\nTime allowed: 2\\(\\frac { 1 }{ 2 }\\) Hours<\/strong>
\nMaximum Marks: 90<\/strong><\/p>\n
General Instructions<\/strong><\/p>\n Section – A<\/strong><\/p>\n Question numbers 1 to 8 is of 1 mark each. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Section – B<\/strong><\/p>\n Question numbers 9 to 14 are of 2 marks each. Question 10. Question 11. Question 12. Question 13. Question 14. Section – C<\/strong><\/p>\n Question numbers 15 to 24 are of 4 marks each. Question 16. Question 17. Question 18. Question 19. Question 20. Question 21. Question 22. Question 23. Question 24. Section – D<\/strong><\/p>\n Question numbers 25 to 29 are of 6 marks each. Question 26. Question 27. Question 28. Question 29.\n
\nQuestion 1.
\nIn the given figure, if \u2220AOC and \u2220COB form a linear pair, then the value of x
\n(a) 60
\n(b) 55
\n(c) 50
\n(d) 45
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q1.1.png?resize=214%2C163&ssl=1\")
\nSolution:
\n\u2220AOC and \u2220COB = 180\u00b0 (Linear pair)
\n\u21d2 2x + x + 15 = 180\u00b0
\n\u21d2 3x = 180\u00b0- 15\u00b0= 165\u00b0
\n\u21d2 x = 55\u00b0 (b)<\/p>\n
\nAn exterior angle of a triangle is 118\u00b0. If one of the two interior opposite angle is 54\u00b0, then the other interior opposite angle is
\n(a) 62\u00b0
\n(b) 54\u00b0
\n(c) 64\u00b0
\n(d) 59\u00b0
\nSolution:
\nOne exterior angle of a triangle = 118\u00b0
\nOne of interior opposite two angles = 54\u00b0
\nSecond inner angle = 118\u00b0 – 54\u00b0 = 64\u00b0 (c)<\/p>\n
\nIn a right-angled triangle, the lengths of two legs are 8 cm and 15 cm. The length of the hypotenuse is
\n(a) 23 cm
\n(b) 20 cm
\n(c) 17 cm
\n(d) 17 m
\nSolution:
\nIn a right angled triangle
\nTwo legs are 8 cm and 15 cm
\nHypotenuse = \\(\\sqrt { { 8 }^{ 2 }+{ 15 }^{ 2 } }\\)
\n= \u221a(64 + 225) = \u221a289 = 17 cm (c)<\/p>\n
\nIf \u0394ABC = \u0394PRQ, the the correct statement is
\n(a) AB = PQ
\n(b) \u2220B = \u2220Q
\n(c) \u2220C = \u2220R
\n(d) AC = PQ
\nSolution:
\n\u2206ABC = \u2206PRQ
\nAC = PQ is correct (d)<\/p>\n
\nThe number of lines that can be drawn parallel to a given line l through a point outside the line l is
\n(a) 0
\n(b) 1
\n(c) 2
\n(d) infinitely many
\nSolution:
\nNumber of lines drawn from a point parallel to a given line is only one. (b)<\/p>\n
\nIf the area of a circle is numerically equal to its circumference, then the diameter of the circle is
\n(a) 2 units
\n(b) 4 units
\n(c) 6 units
\n(d) 8 units
\nSolution:
\nArea of circle = Circumference (numerical)
\n\u03c0r2<\/sup> = \u03c0d = 2\u03c0r \u21d2 r = 2
\nd = 2r = 4 units (b)<\/p>\n
\nA quadrilateral having exactly two lines of symmetry and rotational symmetry of order 2 is a
\n(a) square
\n(b) parallelogram
\n(c) rhombus
\n(d) kite
\nSolution:
\nA quadrilateral has two lines of symmetry
\nand rotational symmetry of order 2 is a rhombus. (c)<\/p>\n
\nA mode is the observation of the data
\n(a) whose position is in the middle
\n(b) having maximum value
\n(c) occurring a maximum number of times
\n(d) occurring a minimum number of times
\nSolution:
\nMode is the observation of the data
\noccurring maximum number of times. (c)<\/p>\n
\nQuestion 9.
\nAn angle is 30\u00b0 more than one-half of its complement. Find the angle.
\nSolution:
\nLet required angle = x
\nIts complement angle = 90\u00b0 – x
\nAccording to the condition,
\nx = \\(\\frac { 90-x }{ 2 }\\) + 30
\n\u21d2 2x = 90 – x + 60
\n\u21d2 2x + x = 150
\n\u21d2 3x = 150 150\u00b0
\n\u21d2 x = 50\u00b0
\nHence required angle = 50\u00b0<\/p>\n
\nIn the given figure, find the value of x.
