{"id":4657,"date":"2020-09-09T10:17:19","date_gmt":"2020-09-09T04:47:19","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=4657"},"modified":"2021-07-02T15:03:35","modified_gmt":"2021-07-02T09:33:35","slug":"kseeb-solutions-class-8-maths-chapter-15-ex-15-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-8-maths-chapter-15-ex-15-4\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4"},"content":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4<\/h2>\n

Question 1.
\nThe sides of a rectangle are in the ratio 2: 1. The perimeter is 30 cm. Calculate the measure of all the sides.
\nAnswer:
\nThe ratio of the sides is 2 : 1
\nLet the sides be 2x and x
\nPerimeter = 30
\n2x + x + 2x + x = 30
\n6x = 30
\nx =\u00a0\\(\\frac{3}{6}\\)\u00a0= 5 cm
\n2x = 2 x 5 = 10cm
\n\u2234 The sides aer 10 cm, 5 cm, 10 cm, 5 cm.<\/p>\n

Question 2.
\nIn the adjacent rectangle ABCD \u2220OCD = 30\u00b0 . Calculate \u2220BOC. What type of triangle is BOC.
\n\"KSEEB
\nAnswer:
\n\u2220BCD = 90\u00b0 [Angle of a rectangle]
\n\u2220OCD + \u2220OCB = 90\u00b0
\n30 + \u2220QCB = 90\u00b0
\n\u2220OCB = 90 \u2013 30
\n\u2220OCB = 60\u00b0
\n\u2220OCB = \u2220OBC = 60\u00b0
\nOC = OB [Diagonals of a rectangle bisect each other]
\n\u2234 \u2220BOC = 180 – (60 + 60) = 180 – 120
\n\u2220BOC = 60\u00b0
\nIn \u0394 BOC, \u2220BOC = \u2220OBC = \u2220OCB = 60\u00b0 & equiangular triangle
\n\u2234 BOC is an equiangular triangle.<\/p>\n

Question 3.
\nAll the rectangles are parallelograms but all the parallelograms are not rectangles. Justify this statement.
\nAnswer:
\nA rectangle has all the properties of a parallelogram, therefore, all rectangles are parallelograms. No angle of a Parallelogram is a right angle but all angles of a rectangle are right angles.
\n\u2234 All parallelograms are not rectangular.<\/p>\n

Question 4.
\nThe side of a rectangular park are in the ratio 4 : 3. If the area is 1728 m find the cost of fencing it at the rate of Rs . 2.50\/m
\nAnswer:
\nThe ratio of the sides = 4 : 3
\nLet the sides be 4x and 3x
\nArea = 1728
\nl \u00d7 b = 1728
\n4x \u00d7 3x = 1728
\n12x2<\/sup> = 1728
\nx2<\/sup> = \\(\\frac{12}{1728}\\)
\nx2<\/sup> = 144
\nx = \u221a144 = 12m
\n4x = 4 \u00d7 12 = 48m
\n3x = 3x \u00d7 12 = 36m
\nPerimeter = 4x + 3x + 4x + 3x
\n= 48 + 36 + 48 + 36
\n= 168m
\nThe cost of fencing 1m is Rs 2.50
\nThe cost of fencing 168m = 168 \u00d7 250 = Rs. 420.<\/p>\n

Question 5.
\nA rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remaining portion is a yard of a trapezoidal shape whose parallel sides have length 15m and 25m. What fraction of the yard is occupied by the flower bed?
\n\"KSEEB
\nAnswer:
\nIn the fig AB = 25 m and FE = 15 m
\n\u2234 DF + EC = 25 \u2013 15 = 10m
\nBut DF = EC
\n\u2234DF = EC = 5m also AD = BC = 5m
\nArea of rectangle \u2206BCD = AB x BC = 1 x b = 25 x 5 = 125 m2<\/sup>
\nArea of \u2206ADF = \\(\\frac{1}{2}\\)\u00a0base \u00d7 height
\n= \\(\\frac{1}{2}\\) 5 \u00d7 5 = \\(\\frac{25}{2}\\)m2<\/sup>
\n\u2206ADF \u2245 ABCE [data]
\nArea of \u2206BCE = \\(\\frac{25}{2}\\) m2<\/sup>
\nSum of Area of the two triangles
\n\u2234 \\(\\frac{1}{5}\\)\u00a0of the field is occupied by the flower bed.<\/p>\n

