c) \, & \, 0.16 \\
(d) \, & \, 1.4 \\
\end{aligned}
\)
Solution
\[ \begin{align*} & = \sqrt{0.2} \times \sqrt{9.8} \\ & = \sqrt{0.2 \times 9.8} \\ & = \sqrt{1.96} \\ & = 1.4 \end{align*} \]
Answer \( \color{red} 1.4 \ \color{orange} (d) \)
10.
The least number by which 72 should be multiplied to make a perfect square is –
\(
\begin{aligned}
(a) \, & \, 3 \\
(b) \, & \, 4 \\
(c) \, & \, 2 \\
(d) \, & \, 1 \\
\end{aligned}
\)
Solution
\[ \begin{array}{c|c} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} 72 &= 2^2 \times 2 \times 3^2 \\ \text{Least number} & \implies 2 \\ & =72 \times 2 \\ & =144 \end{align*} \]
Answer \( \color{red} 2 \ \color{orange} (c) \)
Section - II
11. Write a Pythagorean triplet whose smallest number is 14
Solution
\[ \begin{align*} \color{green} 2m &= \color{green} 14 \\ \\ m &= \frac{14}{2} \\ \\ m &= 7 \\ \\ m^2 - 1 &= 7^2 - 1 \\ & = 49 - 1\\ \color{green} m^2 - 1 &= \color{green} 48 \\ \\ m^2 + 1 &= 7^2 + 1 \\ &= 49 + 1 \\ \color{green} m^2 + 1 &= \color{green} 50 \\ \\ (2m, m^2 - 1, m^2 + 1) &= \color{green} (14,48,50) \end{align*} \]
Answer \( \color{red} (14,48,50) \) are the Pythagorean triplet.
12. Simplify: \( \sqrt{248 + \sqrt{52 +\sqrt{144}}} \)
Solution
\[ \begin{align*} &=\sqrt{248 + \sqrt{52 +\sqrt{144}}} \\ &=\sqrt{248 + \sqrt{52 + 12}} \\ &=\sqrt{248 + \sqrt{64}} \\ &=\sqrt{248 + 8} \\ &=\sqrt{256} \\ &= 16 \end{align*} \]
Answer \( \color{red} 16 \)
13. Find the value of: \( \displaystyle \sqrt{1 + \frac{64}{225}} \)
Solution
\[ \begin{align*} & =\sqrt{1 + \frac{64}{225}} \\ \\ &= \sqrt{\frac{225 + 64}{225}} \\ \\ &= \sqrt{\frac{289}{225}} \\ \\ &= \frac{\sqrt{(17)^2}}{\sqrt{(15)^2}} \\ \\ &= \frac{17}{15} \end{align*} \]
Answer \( \displaystyle \color{red} \frac{17}{15} \)
14. If \( \sqrt{4489} = 67 \), then find the value of \( \sqrt{44.89} + \sqrt{0.4489} \)
Solution
\[ \begin{align*} & = \sqrt{44.89} + \sqrt{0.4489} \\ \\ & = \sqrt{\frac{4489}{100}} + \sqrt{\frac{4489}{10000}} \\ \\ &= \frac{67}{10} + \frac{67}{100} \\ \\ &= 6.7 + 0.67 \\ \\ &= \color{green} 7.37 \end{align*} \]
Answer \( \color{red} 7.37 \)
15. Find the area of the square field, if its perimeter is 96 m.
Solution
\[ \begin{align*} \text{Perimeter of square} &= 96 \ m \\ \text{Perimeter of square} &= 4 \times side \\ 4 \times side &= 96 \\ side &= \frac{\cancel{96}^{24}}{\cancel4_1} \\ side &= 24 \ m \\ \\ \text{Area of square} &= (side)^2 \\ &= (24)^2 \\ &= \color{green} 576 \, \text{m}^2 \end{align*} \]
Answer \( \color{red} 576 \, \text{m}^2 \)
