{"id":1500,"date":"2020-10-30T14:11:20","date_gmt":"2020-10-30T14:11:20","guid":{"rendered":"http:\/\/themindpalace.in\/?p=1500"},"modified":"2021-08-26T05:31:04","modified_gmt":"2021-08-26T05:31:04","slug":"square-square-roots","status":"publish","type":"post","link":"https:\/\/themindpalace.in\/index.php\/2020\/10\/30\/square-square-roots\/","title":{"rendered":"Square ,Square Roots"},"content":{"rendered":"\n

Summary of square and square roots<\/a><\/p>\n\n\n\n

Solved exercise of square roots<\/a><\/p>\n\n\n\n

<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Summary<\/h1>\n\n\n\n

Square Number<\/span><\/strong><\/p>\n\n\n\n

When we multiply the number by itself we will get Square number.A square number is a natural number obtained by multiplying the number it self.<\/p>\n\n\n\n

Ex: 6 x 6 =36<\/p>\n\n\n\n

      5 x 5 = 25<\/p>\n\n\n\n

      7 x 7 = 49<\/p>\n\n\n\n

Area of a Square= side X side<\/span><\/strong><\/p>\n\n\n\n

  • 1 X 1 = 1<\/li>
  • 2 X 2 = 4<\/li>
  • 3 X 3 = 9<\/li>
  • 4 X 4 = 16<\/li>
  • 5 X 5 = 25<\/li><\/ul>\n\n\n\n

    Where 5, 6, 7 are the natural numbers and 25,36,49 are the respective square numbers.<\/p>\n\n\n\n

    If a natural number m can be expressed as n2 , where n is also a natural number, then m is a Square number.<\/p>\n\n\n\n

    Any integer multiplied to itself gives Perfect Squares<\/strong>.<\/span><\/p>\n\n\n\n

    Eg: 25,36,49<\/p>\n\n\n\n

    Non perfect squares: Non perfect squares laying in between perfect squares<\/span><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n
    \"\"<\/figure><\/div>\n\n\n\n

    Properties of Square Numbers<\/span><\/strong><\/p>\n\n\n\n

    We can see that the square numbers are ending with 0, 1, 4, 5, 6 or    <\/strong><\/p>\n\n\n\n

     9 <\/strong>only. <\/strong>None of the square number is ending with 2, 3, 7 or 8.<\/p>\n\n\n\n

    If a number has 1<\/strong> or 9<\/strong> in the units<\/p>\n\n\n\n

    place, then it\u2019s square ends in 1<\/strong>.<\/p>\n\n\n\n

    1<\/strong><\/td>1<\/strong><\/td><\/tr>
    9<\/strong><\/td>81<\/td><\/tr>
    1<\/strong>1<\/strong><\/td>121<\/td><\/tr>
    1<\/strong>9<\/strong><\/td>361<\/td><\/tr>
    2<\/strong>1<\/strong><\/td>441<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

    Some More Interesting Patterns<\/span><\/strong><\/em><\/p>\n\n\n\n

    Adding Triangular Numbers<\/span><\/strong><\/p>\n\n\n\n

    If we could arrange the dotted pattern of the numbers in a triangular form then these numbers are called Triangular Numbers<\/strong>.<\/p>\n\n\n\n

    Eg: 1, 3, 6, 10\u2026\u2026\u2026\u2026<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    If we add two consecutive triangular numbers then we can get the square number.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    2. Numbers between Square Numbers<\/span><\/strong><\/p>\n\n\n\n

    If we take two consecutive numbers n and n + 1, then there will be (2n) non-perfect square numbers between their squares numbers.<\/p>\n\n\n\n

    Example<\/strong><\/p>\n\n\n\n

    Let\u2019s take n = 5 and 52<\/sup> = 25<\/p>\n\n\n\n

    n + 1 = 5 + 1 = 6 and 62<\/sup> = 36<\/p>\n\n\n\n

    2n = 2(5) = 10<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Adding Odd Numbers<\/span><\/strong><\/p>\n\n\n\n

    Sum of first n natural odd numbers is n2<\/sup>.<\/p>\n\n\n\n

    Any square number must be the sum of consecutive odd numbers starting from 1.<\/p>\n\n\n\n

