{"id":1735,"date":"2024-02-16T12:14:26","date_gmt":"2024-02-16T06:44:26","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=1735"},"modified":"2024-02-16T16:41:41","modified_gmt":"2024-02-16T11:11:41","slug":"samacheer-kalvi-6th-maths-term-3-chapter-3-ex-3-1","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-6th-maths-term-3-chapter-3-ex-3-1\/","title":{"rendered":"Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1"},"content":{"rendered":"

You can Download Samacheer Kalvi 6th Maths Book Solutions<\/a>\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
\n\"Samacheer<\/p>\n

Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1<\/h2>\n

Question 1.
\nThe table given below contains some measures of the rectangle. Find the unknown values.
\n\"Samacheer
\nSolution:
\n(i) Area of the rectangle = (length \u00d7 breadth) sq unit.
\nPerimeter of a rectangle = 2(1 + b) units.
\nl = 5 cm
\nb = 8 cm
\n\u2234 p = 2 (l + b) cm = 2 (5 + 8) cm = 2 \u00d7 13 cm
\np = 26 cm
\nArea = (l \u00d7 b) cm2<\/sup> = (5 \u00d7 8) cm2<\/sup>
\nA = 40 cm2<\/sup><\/p>\n

(ii) l = 13 cm
\np = 54 cm
\nPerimeter = 2 (l + b) units
\n54 = 2 (13 + b) cm
\n\\(\\frac{54}{2}\\) = 13 + b
\n27 = 13 + b
\nb = 27 – 13
\nb = 14 cm
\nArea = l \u00d7 b sq. unit = 13 \u00d7 14 cm2<\/sup>
\nA = 182 cm2<\/sup><\/p>\n

(iii) b = 15 cm
\np = 60 cm
\np = 2 (l + b) units
\n60 = 2 (l + 15) cm
\n\\(\\frac{60}{2}\\) = l + 15
\n30 = l + 15
\nl = 30 – 15 .
\nl = 15 cm
\nArea = l \u00d7 b unit2<\/sup> = 15 \u00d7 15 cm2<\/sup> = 225 cm2<\/sup>
\nA = 225 cm2<\/sup><\/p>\n

(iv) l = 10 m
\nArea = 120 sq metre
\nArea = l \u00d7 b sq.m
\n120 = 10 \u00d7 6
\nb = \\(\\frac{120}{10}\\)
\nb = 12 m
\nPerimeter =2 (l + b) units = 2(10 + 12) units = 2 \u00d7 22 m
\nA = 44 m<\/p>\n

(v) b = 4 feet.
\nArea = 20 sq. feet
\nArea = l \u00d7 b sq .feet
\n20 = l \u00d7 4
\nl = \\(\\frac{20}{4}\\) feet
\nl = 5 feet
\nPerimeter = 2 (l + b) units.
\np = 2 (5 + 4) feet = 2 \u00d7 9
\np = 18 feet
\nCompleting the unknown values in the table.
\n\"Samacheer<\/p>\n

Question 2.
\nThe table given below contains some measures of the square. Find the unknown values.
\n\"Samacheer
\nSolution:
\nPerimeter of a square = (4 \u00d7 side) units
\nArea of a square = (side \u00d7 side) unit2<\/sup>
\n(i) s = 6 cm
\nPerimeter = 4s units = 4 \u00d7 6 cm = 24 cm
\nP = 24 cm
\nArea = s \u00d7 s unit2<\/sup> = 6 \u00d7 6 cm2<\/sup> = 36 cm2<\/sup>
\nA = 36 cm2<\/sup><\/p>\n

(ii) Perimeter = 4 \u00d7 s unit
\n100 = (4 \u00d7 s) m
\n\\(\\frac{100}{4}\\) = s
\ns = 25 m
\nArea = s \u00d7 s unit2<\/sup>= 25 \u00d7 25 m2<\/sup> = 625m2<\/sup>
\nA = 625m2<\/sup><\/p>\n