16. Find the greatest 4-digit number which is a perfect square.
Solution
\[ \begin{align*} \text{Greatest 4-digit number} &= 9999 \\ \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}99 \\ \hline 9 & \phantom{-}\overline{99} \,\, \overline{99} \\ & -81 \\ \hline 189 & \phantom{-}1899 \\ & -1701 \\ \hline & \phantom{-0}198 \\ \hline \end{array} \] \[ \begin{align*} \text{Perfect square number} &= 9999 - 198 \\ &= 9801 \\ \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}99 \\ \hline 9 & \phantom{-}\overline{98} \,\, \overline{01} \\ & -81 \\ \hline 189 & \phantom{-}1701 \\ & -1701 \\ \hline & \phantom{0000}0 \\ \end{array} \] \[ \sqrt{9801} = 99 \]
Answer \( \color{red} 9801 \)
17. Find the value of \( \displaystyle x : \frac{7568}{\sqrt{x}} = 75.68 \)
Solution
\[ \begin{align*} \frac{7568}{\sqrt{x}} &= 75.68 \\ \\ 7568 &= \sqrt{x} \times 75.68 \\ \\ \frac{7568}{75.68} &= \sqrt{x} \\ \\ \frac{7568 \times 100}{75.68 \times 100} &= \sqrt{x} \\ \\ \frac{\cancel{7568}^1 \times 100}{\cancel{7568}_1} &= \sqrt{x} \\ \\ 100 &= \sqrt{x} \\ \\ (100)^2 &= x \\ \\ x & = \color{green} 10000 \end{align*} \]
Answer \( x = \color{red} 10000 \)
18. Find the smallest number by which 288 should be divided so that the quotient is a perfect square.
Solution
\[ \begin{array}{c|c} 2 & 288 \\ \hline 2 & 144 \\ \hline 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} 288 &= 2^2 \times 2^2 \times 3^2 \times 2 \\ \\ \text{Smallest number } &= \color{red}\boxed{2} \\ \frac{288}{2} &= 144 \\ \sqrt{144} & = 12 \\ \end{align*} \]
Answer Smallest number to be divided \( = \color{red}\boxed{2} \).
19. Find the square root of 500 by estimation method.
Solution
\[ \begin{aligned} &\text{The perfect squares near 500 are 484 and 529.} \\ &484 < 500 < 529 \\ &22^2 < 500 < 23^2 \\ &\text{So, } \sqrt{500} \text{ lies between 22 and 23.} \\ \\ & (22.3)^2 = 497.29 < 500 \\ & (22.4)^2 = 501.76 > 500 \\ \\ & 497.29 < 500 < 501.76 \\ \\ & \implies 500 \text{ is closer to 501.76} \\ \\ &\therefore \color{green} \sqrt{500} \text{ is approximately } 22.4 \end{aligned} \]
Answer \( \color{red} \sqrt{500} = 22.4 \, \text{approximately} \)
20. Find the square root of 19881 by prime factorisation method.
Solution
\[ \begin{array}{c|c} 3 & 19881 \\ \hline 3 & 6627 \\ \hline 47 & 2209 \\ \hline 47 & 47 \\ \hline & 1 \\ \end{array} \] \[ \begin{align*} \sqrt{19881} &= \sqrt{3^2 \times 47^2} \\ &= 3 \times 47 \\ &= 141 \\ \end{align*} \]
Answer \( \color{red} \sqrt{19881} = 141 \)
Section - III
21. The product of two numbers is 1296. If one number is 16 times the other, find the number.
Solution
\[ \begin{align*} \text{Let first number be} & = x\\ \text{Let other number be} & = 16x\\ \\ x \times 16 x & = 1296 \\ 16 x^2 & = 1296 \\ \\ x^2 & = \frac{\cancel{1296}^{81}}{\cancel{16}_1} \\ \\ x^2 & = 81 \\ x & = \sqrt{81} \\ x & = \sqrt{9 \times 9} \\ x & = 9 \\ \\ \color{green} \text{First number} & = \color{green} 9 \\ \\ \text{Other number} & = 16x \\ & = 16 \times 9 \\ \color{green} \text{Other number} & = \color{green} 144 \\ \\ \color{magenta} \text{Check} \implies 9 \times 144 &= \color{magenta}1296 \end{align*} \]
Answer The numbers are \( = \color{red} 9 \text{ and } 144 \)
22. Find the least number that should be added to 9210 to make it a perfect square. Also, find the square root of the number obtained.
Solution
\[ \begin{array}{r|l} & \phantom{0}95 \\ \hline 9 & \phantom{-}\overline{92}\ \overline{10} \\ & -81 \\ \hline 185 & \phantom{-}1110 \\ & \phantom{}-925 \\ \hline & \phantom{-0}185 \\ \end{array} \]
\[ \begin{aligned} 95^2 < 9210 & < 96^2 \\ \\ 96^2 &= 9216 \\ \\ \text{Required Number} &= 9216 - 9210 \\ &= \color{green} 6 \\ \sqrt{9216} &= \color{green} 96 \end{aligned} \]
Answer Least number = \( \color{red} 6\), \( \sqrt{9216} = \color{red}96\)
23. Find the side of the square whose area is equal to the area of the rectangle with sides 6.4 m and 2.5 m.
Solution
\[ \begin{align*} \text{Length} &= 6.4\ \text{m}\\ \text{Breadth} &= 2.5\ \text{m}\\ \text{Area of square} &= \text{Area of rectangle} \\ (Side)^2 &= Length \times Breadth \\ (Side)^2 &= 6.4 \ m \times 2.5 \ m \\ (Side)^2 &= 16\ m^2 \\ Side &= \sqrt{16 \ m^2} \\ Side &= 4m \end{align*} \]
Answer Side of the square \(= \color{red}4m \)
24. A gardener has 2025 plants. He wants to plant these in such a way that each row contains as many plants as the number of rows. Find the number of rows and number of plants in each row.