    And if any natural number which is not a sum of successive odd natural numbers starting with 1, then it will not be a perfect square.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    4. A Sum of Consecutive Natural Numbers<\/span><\/strong><\/p>\n\n\n\n

    Every square number is the summation of two consecutive positive natural numbers.<\/p>\n\n\n\n

    If we are finding the square of n the to find the two consecutive natural numbers we can use the formula.<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    5. The Product of Two Consecutive Even or Odd Natural Numbers<\/span><\/strong><\/p>\n\n\n\n

    If we have two consecutive odd or even numbers (a + 1) and (a -1) then their product will be (a2<\/sup>– 1)<\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    6. Some More Interesting Patterns about Square Numbers<\/span><\/strong><\/p>\n\n\n\n

    \"\"<\/figure><\/div>\n\n\n\n

    Exercise 5.1<\/span><\/strong><\/h1>\n\n\n\n

    1. What will be the unit digit of the squares of the following numbers?<\/span><\/p>\n\n\n\n

    (i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 20387 (vii) 52698 (viii) 99880 (ix) 12796
    (x) 55555<\/p>\n\n\n\n

    Solution<\/span><\/strong><\/p>\n\n\n\n

    (i) Unit digit of 812<\/sup> = 1<\/p>\n\n\n\n

    (ii) Unit digit of 2722<\/sup> = 4
    (iii) Unit digit of 7992<\/sup> = 1
    (iv) Unit digit of 38532<\/sup> = 9<\/p>\n\n\n\n

    v) Unit digit of 12342<\/sup> = 6
    (vi) Unit digit of 263872<\/sup> = 9
    (vii) Unit digit of 526982<\/sup> = 4
    (viii) Unit digit of 998802<\/sup> = 0<\/p>\n\n\n\n

    ix) Unit digit of 127962<\/sup> = 6
    (x) Unit digit of 555552<\/sup> = 5<\/p>\n\n\n\n

    2. The following numbers are not perfect squares. Give reason.<\/span><\/p>\n\n\n\n

    i) 1057 (ii) 23453  (iii) 7928 (iv) 222222 (v) 64000 (vi) 89722 (vii) 222000 (viii) 505050<\/p>\n\n\n\n

    (i)1057 ends with 7 at unit place. So it is not a perfect square number.<\/p>\n\n\n\n


    (ii) 23453 ends with 3 at unit place. So it is not a perfect square number.<\/p>\n\n\n\n


    (iii) 7928 ends with 8 at unit place. So it is not a perfect square number.<\/p>\n\n\n\n


    (iv) 222222 ends with 2 at unit place. So it is not a perfect square number.<\/p>\n\n\n\n


    (v) 64000 ends with 3 zeros. So it cannot a perfect square number.<\/p>\n\n\n\n


    (vi) 89722 ends with 2 at unit place. So it is not a perfect square number.<\/p>\n\n\n\n


    (vii) 22000 ends with 3 zeros. So it can not be a perfect square number.<\/p>\n\n\n\n


    (viii) 505050 ends with 1 zero. So it is not a perfect square number.<\/p>\n\n\n\n

    3. The squares of which of the following would be odd numbers?<\/span><\/p>\n\n\n\n

    (i) 431 (ii) 2826 (iii) 7779  (iv) 82004<\/p>\n\n\n\n

    (i)4312<\/sup> is an odd number.<\/p>\n\n\n\n


    (ii) 28262<\/sup> is an even number.<\/p>\n\n\n\n


    (iii) 77792<\/sup> is an odd number.<\/p>\n\n\n\n


    (iv) 820042<\/sup> is an even number.<\/p>\n\n\n\n

    4. Observe the following pattern and find the missing digits<\/span><\/p>\n\n\n\n

    112<\/sup> = 121
    1012<\/sup> = 10201
    10012<\/sup> = 1002001
    1000012<\/sup> = 1\u20262\u20261
    100000012<\/sup> = \u2026\u2026\u2026<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    According to the above pattern, we have
    1000012<\/sup> = 10000200001
    100000012<\/sup> = 100000020000001<\/p>\n\n\n\n