(iii) Area = s \u00d7 s unit2<\/sup>
\n49 = s \u00d7 s square feet
\ns2<\/sup> = 72<\/sup>
\ns = 7 feet
\nPerimeter = 4 \u00d7 s unit = 4 \u00d7 7 feet = 28 feet
\nPerimeter = 28 feet
\nCompleting the unknown values in the table
\n\"Samacheer<\/p>\n

Question 3.
\nThe table given below contains some measures of the triangle. Find the unknown values.
\n\"Samacheer
\nSolution:
\nArea of the right triangle = \\(\\frac{1}{2}\\) \u00d7 (base \u00d7 height) unit2<\/sup>
\n(i) b = 20 cm
\nh = 40 cm
\nArea = \\(\\frac{1}{2}\\) (b \u00d7 h) cm2<\/sup> = \\(\\frac{1}{2}\\) \u00d7 20 \u00d7 40 = 400 cm2<\/sup>
\nA = 400 cm2<\/sup><\/p>\n

(ii) b = 5 feet
\nArea = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h unit2<\/sup>
\n= 20 = \\(\\frac{1}{2}\\) \u00d7 5 \u00d7 h sq. feet
\n\\(\\frac{20 \\times 2}{5}\\) = h
\nh = 8 feet<\/p>\n

(iii) Area = \\(\\frac{1}{2}\\) \u00d7 (base \u00d7 height) unit2<\/sup>
\n24 = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 12 m2<\/sup>
\nbase = \\(\\frac{24 \\times 2}{12}\\) m = 4 m
\nBase = 4m
\nTabulating the unknown values
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nThe table given below contains some measures of the triangles. Find the unknown values.
\n\"Samacheer
\nSolution:
\nPerimeter of a triangle = sum of three sides.
\n(i) Perimeter = 6 + 5 + 2 cm = 13 cm
\np = 13 cm<\/p>\n

(ii) Perimeter = (side 1 + side 2 + side 3) m
\n17 = (side 1 + 8 + 3) m
\n17 m = (side 1 + 11) m
\nside 1 = 17 – 11 = 6m<\/p>\n

(iii) Perimeter = side 1 + side 2 + side 3
\n28 feet = 11 feet + side 2 + 9 feet
\n28 ft = 20 feet + side 2
\n28 – 20 = side 2
\nside = 8 feet
\nTabulating the unknowns.
\n\"Samacheer<\/p>\n

Question 5.
\nFill in the blanks.
\ni) 5 cm2<\/sup> = mm2<\/sup>
\nHint: 1 cm2<\/sup> = 100 mm2<\/sup>
\nii) 26 m2<\/sup> = cm2<\/sup>
\nHint: 1 m2<\/sup> = 10000
\niii) 8 km2<\/sup> = m2<\/sup>
\nHint 4 1 km2<\/sup>– 1000000 m2<\/sup>
\nSolution:
\n(i) 500
\n(ii) 2,60,000
\n(iii) 80,00,000<\/p>\n

Question 6.
\nFind the perimeter and area of the following shapes.
\n\"Samacheer
\nSolution:
\n(i) Perimeter = (4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4 + 4) cm = 48 cm
\nPerimeter = 48 cm
\nArea of 5 squares of side 4 cm
\nArea of a square = (side \u00d7 side) unit2<\/sup>
\n\u2234 A = 5 \u00d7 (4 \u00d7 4) cm2<\/sup> = 5 \u00d7 16 cm2 = 80 cm2<\/sup>
\n80 cm2<\/sup>
\n\"Samacheer
\n(ii) Perimeter = (4 + 5 + 4 + 5 + 4 + 5 + 4 + 5)cm = 36cm
\nPerimeter = 36 cm
\nArea of a square of side 3cm + Area of 4 right triangles
\n= (3 \u00d7 3) + [4 \u00d7 \\(\\frac{1}{2}\\) \u00d7 4 \u00d7 3] cm2<\/sup> = (9 + 24) cm2<\/sup> = 33 cm2<\/sup>
\nArea = 33 cm2<\/sup><\/p>\n