Solution
\[ \begin{align*} \text{No. of rows} &= x\\ \text{No. of plants in each row} &= x\\ \text{Total no. of plants} &= 2025\\ x\times x &= 2025 \\ x^2 &= 2025\\ x &= \sqrt{2025} \end{align*} \]
\[ \begin{array}{r|l} & \phantom{0}45 \\ \hline 4 & \phantom{-} \overline{20} \, \overline{25} \\ & -16 \\ \hline 85 & \phantom{0-}425 \\ & \phantom{}-425 \\ \hline & \phantom{0000}0 \end{array} \]
\[ \begin{align*} x &= 45 \end{align*} \]
Answer Number of rows \(=\color{red}{45}\), Plants in each row \(=\color{red}{45}\).
25. Simplify: \( (\sqrt{81} + \sqrt{0.81} + \sqrt{0.0081}) \times \sqrt{10000}\)
Solution
\[ \begin{aligned} &=(\sqrt{81} + \sqrt{0.81} + \sqrt{0.0081}) \times \sqrt{10000} \\ \\ &= \left(9 + \sqrt{\frac{81}{100}} + \sqrt{\frac{81}{1000}} \right) \times 100 \\ \\ &= \left(9 + \frac{9}{10} + \frac{9}{100} \right) \times 100 \\ \\ &= \left(9 + 0.9 + 0.09 \right) \times 100 \\ &= 9.99 \times 100 \\ &= 999 \end{aligned} \]
Answer \( \color{red}{999}\)
26. Find the square root of the following decimal numbers:
a. 1056.25
Solution
\[ \begin{array}{r|l} & \phantom{00}32.5 \\ \hline 3 & \phantom{-}\overline{10}\,\overline{56}\,\overline{.25} \\ & \phantom{}-9 \\ \hline 62 & \phantom{-0}156 \\ & \phantom{}-124 \\ \hline 645 & \phantom{-0}3225 \\ & \phantom{}-3225 \\ \hline & \phantom{--0} 0 \end{array} \] \[ \sqrt{1056.25} = 32.5 \]
Answer \( \sqrt{1056.25} = \color{red} {32.5}\)
b. 10020.1
Solution
\[ \begin{array}{r|l} & \phantom{00}100.1 \\ \hline 1 & \phantom{-}\overline{1}\,\overline{00}\,\overline{20}\,\overline{.01} \\ & -1 \\ \hline 20 & \phantom{00}000 \\ & \phantom{0} -0 \\ \hline 200 & \phantom{0000}020 \\ & \phantom{000}-0 \\ \hline 2001 & \phantom{00000}2001 \\ & \phantom{0-} -2001 \\ \hline & \phantom{----}0 \end{array} \] \[ \sqrt{10020.01} = 100.1 \]
Answer \(\sqrt{10020.01} = \color{red}{100.1}\)
27. Find the square root of 8 correct to 2 decimal places.
Solution
\[ \begin{array}{r|l} & \phantom{0} \ 2 \ . \ 8 \ 2 \ 8 \ \\ \hline 2 & \phantom{-}8 . \overline{00} \, \overline{00} \, \overline{00} \\ & - 4 \\ \hline 48 & \phantom{-}400 \\ & -384 \\ \hline 562 & \phantom{-0}1600 \\ & \phantom{}-1124 \\ \hline 5648 & \phantom{-00}47600 \\ & \phantom{0}-45184 \\ \hline & \phantom{--0}2416 \\ \hline \end{array} \] \[ \begin{aligned} \sqrt{8} & = 2.828 \\ \color{green}\sqrt{8} & \approx \color{green}2.83 \end{aligned} \]
Answer \( \color{red}\sqrt{8} \approx 2.83 \)