    5. Observe the following pattern and supply the missing numbers.<\/span><\/p>\n\n\n\n

    112<\/sup> = 121
    1012<\/sup> = 10201
    101012<\/sup> = 102030201
    10101012<\/sup> = \u2026\u2026\u2026.
    \u2026\u2026\u2026.2<\/sup> = 10203040504030201<\/p>\n\n\n\n

    Solution<\/p>\n\n\n\n

    According to the above pattern, we have
    <\/p>\n\n\n\n

    10101012<\/sup> = 1020304030201<\/p>\n\n\n\n


    1010101012<\/sup> = 10203040504030201<\/p>\n\n\n\n

    6. Using the given pattern, find the missing numbers<\/span><\/p>\n\n\n\n

    12<\/sup> + 22<\/sup> + 22<\/sup> = 32<\/sup>
    22<\/sup> + 32<\/sup> + 62<\/sup> = 72<\/sup>
    32<\/sup> + 42<\/sup> + 122<\/sup> = 132<\/sup>
    42<\/sup> + 52<\/sup> + \u2026.2<\/sup> = 212<\/sup>
    52<\/sup> + \u2026.2<\/sup> + 302<\/sup> = 312<\/sup>
    62<\/sup> + 72<\/sup> + \u2026..2<\/sup> = \u2026\u20262<\/sup><\/p>\n\n\n\n

    Solution<\/span><\/p>\n\n\n\n

    According to the given pattern, we have<\/p>\n\n\n\n


    42<\/sup> + 52<\/sup> + 202<\/sup> = 212<\/sup><\/p>\n\n\n\n


    52<\/sup> + 62<\/sup> + 302<\/sup> = 312<\/sup><\/p>\n\n\n\n


    62<\/sup> + 72<\/sup> + 422<\/sup> = 432<\/sup><\/p>\n\n\n\n

    7. Without adding, find the sum.<\/span><\/p>\n\n\n\n

    (i) 1 + 3 + 5 + 7 + 9
    (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
    (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    We know that the sum of n odd numbers = n2<\/sup><\/p>\n\n\n\n

    (i)1 + 3 + 5 + 7 + 9 = (5)2<\/sup> = 25 [\u2235 n = 5]<\/p>\n\n\n\n


    (ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2<\/sup> = 100 [\u2235 n = 10]<\/p>\n\n\n\n


    (iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)2<\/sup> = 144 [\u2235 n = 12]<\/p>\n\n\n\n

    8. (i) Express 49 as the sum of 7 odd numbers.
    (ii) Express 121 as the sum of 11 odd numbers.<\/span><\/p>\n\n\n\n

    Solution:
    (i) 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)<\/p>\n\n\n\n


    (ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)<\/p>\n\n\n\n

    9. How many numbers lie between squares of the following numbers?<\/span>
    (i) 12 and 13
    (ii) 25 and 26
    (iii) 99 and 100.<\/p>\n\n\n\n

    Solution:<\/p>\n\n\n\n

    i)We know that numbers between n2<\/sup> and (n + 1)2<\/sup> = 2n<\/p>\n\n\n\n


    Numbers between 122<\/sup> and 132<\/sup> = (2n) = 2 \u00d7 12 = 24<\/p>\n\n\n\n

    ii) Numbers between 252<\/sup> and 262<\/sup> = 2 \u00d7 25 = 50 (\u2235 n = 25)<\/p>\n\n\n\n


    (iii) Numbers between 992<\/sup> and 1002<\/sup> = 2 \u00d7 99 = 198 (\u2235 n = 99)<\/p>\n","protected":false},"excerpt":{"rendered":"

    Summary of square and square roots Solved exercise of square roots Summary Square Number When we multiply the number by itself we will get Square number.A square number is a natural number obtained by...<\/p>\n","protected":false},"author":3,"featured_media":1514,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[110,475,14],"tags":[186,184],"class_list":["post-1500","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8th-class-ncert-syllabus","category-posts-in-english","category-mathematics","tag-8th-std-maths-ncert-square","tag-8th-std-maths-square-and-square-roots"],"cp_meta_data":{"_edit_lock":["1629955864:2"],"_edit_last":["2"],"_layout":["inherit"],"_thumbnail_id":["1514"],"_oembed_95287caaddeb112cd4edfcbd8e525566":["