Question 6.
\nIn a rhombus ABCD \u00a0\u2220C = 70\u00b0. Find the other angles of the rhombus.
\nAnswer:
\n\u2220A = \u2220C = 70\u00b0 [Opposite angles]
\n\u2220A +\u2220B = 180\u00b0 [Adjacent angles of a rhombus]
\n70 + \u2220B = 180\u00b0
\n\u2220B = 180 \u2013 70
\n\u2220B = 110\u00b0
\n\u2220B = \u2220D = 110\u00b0 [Opposite angles are equal].
\nAngles are 70\u00b0, 110\u00b0, 70\u00b0 & 110\u00b0<\/p>\n

Question 7.
\nIn a rhombus PQRS, if PQ = 3x – 7 and QR = x + 3 find PS.
\n\"KSEEB
\nAnswer:
\nPQ = QR [Sides of a rhombus]
\n3x \u2013 7 = x + 3
\n3x \u2013 x = 3 + 7
\n2x = 10
\n\\(\\frac{10}{2}\\)
\nx = 5
\nPQ = 3x \u2013 7 = 3 \u00d7 5 \u2013 7 = 15 \u2013 7
\nPQ = 8 cm
\n\u2234 SP = 8<\/p>\n

Question 8.
\nRhombus is a parallelogram. Justify.
\nAnswer:
\nIn a rhombus opposite sides are equal and parallel to each other therefore it is a parallelogram.<\/p>\n

Question 9.
\nIn a given square ABCD, if the area of triangle ABD is 36cm2 Find
\n1. The area of the triangle BCD and
\n2. Area of the square ABCD
\nAnswer:
\nThe diagonal of a square divides the square into two congruent triangles.
\n\"KSEEB
\n\u2234 Area of \u2206ABD = \u2206BCD = 36cm\u00b2
\nArea of square ABCD = 36 + 36 = 72 cm\u00b2.<\/p>\n

Question 10.
\nThe side of square ABCD is 5cm and another square PQRS has a perimeter equal to 40cm. Find the ratio of perimeter ABCD to the perimeter of PQRS. Find the ratio of the area of ABCD to the area of PQRS.
\nAnswer:
\nSide of ABCD = 5 cm
\nPerimeter of ABCD = 5 \u00d7 4 = 20 cm
\nPerimeter of PQRS = 40 cm 4 \u00d7 side = 40
\nside = \\(\\frac { 40 }{ 4 }\\)
\nside = 10cm
\nPerimeter of ABCD : Perimeter PQRS
\n= 20 : 40 = 1 : 2
\nArea of ABCD:Area of PQRS
\n= 52 : 102 = 25 : 100 = 1 : 4<\/p>\n

Question 11.
\nA square field has a side of 20m. Find the length of the wire required to fence it four times.
\nAnswer:
\nPerimeter = 4a = 4 \u00d7 20 = 80m
\nLength of the wire required to fence 4 times. = 4 \u00d7 80 = 320 m.<\/p>\n

Question 12.
\nList out the difference between square and rhombus
\nAnswer:<\/p>\n\n\n\n\n\n\n
Square<\/td>\nRhombus<\/td>\n<\/tr>\n
i. All the angles are equal.<\/td>\ni. Opposite angles are equal<\/td>\n<\/tr>\n
ii. All the angles are right angles<\/td>\nii. No angle is a right angle<\/td>\n<\/tr>\n
iii. Diagonals are equal<\/td>\niii. Diagonals are not equal.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

<\/h4>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 15 Quadrilaterals Ex 15.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.4 Question 1. The sides of a rectangle are …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4657"}],"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=4657"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/4657\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=4657"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=4657"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=4657"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}