28. Find the square root of 5 upto 3 decimal places.
Solution
\[ \begin{array}{r|l} & \phantom{0}2 \ .\ 2\ 3\ 6\ \\ \hline 2 & \phantom{-}5 . \overline{00}\,\overline{00}\,\overline{00} \\ & -4 \\ \hline 42 & \phantom{-}100 \\ & \phantom{}-84 \\ \hline 443 & \phantom{0-}1600 \\ & \phantom{}-1329 \\ \hline 4466 & \phantom{-00}27100 \\ & \phantom{0}-26796 \\ \hline & \phantom{--0}304 \\ \hline \end{array} \] \[ \begin{aligned} \sqrt{5} &= 2.236\\ \end{aligned} \]
Answer \( \color{red}\sqrt{5} = 2.236\)
29. Using long division method, find the square root of the following numbers:
a. 77841
Solution
\[ \begin{array}{r|l} & \phantom{0}279 \\ \hline 2 & \phantom{-}\overline{7}\,\overline{78}\,\overline{41} \\ & -4 \\ \hline 47 & \phantom{-}378 \\ & -329 \\ \hline 549 & \phantom{-0}4941 \\ & \phantom{}-4941 \\ \hline & \phantom{00000}0 \end{array} \] \[ \sqrt{77841} = 279 \]
Answer \( \displaystyle \sqrt{77841} = \color{red}{279}\)
b. 2044900
Solution
\[ \begin{array}{r|l} & \phantom{0}1430 \\ \hline 1 & \phantom{-}\overline{2}\,\overline{04}\,\overline{49}\,\overline{00} \\ & -1 \\ \hline 24 & \phantom{-}104 \\ & \phantom{}-96 \\ \hline 283 & \phantom{-00}849 \\ & -\phantom{00}849 \\ \hline 2860 & \phantom{-0000}000 \\ & \phantom{000}-000 \\ \hline & \phantom{000000}0 \end{array} \] \[ \sqrt{2044900} = 1430 \]
Answer \( \displaystyle \sqrt{2044900} = \color{red}{1430}\)
30. Find the smallest 6-digit number which is a perfect square. Find the square root of this number.
Solution
\[ \begin{align*} \text{Least number of 6 digits} &= 100000 \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}317 \\ \hline 3 & \phantom{-}\overline{10}\,\overline{00}\,\overline{00} \\ & -\phantom{1}9 \\ \hline 61 & \phantom{0-}100 \\ & -\phantom{00}61 \\ \hline 627 & \phantom{-00}3900 \\ & -\phantom{00}4389 \\ \hline & \phantom{00}-489 \end{array} \] \[ \begin{align*} \\ & = 100000 + 489 \\ & = 100489 \\ \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}317 \\ \hline 3 & \phantom{-}\overline{10}\,\overline{04}\,\overline{89} \\ & -\phantom{1}9 \\ \hline 61 & \phantom{0-}104 \\ & -\phantom{00}61 \\ \hline 627 & \phantom{-00}4389 \\ & -\phantom{00}4389 \\ \hline & \phantom{0--}0 \end{array} \] \[ \sqrt{100489} = 317 \]
Answer Perfect square number \(= \color{red}{100489}\), \(\sqrt{100489} = \color{red}{317}\)
Section - IV
31. The area of rectangular field whose length is twice its breadth is \( 2450 \ m^2 \). Find the perimeter of the field.
Solution
\[ \begin{align*} \text{Let breadth} &= x \\ \text{length} &= 2x \\ \text{Area of rectangle} &= 2450 \ m^2 \\ L \times B &= 2450 \ m^2 \\ 2x \times x &= 2450 \ m^2 \\ 2x^2 &= 2450 \\ x^2 &= \frac{\cancel{2450}^{1225}}{\cancel2_1} \\ x^2 &= 1225 \\ x &= \sqrt{1225} \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}35 \\ \hline 3 & \phantom{-}\overline{12}\,\overline{25} \\ & -\phantom{1}9 \\ \hline 65 & \phantom{0-}325 \\ & -\phantom{0}325 \\ \hline & \phantom{-1-}0 \end{array} \] \[ \begin{align*} x &= 35 \\ \text{Breadth} &= 35 m \\ \\ \text{Length} &= 2 \times 35 \\ &= 70m \\ \\ \text{Perimeter of rectangle} &= 2 \times (L + B) \\ & = 2 \times (70 + 35) \\ & = 2 \times 105 \\ &= 210 \ m \end{align*} \]
Answer The perimeter of the field \( = \color{red}210 \ m \)
