{"id":17511,"date":"2020-01-03T12:30:54","date_gmt":"2020-01-03T07:00:54","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17511"},"modified":"2021-07-02T15:31:27","modified_gmt":"2021-07-02T10:01:27","slug":"kseeb-solutions-for-class-6-maths-chapter-10-ex-10-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-6-maths-chapter-10-ex-10-3\/","title":{"rendered":"KSEEB Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3"},"content":{"rendered":"

Students can Download Chapter 10 Mensuration Ex 10.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 6 Maths Chapter 10 Mensuration Ex 10.3<\/h2>\n

Question 1.
\nFind the areas of the rectangles whose sides are:
\na) 3 cm and 4 cm
\nb) 12 m and 21 m
\nc) 2 km and 3 km
\nd) 2m and 70cm.
\nSolution:
\nIt is known that
\nArea of rectangle = length \u00d7 Breadth
\na) 1 = 3 cm, b = 4 cm Area = l \u00d7 b = 3 \u00d7 4 = 12 cm2<\/sup>
\nb) 1 = 12 m, b = 21 m
\nArea = l \u00d7 b = 12 \u00d7 21 = 252 m2<\/sup>
\nc) l = 2 km, b = 3 km
\nArea = l \u00d7 b = 2 \u00d7 3 = 6km2<\/sup>
\nd) l = 2 m , b = 70 cm = 0.70 m
\nArea = l \u00d7 b = 2 \u00d7 0.70 = 1.40 m2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the areas of the squares whose sides are:-
\na) 10 cm
\nb) 14 cm
\nc) 5 m
\nSolution:
\nIt\u2019s known that, Area o f square = side \u00d7 side
\na) 10cm
\nside = 10 cm
\nArea of the square =10 \u00d7 10= 100 cm2<\/sup><\/p>\n

b) 14 cm
\nside = 14 cm
\nArea of the square = 14 \u00d7 14 = 196 cm2<\/sup><\/p>\n

c) 5m
\nSide = 5 m
\nArea of the square = 5 \u00d7 5 = 25 m2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nThe length and breadth of three rectangles are as given below :
\na) 9 m and 6 m
\nb) 17 m and 3 m
\nd) 4 m and 14 m
\nWhich one has the largest area and which one has the smallest?
\nSolution:
\nIt is known that
\nArea of rectangle = length \u00d7 Breadth
\na) l = 9 m b = 6 m
\nArea = l \u00d7 b = 9 \u00d7 6 = 54 m2<\/sup>
\nb) l = 17 m b = 3 m
\nArea = l \u00d7 b = 17 \u00d7 3 = 51 m2<\/sup>
\nc) l = 4 m b = 14 m
\nArea = l \u00d7 b = 4 \u00d7 14 = 56 m2<\/sup>
\nIt can be see that rectangle (c) has the largest area and rectangle (b) has the smallest area.<\/p>\n

Question 4.
\nThe areas of a rectangular garden 50m long is 300 sq m. Find the width of the garden.
\nSolution:
\nGiven:-
\nLength of the garden = 50 m = l
\nand Areas of the garden = 300 sq m = A
\nWidth or breadth = ? = b
\nA = l \u00d7 b
\n300 = 50 \u00d7 b
\n\"KSEEB
\nb = 6 m
\nThe breadth or width of the garden = 6 m<\/p>\n

Question 5.
\nWhat is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?
\nSolution:
\nArea of rectangular plot = 500 \u00d7 200 = 100000 m2<\/sup>
\ncost of tiling per 100 m2<\/sup> = Rs. 8
\n\"KSEEB<\/p>\n

Question 6.
\nA table – top measures 2 m by 1 m 50 cm. What is its area in square metres?
\nSolution:
\n\"KSEEB
\nAreas = 1 \u00d7 6 = 2 \u00d7 1.5 = 3m2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nA room is 4 m long and 3 m 50 cm wide . How many square metres of carpet is needed to cover the floor of the room?
\nSolution:
\nLength (l) = 4 m,
\nBreadth (b) = 3 m 50 cm = 3.5 m
\nArea = l \u00d7 b = 4 \u00d7 3.5 = 14 m2<\/sup><\/p>\n

Question 8.
\nA floor is 5 m long and 4 m wide. A square carpet of sides 3m is laid on the floor. Find the area of the floor that is not carpeted.
\nSolution:
\nLength (l) = 5 m,
\nBreadth (b) = 4 m
\nArea = l \u00d7 b = 5 \u00d7 4 = 20 m2<\/sup>
\nArea covered by the carpet = (side)2<\/sup> = (3)2<\/sup> = 9 m2<\/sup>
\nArea not covered by the carpet = 20 – 9 = 11 m2<\/sup><\/p>\n

