{"id":43256,"date":"2023-04-10T14:19:58","date_gmt":"2023-04-10T08:49:58","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=43256"},"modified":"2023-04-11T10:16:54","modified_gmt":"2023-04-11T04:46:54","slug":"ml-aggarwal-class-7-icse-maths-model-question-paper-4","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-7-icse-maths-model-question-paper-4\/","title":{"rendered":"ML Aggarwal Class 7 ICSE Maths Model Question Paper 4"},"content":{"rendered":"
(Based on Chapters 10 to 12) General Instructions<\/strong><\/p>\n Choose the correct answer from the given four options (1-2): Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Solution: ML Aggarwal Class 7 ICSE Maths Model Question Paper 4 (Based on Chapters 10 to 12) Time allowed: 1 Hour Maximum Marks: 25 General Instructions Questions 1-2 carry 1 mark each. Questions 3-5 carry 2 marks each. Questions 6-8 cany 3 marks each Questions 9-10 carry 4 marks each. Choose the correct answer from the …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43256"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=43256"}],"version-history":[{"count":1,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43256\/revisions"}],"predecessor-version":[{"id":159045,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/43256\/revisions\/159045"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=43256"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=43256"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=43256"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nTime allowed: 1 Hour<\/strong>
\nMaximum Marks: 25<\/strong><\/p>\n\n
\nQuestion 1.
\nIn the given figure, if l || m then the value of x is
\n(a) 65\u00b0
\n(b) 105\u00b0
\n(c) 115\u00b0
\n(d) 125\u00b0
\n
\nSolution:
\nIn the given figure, if l || m
\n
\n\u22201 = 65\u00b0 (Vertically opposite angles)
\nBut \u22201 + x = 180\u00b0 (Co-interior angles)
\n\u21d2 65\u00b0 + x = 180\u00b0
\n\u21d2 x = 180\u00b0 – 65\u00b0 = 115\u00b0 (c)<\/p>\n
\nA triangle whose two angles measure 40\u00b0 and 100\u00b0 is
\n(a) scalene
\n(b) isosceles
\n(c) equilateral
\n(d) right-angled
\nSolution:
\nIn a triangle, two angles are 40\u00b0 and 100\u00b0
\nThird angle = 180\u00b0 – (40\u00b0 + 100\u00b0) = 180\u00b0 – 140\u00b0 = 40\u00b0
\nThe triangles is an isosceles as two angles are equal,
\nthen their opposite sides are equal. (b)<\/p>\n
\nIn the given figure, lines l and m intersect at O. If \u22201 + \u22203 = 222\u00b0, then find the measure of \u22202.
\n
\nSolution:
\nIn the given figure, two lines l and m intersect each other at O.
\n\u22201 + \u22203 = 222\u00b0
\n\u21d2 \u22201 = \u22203 (Vertically opposite angles)
\n\u21d2 \u22201 = \u22203 = \\(\\frac { 222 }{ 2 }\\) = 111\u00b0
\nBut \u22201 + \u22202 = 180\u00b0 (Linear pair)
\n\u21d2 111\u00b0 + \u22202 = 180\u00b0
\n\u21d2 \u22202 = 180\u00b0 – 111\u00b0
\n\u21d2 \u22202 = 69\u00b0<\/p>\n
\nIn the given figure, AB = AC. Find the value of x.
\n
\nSolution:
\nIn the given figure,
\nAB = AC
\n\u2220B = \u2220C (Angles opposite to equal sides)
\nBut \u2220A = 70\u00b0
\n\u2220B + \u2220C = 180\u00b0 – 70\u00b0 = 110\u00b0
\n\u21d2 \u2220B = \\(\\frac { 110 }{ 2 }\\)
\n\u21d2 \u2220B = x = 55\u00b0<\/p>\n
\nIf \u2206PQR = \u2206EFD, write the parts of \u2206EFD that correspond to:
\n(i) \u2220Q
\n(ii) \\(\\bar { PR }\\)
\n(iii) \u2220P
\n(iv) \\(\\bar { QR }\\)
\nSolution:
\n\u2206PQR = \u2206EFD, then corresponding parts are
\n(i) \u2220Q = \u2220F
\n(ii) \\(\\bar { PR }\\) = ED
\n(iii) \u2220P = \u2220E
\n(iv) \\(\\bar { QR }\\) = FD<\/p>\n
\nIn the given figure, l || m and p || q. Find the value of x, y, and z.
\n
\nSolution:
\nIn the given figure,
\n
\nl || m and p || q
\nx = 70\u00b0 (Corresponding angles)
\n\u22201 = x = 75\u00b0 (Corresponding angles)
\nBut \u22201 + y = 180\u00b0 (Co-interior angles)
\ny = 180\u00b0 – \u22201 = 180\u00b0 – 75\u00b0 = 105\u00b0
\nz = \u22201 (Vertically opposite angles)
\nz = 75\u00b0
\nHence, x = 75\u00b0, y = 105\u00b0, z = 75\u00b0<\/p>\n
\nIn the given figure, AB = AC. Find the values of x and y.
\n
\nSolution:
\nIn the given figure,
\n
\nAB = AC
\n\u2220EAF = 80\u00b0
\n\u2220BAC = \u2220EAF (Vertically opposite angles)
\n\u22201 = 80\u00b0
\n\u2220B + \u2220C = 180\u00b0 – \u2220A = 180\u00b0 – 80\u00b0 = 100\u00b0
\nBut \u2220B = \u2220C (Angles opposite to equal sides)
\n\u2220B = \u2220C = \\(\\frac { 100 }{ 2 }\\) = 50\u00b0
\nx = 50\u00b0
\nNow Ext. \u2220ACD = \u2220B + \u2220BAC = x + \u22201
\ny = 50\u00b0 + 80\u00b0 = 130\u00b0
\nx = 50\u00b0, y = 130\u00b0<\/p>\n
\nIf the lengths of two sides of a triangle are 5 cm and 7.5 cm then what can be the length of the third?
\nSolution:
\nLenghts of two sides of a triangle are 5 cm, 7.5 cm
\nThen length of third side < (5 + 7.5) = 12.5 cm
\nor greater than 7.5 – 5.0 = 2.5 cm<\/p>\n
\nAn apple orchard is in the shape of a rectangle. If its length is 60 m and the length of one diagonal is 75 m, then find:
\n(i) the breadth of the orchard.
\n(ii) the perimeter of the orchard.
\nSolution:
\nAn apple orchard is in shape of a rectangle Length = 60 cm
\nOne diagonal = 75 m
\n
\nand perimeter = 2(Length + Breadth)
\n= 2(60 + 45) m = 2 \u00d7 105 = 210 m
\nApples keep our body hale and healthy.<\/p>\n
\nIn the given figure, measures of some parts are given.
\n(i) State the three pairs of equal parts in \u2206ABD and \u2206ACD.
\n(ii) Is \u2206ABD = \u2206ACD? Give reason.
\n(iii) Is D mid-point of BC? Why?
\n<\/p>\n
\nIn the given figure,
\nIn right \u2206ABD and \u2206ACD
\nSide AD = AD (Common)
\nHypotenuse, AB = AC (Each = 3.6 cm)
\n\u2206ABD = \u2206ACD (RHS criterion)
\nBD = DC (c.p.c.t.)
\nD is mid-point of BC.<\/p>\nML Aggarwal Class 7 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"