32. Find the square root of \(2 \displaystyle \frac{2}{5}\) correct to three decimal places.
Solution
\[ \begin{align*} & = \sqrt{2\frac{2}{5}} \\ \\ &= \sqrt{\frac{12}{5}} \\ \\ &= \sqrt{2.4} \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}1.5491 \\ \hline 1 & \phantom{-}\overline{2}\,\overline{.40}\,\overline{00}\,\overline{00} \, \overline{00}\\ & -1 \\ \hline 25 & \phantom{-}140 \\ & -125 \\ \hline 304 & \phantom{-0}1500 \\ & \phantom{}-1216 \\ \hline 3089 & \phantom{-00}28400 \\ & \phantom{0}-27801 \\ \hline 30981 & \phantom{---}59900 \\ & \phantom{000}-30981 \\ \hline & \phantom{---}28991 \\ \end{array} \] \[ \begin{align*} \sqrt{2\frac{2}{5}} &= 1.5491 \\ \\ \sqrt{2\frac{2}{5}} & \approx {1.549} \end{align*} \]
Answer \(\displaystyle \sqrt{2\frac{2}{5}} \approx \color{red}{1.549}\)
33. The cost of levelling the square lawn at ₹12 per square metre is ₹35,643. Find the cost of fencing the lawn at ₹25 per metre.
Solution
\[ \begin{align*} \text{Total cost of levelling} &= \text{₹}35643\\ \text{Cost of levelling per square metre} &= \text{₹}12\\ \text{Area of square lawn} &= \frac{35643}{12}\\ \color{green}\text{Area of square lawn} &= \color{green}2970.25 \ m^2\\ (\text{Side})^2 &= 2970.25\\ \text{Side} &= \sqrt{2970.25} \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}54.5 \\ \hline 5 & \phantom{-}\overline{29}\,\overline{70}\,\overline{.25} \\ & -25 \\ \hline 104 & \phantom{-0}470 \\ & -\phantom{0}416 \\ \hline 1085 & \phantom{-00}5425 \\ & -\phantom{00}5425 \\ \hline & \phantom{---}0 \end{array} \] \[ \begin{align*} \sqrt{2970.25} &= 54.5\\ Side &= 54.5 \\ \text{Side of square lawn} &= 54.5 \ m \\ \\ \text{Perimeter of square lawn} &= 4 \times side \\ &= 4 \times 54.5 \\ &= 218 \ m \\ \\ \text{Cost of fencing per metre} &= \text{₹}25\\ \text{Total cost of fencing} &= 25 \times 218 \\ & = \color{green}\text{₹}5450 \end{align*} \]
Answer Total cost of fencing \(= \color{red}\text{₹}5450\)
34. During a mass drill exercise 6250 students of different schools are arranged in rows such that the number of students in each row is equal to the number of rows. In doing so, the instructor finds out that 9 children are left out. Find the number of children in each row.
Solution
\[ \begin{align*} \text{Total students} &= 6250 \\ \text{Left-out students} &= 9 \\ \text{Remaining students} &= 6241 \\ \text{No. of rows} &= x \\ \text{No. of students in each row} &= x \\ x^2 &= 6241 \\ x &= \sqrt{6241} \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}79 \\ \hline 7 & \phantom{-}\overline{62}\,\overline{41} \\ & -49 \\ \hline 149 & \phantom{-}1341 \\ & -1341 \\ \hline & \phantom{00-}0 \end{array} \] \[ \begin{align*} \sqrt{6241} = 79 \\ x = 79 \end{align*} \]
Answer No. of children in each row \(= \color{red}79\)
35. The area of a square field is 6084 m². If the perimeter of a rectangle field, whose length is twice its breadth, is equal to the perimeter of the square field, find the dimensions of the rectangle field.
Solution
\[ \begin{align*} \text{Area of square field} &= 6084 \ m^2 \\ (\text{Side})^2 &= 6084 \\ \text{Side} &= \sqrt{6084} \end{align*} \] \[ \begin{array}{r|l} & \phantom{0}78 \\ \hline 7 & \phantom{-}\overline{60}\,\overline{84} \\ & -49 \\ \hline 148 & \phantom{-}1184 \\ & -1184 \\ \hline & \phantom{00-}0 \end{array} \] \[ \begin{align*} Side &= 78 \ m \\ \text{Perimeter of square field} &= 4 \times side \\ &= 4 \times 78 \\ &= 312 \ m \\\\ Breadth &= x \\ Length &= 2x \\ \text{Perimeter of rectangle} &= 312 \ m \\ 2(l + b) & = 312 \\ 2(2x +x) & = 312 \\ 2(3x) & = 312 \\ 6x &= 312 \\ x &= \frac{\cancel{312}^{52}}{\cancel6_1} \\ x &= 52 \ m \\ \\ Breadth &= 52 \ m \\ Length &= 2x \\ & = 2 \times 52 \\ &= 104 \ m \end{align*} \]
Answer Length \(= \color{red}104 \ m,\) Breadth \(= \color{red}52 \ m \)