(iii) Perimeter = (50 + 12 + 13 + 40 + 10 + 10 + 10 + 5) cm = 150 cm
\nPerimeter = 150 cm
\nArea = Area of a rectangle + Area of a square + Area of a right triangle.
\n= (l \u00d7 b) + (s \u00d7 s) + ( \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h) cm2<\/sup>
\n= (50 \u00d7 5) + (10 \u00d7 10) + \\(\\frac{1}{2}\\) \u00d7 12 \u00d7 5) cm2<\/sup>
\n= (250 + 100 + 30) cm2<\/sup> = 380 cm2<\/sup>
\nArea = 380 cm2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 7.
\nFind the perimeter and area of the rectangle whose length is 6 m and breadth 4 m.
\nSolution:
\nl = 6 m, b = 4 m Perimeter of the rectangle
\n= 2 (l + b) units
\n= 2 (6 + 4) m
\n= 2 (10) m
\n= 20 m
\nArea of the rectangle = l \u00d7 b sq units
\n= 4 \u00d7 6 m\u00b2
\n= 24 m\u00b2<\/p>\n

Question 8.
\nFind the perimeter and the area of the square whose side is 8 cm.
\nSolution:
\nPerimeter of a square = (4 \u00d7 side) units
\nSide = 8 cm
\n\u2234 Perimeter = 4 \u00d7 8 cm = 32 cm
\nPerimeter = 32 cm
\nArea of a square = (side \u00d7 side) unit2<\/sup> = (8 \u00d7 8) cm2<\/sup> = 64 cm2<\/sup>
\nArea = 64 cm2<\/sup><\/p>\n

Question 9.
\nFind the perimeter and area of the right angled triangle whose sides are 6 feet, 8 feet and 10 feet.
\nSolution:
\nPerimeter of the triangle
\n= (a + b + c) units
\n= (6 + 8 + 10) feet
\n= 24 feet
\nArea of the triangle = \\(\\frac{1}{2}\\) \u00d7 b \u00d7 h sq units
\n\\(\\frac{1}{2}\\) \u00d7 6\u00b3\u00d7 8 feet square
\n= 24 sq. feet<\/p>\n

Question 10.
\nFind the perimeter of
\ni) A scalene triangle with sides 7 m, 8 m, 10 m
\nii) An isosceles triangle with equal sides 10 cm each and third side is 7 cm.
\niii) An Equilateral triangle with side 6 cm.
\nSolution:
\ni) Perimeter of a scalene triangle = (7 + 8 + 10) m = 25 m
\nii) The three sides of the isosceles triangle are 10 cm, 10 cm and 7 cm
\n\u2234 Perimeter = (10 + 10 + 7) cm = 27 cm
\niii) An equilateral triangle with side 6 cm.
\nThe sides of equilateral triangle are 6 cm, 6 cm and 6 cm
\n\u2234 Perimeter = (6 + 6 + 6) cm = 18 cm<\/p>\n

\"Samacheer<\/p>\n

Question 11.
\nThe area of a rectangular shaped photo is 820 sq. cm. and its width is 20 cm. What is its length? Also find its perimeter.
\nSolution:
\nGiven Area = 820 cm\u00b2
\nWidth = 20 cm
\nArea of the rectangle
\n= l \u00d7 b sq. units
\n820 = l \u00d7 20
\n\\(\\frac{820}{20}\\) = l
\n41 = l
\nlength l = 41 cm
\nPerimeter = 2(l + b) units
\n= 2(41 + 20) cm
\n= 2(61) cm
\n= 122 cm<\/p>\n