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q10.1.png?resize=219%2C160&ssl=1\")
\nSolution:
\nIn the given figure,
\n\u2220BAC = \u2220EAF (Vertically opposite angles) = 70\u00b0
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q10.2.png?resize=234%2C182&ssl=1\")
\nBut Ext. \u2220ACD = \u2220BAC + \u2220B
\n125\u00b0 = 70\u00b0 + x
\n\u21d2 x = 125\u00b0 – 70\u00b0 = 55\u00b0
\nx = 55\u00b0<\/p>\n
\nA die is thrown once. Find the probability of getting
\n(i) a prime number
\n(ii) a composite number
\nSolution:
\nA die is thrown
\nTotal number of outcomes = 6
\n(i) Probability of a prime number = (2, 3, 5) = \\(\\frac { 3 }{ 6 }\\) = \\(\\frac { 1 }{ 2 }\\)
\n(ii) Probability of a composite number= (4, 6) = \\(\\frac { 2 }{ 6 }\\) = \\(\\frac { 1 }{ 3 }\\)<\/p>\n
\nYou want to show that \u0394DEF = \u0394PQR by SAS congruence rule.
\nIt is given that \u2220E = \u2220Q, you need to have
\n(i) EF = …….
\n(ii) PQ = ………
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q12.1.png?resize=305%2C143&ssl=1\")
\nSolution:
\n\u2206DEF = \u2206PQR (SAS criterion)
\n\u2220E = \u2220Q
\nEF = QR
\nPQ = DE
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q12.2.png?resize=312%2C141&ssl=1\")
<\/p>\n
\nIn the given figure, two adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 8 cm, find the
\n(i) area of parallelogram
\n(ii) the distance between the shorter sides.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q13.1.png?resize=264%2C153&ssl=1\")
\nSolution:
\nIn ||gm ABCD
\nAB = 15 cm, BC = 10 cm
\nPerpendicular DE = 8 cm and DF = ?
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q13.2.png?resize=269%2C154&ssl=1\")
\nNow area of ||gm ABCD = Base \u00d7 Height
\n= AB \u00d7 DE = 15 \u00d7 8 cm2\u00a0<\/sup>= 120 cm2<\/sup>
\nIf base BC = 10 cm
\nThen height DF = \\(\\frac { Area }{ Base }\\) = \\(\\frac { 120 }{ 10 }\\) = 12 cm<\/p>\n
\nFind the mean age of six students whose ages (in years) are:
\n15, 13, 16, 13, 14, 16.
\nSolution:
\nMean age of 15, 13, 16, 13, 14, 16
\n= \\(\\frac { 15+13+16+13+14+16 }{ 6 }\\)
\n= \\(\\frac { 87 }{ 6 }\\)
\n= 14.5 years<\/p>\n
\nQuestion 15.
\nIn the given figure, lines l and m are parallel. Find the values of x, y, and z.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q15.1.png?resize=217%2C148&ssl=1\")
\nSolution:
\nIn the given l || m
\nx = 65\u00b0 (Alternate Angles)
\nSimilarly z = y
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q15.2.png?resize=229%2C169&ssl=1\")
\nBut x + 45\u00b0 + z = 180\u00b0 (Angle on one side of a line)
\n\u21d2 65\u00b0 + 45\u00b0 + z = 180\u00b0
\n\u21d2 110\u00b0 + z = 180\u00b0
\n\u21d2 z = 180\u00b0- 110\u00b0 = 70\u00b0
\ny = z = 70\u00b0
\nx = 65\u00b0, y = 70\u00b0, z = 70\u00b0<\/p>\n
\nIn the given figure, BC = AC. Find the value of x.
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q16.1.png?resize=247%2C118&ssl=1\")
\nSolution:
\nIn the given figure,
\nBC = AC, \u2220C = x, \u2220ABD = 155\u00b0
\n\u2220CAB = \u2220CBA
\nBut \u2220ABC + \u2220ABD = 180\u00b0 (Linear pair)
\n\u21d2\u2220ABC + 155\u00b0= 180\u00b0
\n\u21d2 \u2220ABC = 180\u00b0 – 155\u00b0 = 25\u00b0
\n\u21d2 \u2220CAB = \u2220ABC = 25\u00b0
\nBut \u2220CAB + \u2220ABC + \u2220ACB = 180\u00b0 (Angles of a triangle)
\n\u21d2 25\u00b0 + 25\u00b0 + x\u00b0 = 180\u00b0
\n\u21d2 x + 50\u00b0 = 180\u00b0
\n\u21d2 x = 180\u00b0 – 50\u00b0 = 130\u00b0<\/p>\n
\nIf the lengths of the two sides of a triangle are 6 cm and 8.5 cm, then what can be the length of the third side?