Question 9.
\nFive square flower beds each of sides 1 m are dug on a piece of land 5m long and 4m wide. What is the area of the remaining part of the land?
\nSolution:
\nArea of the land = 5 \u00d7 4 = 20 m2<\/sup>
\nArea occupied by 5 flower beds = 5 \u00d7 (side)2<\/sup> = 5 \u00d7 (1 )2<\/sup> = 5m2<\/sup>
\nAreas of the remaining part = 20 – 5 = 15 m2<\/sup><\/p>\n

\"KSEEB<\/p>\n

Question 10.
\nBy splitting the following figures into rectangles, find their areas ( The measures are given in centimetres).
\n\"KSEEB
\nSolution :
\n(a) The given figure can be broken into rectangles as follows.
\n\"KSEEB
\nArea of 1st<\/sup> rectangle = 4 \u00d7 2 = 8 cm2<\/sup>
\nArea of 2nd<\/sup> rectangle = 6 \u00d7 1 = 6 cm2<\/sup>
\nArea of 3rd<\/sup> rectangle = 3 \u00d7 2 = 6 cm2<\/sup>
\nArea of 4th<\/sup> rectangle = 4 \u00d7 2 = 8 cm2<\/sup>
\nTotal area of the complete figure = 8 + 6 + 6 + 8 = 28 cm2<\/sup><\/p>\n

(b) The given figure can be broken into rectangles as follows.
\n\"KSEEB
\nArea of 1st<\/sup> rectangle = 3 \u00d7 1 = 3 cm2<\/sup>
\nArea of 2nd<\/sup> rectangle = 3 \u00d7 1 = 3 cm2<\/sup>
\nArea of 3rd<\/sup> rectangle = 3 \u00d7 1 = 3 cm2<\/sup>
\nTotal area of the complete figure = 3 + 3 + 3 = 9 cm2<\/sup><\/p>\n

Question 11.
\nSplit the following shapes into rectangles and find their areas.( The measures are given in centimetres)
\n\"KSEEB
\nSolution:
\n(a) The given figure can be broken into rectangles as follows.
\n\"KSEEB
\nArea of 1st<\/sup> rectangle = 12 \u00d7 2 = 24 cm
\nArea of 2nd<\/sup> rectangle = 8 \u00d7 2 = 16 cm
\nTotal area of the complete figure = 24 + 16 = 40 cm<\/p>\n

(b) The given figure can be broken into rectangles as follows.
\n\"KSEEB
\nArea of 1st<\/sup> rectangle = 21 \u00d7 7 = 147 cm
\nArea of 2nd<\/sup> square = 7 \u00d7 7 = 49 cm
\nArea of 3rd<\/sup> square = 7 \u00d7 7 = 49 cm
\nTotal area of the complete figure = 147 + 49 + 49 = 245 cm<\/p>\n

\"KSEEB<\/p>\n

(c) The given figure can be broken into rectangles as follows:
\n\"KSEEB
\nArea of 1st<\/sup> rectangle = 5 \u00d7 1 = 5 cm
\nArea of 2nd<\/sup> rectangle = 4 \u00d7 1 = 4 cm
\nTotal area of the complete figure = 5 + 4 = 9 cm<\/p>\n

\"KSEEB<\/p>\n

Question 12.
\nHow many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
\na) 100 cm and 144 cm
\nb) 70 cm and 36 cm.
\nSolution:
\n(a) Total area of the region = 100 \u00d7 144 = 14400 cm2<\/sup>
\nArea of one tile = 12 \u00d7 5 = 60 cm2<\/sup>
\nimg class=”alignnone size-full wp-image-45590″ src=”https:\/\/kseebsolutions.guru\/wp-content\/uploads\/2020\/10\/KSEEB-Solutions-for-Class-6-Maths-Chapter-10-Mensuration-Ex-10.3-60.png” alt=”KSEEB Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 60″ width=”433″ height=”57″ \/>
\nTherefore, 240 tiles are required.<\/p>\n

(b) Total area of the region = 70 \u00d7 36 = 2520 cm2<\/sup>
\nArea of one tile = 60 cm2<\/sup>
\n\"KSEEB
\nTherefore, 42 tiles are required.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Chapter 10 Mensuration Ex 10.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 6 Maths Chapter 10 Mensuration Ex 10.3 Question 1. Find the areas of the rectangles …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[83],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17511"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=17511"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17511\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17511"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17511"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17511"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}