Question 12.
\nA square park has 40 m as its perimeter. What is the length of its side? Also find its area.
\nSolution:
\nGiven perimeter = 40 m
\nPerimeter of a square = 4 \u00d7 Length of a side
\n40 = 4 \u00d7 Length of a side
\n\u2234 Length of its side = \\(\\frac{40}{4}\\) m = 0 m
\n\u2234 Side of the park = 10m
\nArea of a square = (Side \u00d7 side) unit2<\/sup> = (10 \u00d7 10) m2<\/sup> = 100 m2<\/sup>
\n\u2234 Area of the Park = 100 m2<\/sup><\/p>\n

Question 13.
\nThe scalene triangle has 40 cm as its perimeter and whose two sides are 13 cm and 15 cm, find the third side.
\nSolution:
\nLet the third side be C
\nperimeter = (a + b + c) units
\n40 = 13 + 15 + C
\n40 = 28 + C
\nC = 40 – 28
\nC = 12 units
\nC = 12 cm<\/p>\n

Question 14.
\nA field is in the shape of right angled triangle whose base is 25 m and height 20 m. Find the cost of levelling the field at the rate of \u20b9 45\/- per sq. m.
\nSolution:
\nArea of a right angled triangle = \\(\\frac{1}{2}\\) \u00d7 (base \u00d7 height) unit2<\/sup>
\nbase = 25 m
\nheight = 20 m
\n\u2234 Area = \\(\\frac{1}{2}\\) \u00d7 (25 \u00d7 20)
\nArea = 250 m2<\/sup>
\nCost of levelling per m2<\/sup> = \u20b9 45.
\n\u2234 Cost of levelling 250 m2<\/sup> = 250 \u00d7 45 = \u20b9 11,250
\nCost of levelling = \u20b9 11,250<\/p>\n

Question 15.
\nA square of side 2 cm is joined with a rectangle of length 15 cm and breadth 10 cm. Find the perimeter of the combined shape.
\nSolution:
\nPerimeter of the combined shape = Lengths of the outer boundaries
\n= (15 + 10 + 2 + 2 + 2 + 13 + 10) cm = 54 cm
\nPerimeter = 54 cm
\n\"Samacheer<\/p>\n

Objective Type Questions<\/span><\/p>\n

Question 16.
\nThe following figures are of equal area. Which figure has the least perimeter?
\n\"Samacheer
\nSolution:
\n\"Samacheer
\nHint:
\n(a) 12 units
\n(b) 10 units
\n(c) 12 units
\n(d) 12 units<\/p>\n

Question 17.
\nIf two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
\n(a) equal to 60 cm
\n(b) less than 60 cm
\n(c) greater than 60 cm
\n(d) equal to 45 cm
\nSolution:
\n(b) less than 60 cm
\nHint:
\n\"Samacheer<\/p>\n

Question 18.
\nIf every side of a rectangle is doubled, then its area becomes times.
\n(a) 2
\n(b) 3
\n(c) 4
\n(d) 6
\nSolution:
\n(c) 4<\/p>\n

\"Samacheer<\/p>\n

Question 19.
\nThe side of a square is 10 cm. If its side is tripled, then by how many times will its perimeter increase?
\n(a) 2 times
\n(b) 4 times
\n(c) 6 times
\n(d) 3 times
\nSolutionL
\n(d) 3 times
\n30 \u00d7 4 = 120 = 3 \u00d7 40<\/p>\n

Question 20.
\nThe length and breadth of a rectangular sheet of a paper are 15 cm and 12 cm respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
\n(a) Perimeter remains the same but the area changes
\n(b) Area remains the same but the perimeter changes
\n(c) There will be a change in both area and perimeter
\n(d) Both the area and perimeter remains the same
\nSolution:
\n(c) There will be a change in both area and perimeter
\nHint:
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 6th Maths Book Solutions\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 3 Chapter 3 Perimeter and Area Ex 3.1 Question 1. The table given below contains some measures of the rectangle. Find […]<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[6],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1735"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=1735"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1735\/revisions"}],"predecessor-version":[{"id":55741,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/1735\/revisions\/55741"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=1735"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=1735"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=1735"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}