\nSolution:
\nLength of two sides of a triangle are 6 cm and 8.5 cm.
\nThird side will be less than (6 + 8.5) = 14.5 cm
\nand greater than (8.5 – 6) = 2.5 cm<\/p>\n
\nIn the given figure, AD is perpendicular bisector of \\(\\bar { BC }\\).
\n(i) State three pairs of equal parts in \u2206ABD and \u2206ACD.
\n(ii) Is \u2206ABD = \u2206ACD? Give reasons.
\n(iii) Is ABC an isosceles triangle? Give reasons.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q18.1.png?resize=148%2C186&ssl=1\")
\nSolution:
\nIn \u2206ABC, AD is perpendicular bisector of BC
\nBD = DC and \u2220ADB = \u2220ADC = 90\u00b0
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q18.2.png?resize=152%2C185&ssl=1\")
\nNow in \u2206ABD and \u2206ACD
\nBD = DC (D is the mid-point of BC)
\n\u2220ADB = \u2220ADC (Each 90\u00b0)
\nAB = AD (Common)
\n\u2206ABD = \u2206ACD (SAS criterion)
\nAB = AC (c.p.c.t.)
\n\u2206ABC is an isosceles triangle.<\/p>\n
\nDraw a line, say AB, take a point P outside line AB. Through P, draw a line parallel to line AB using ruler and compasses only.
\nSolution:
\nSteps of construction :
\n(i) Draw a line AB.
\n(ii) Take a point P outside AB.
\n(iii) Take a point Q on AB and join PQ.
\n(iv) From P, construct \u2220OPX = \u2220PQB and product XP to Y.
\nThen XY is parallel to AB.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q19.1.png?resize=243%2C161&ssl=1\")
<\/p>\n
\nIn the given figure, ABCD is a rectangle with AB = 20 cm and BC = 14 cm. Two semicircles are cut from each of two breadths as diameters. Find
\n(i) the area of the shaded region
\n(ii) the perimeter of the shaded region.
\nTake \u03c0 = \\(\\frac { 22 }{ 7 }\\).
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q20.1.png?resize=194%2C158&ssl=1\")
\nSolution:
\nIn the given figure,
\nABCD is a rectangle whose length (l) = 20 cm
\nBreadth (b) = 14 cm
\nTwo semicircles have been cut out from
\nthe breadth sides whose radius = \\(\\frac { 14 }{ 2 }\\) = 7 cm
\n(i) Area of shaded portion
\n= Area of rectangle – Area of two semicircle
\n= l \u00d7 b – 2 \u00d7 \\(\\frac { 1 }{ 2 }\\) \u03c0r2<\/sup>
\n= 20 \u00d7 14 – 2 \u00d7 \\(\\frac { 1 }{ 2 }\\) \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 7 \u00d7 7
\n= 280 – 154= 126 cm2<\/sup>
\n(ii) Perimeter of shaded portion = 2 \u00d7 l + 2\u03c0r
\n= 2 \u00d7 20 + 2 \u00d7 \\(\\frac { 22 }{ 7 }\\) \u00d7 7
\n= 40 + 44
\n= 84 cm<\/p>\n
\nDraw the reflection of the letter E in the given mirror line shown dotted.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q21.1.png?resize=139%2C150&ssl=1\")
\nSolution:
\nReflection of the letter E has been shown dotted.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q21.2.png?resize=212%2C154&ssl=1\")
<\/p>\n
\nThree cubes each with edge 2 units are placed side by side to form a cuboid. Find the dimensions of the cuboid so formed and draw an isometric sketch of this cuboid.
\nSolution:
\nThree cubes, every 2 units are placed
\ntogether side by side forming a cuboid.
\nThen length = 2 \u00d7 3 = 6 units
\nBreadth = 2 units
\nHeight = 2 units
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q22.1.png?resize=231%2C205&ssl=1\")
<\/p>\n
\nDraw two nets of a regular tetrahedron.
\nSolution:
\nTwo nets of the regular tetrahedron are given below:
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q23.1.png?resize=267%2C159&ssl=1\")
<\/p>\n
\nA boy scored the following marks in various class tests during a year, each test is marked out of 20:
\n15, 17, 16, 7, 9, 12, 14, 16, 3, 19, 12, 16
\n(i) Arrange the marks in ascending order.
\n(ii) What are his modal marks?
\n(iii) What are his median marks?
\n(iv) What are his mean marks?
\nSolution:
\nMarks scored by a boy during a year:
\n15, 17, 16, 7, 9, 12, 14, 16, 3, 19, 12, 16
\n(i) Arranging in ascending order:
\n3, 7, 9, 12, 12, 14, 15, 16, 16, 16, 17, 19
\n(ii) Model marks are 16
\n(iii) Median: Here n = 12 which is even
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q24.1.png?resize=334%2C136&ssl=1\")
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q24.2.png?resize=464%2C241&ssl=1\")
<\/p>\n
\nQuestion 25.
\nIn the given figure, all measurements are in centimeters. If AD is perpendicular to BC, find the length of B.
\n![\"ML](\"https:\/\/i2.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q25.1.png?resize=215%2C169&ssl=1\")
\nSolution:
\nIn the figure, AC = 25, BC = 28,
\nAD \u22a5 BC and AD = 15
\nNow in right \u2206ADC
\nAC2<\/sup> = AD2<\/sup> + DC2<\/sup> (Pythagoras Theorem)
\n\u21d2 (25)2<\/sup> = 152<\/sup> + DC2<\/sup>
\n\u21d2 625 = 225 + DC2<\/sup>
\n\u21d2 DC2<\/sup> = 625 – 225 = 400
\n\u21d2 DC2<\/sup> = (20)2<\/sup>
\n\u21d2 DC = 20 units
\nBC = 28 units
\nBD = BC – CD = 28 – 20 = 8 units
\nNow in right \u2206ABD
\nAB2<\/sup> = AD2<\/sup> + BD2<\/sup> = 152<\/sup> + 82<\/sup> = 225 + 64 = 289 = (17)2<\/sup>
\n\u21d2 AB = 17 units<\/p>\n
\nIn the adjoining figure, show that \u0394ABC = \u0394DBC. Hence, find the values of x and y.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q26.1.png?resize=173%2C181&ssl=1\")
\nSolution:
\nIn the given figure
\n\u2206ABC = \u2206DBC
\n\u2220ABC = \u2220DBC
\n\u21d2 2y +6 = 56\u00b0
\n\u21d2 2y = 56\u00b0 – 6\u00b0 = 50\u00b0
\ny = \\(\\frac { 50 }{ 2 }\\) = 25\u00b0
\nand \u2220ABC = 56\u00b0
\n\u2220ACB = \u2220BCD = 40\u00b0
\nBut in \u2206DBC
\n\u2220CBD + \u2220BCD + \u2220BDC = 180\u00b0 (Angle of a triangle)
\n56\u00b0 + x + 40\u00b0 = 180\u00b0
\n\u21d2 x + 96\u00b0 = 180\u00b0
\n\u21d2 x = 180\u00b0 – 96\u00b0 = 84\u00b0
\nx = 84\u00b0, y = 25\u00b0<\/p>\n
\nBy using ruler and compasses only, construct a triangle ABC with BC = 7.5 cm, \u2220B = 60\u00b0 and \u2220A = 90\u00b0.
\nSolution:
\nSteps of construction:
\nIn \u2206ABC
\n\u2220B = 60\u00b0, \u2220A = 90\u00b0;
\nthen \u2220C = 180\u00b0 – (60\u00b0 + 90\u00b0)
\n\u21d2\u2220C = 180\u00b0 – 150\u00b0 = 30\u00b0
\n(i) Take a line segment BC = 7.5 cm
\n(ii) At B, draw a ray BX making an angle of 60\u00b0 and at C,
\ndraw a ray CY making an angle of 30\u00b0
\nwhich intersect each other at A.
\n\u2206ABC is the required triangle whose \u2220A will be 90\u00b0.
\n![\"ML](\"https:\/\/i0.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q27.1.png?resize=225%2C207&ssl=1\")
<\/p>\n
\nAnjali took a wire of length 88 cm and bent it into the shape of a circle. Find the area enclosed by that circle. If the same wire is bent in the shape of a square, then find the area enclosed by that square. Which shape encloses more area and by how much?
\nTake \u03c0 = \\(\\frac { 22 }{ 7 }\\)
\nSolution:
\nLength of wire = 88 cm
\nBy bending it into a shape of a circle, the circumference of circle = 88 cm
\n![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q28.1.png?resize=411%2C117&ssl=1\")
\nBy bending the wire into the shape of a square.
\nPerimeter of square = 88 cm
\nThen side = \\(\\frac { Perimeter }{ 4 }\\) = \\(\\frac { 88 }{ 4 }\\) = 22 cm
\nand area = (Side)2<\/sup> = (22)2<\/sup> cm2<\/sup> = 484 cm2<\/sup>
\nWe see that area of circle is greater than that of square.
\nDifference = 616 – 484 = 132 cm2<\/sup><\/p>\n
\nGiven below is the data of school going students (boys and girls):<\/p>\n
![\"ML](\"https:\/\/i1.wp.com\/icsesolutions.com\/wp-content\/uploads\/2019\/06\/ML-Aggarwal-Class-7-ICSE-Maths-Model-Question-Paper-6-Q29.1.png?resize=310%2C337&ssl=1\")