Exercise Problems With Solutions

EXERCISE 1. How many lines can be drawn through one point? (a) One (b) Infinitely many (c) No line can be drawn (d) Two Sol: We know that through one point infinitely many lines can be drawn, so (b) is the right choice. 2. How many lines can pass through two points? (a) Two (b) Infinitely many (c) One (d) Four Sol: We know that through two points one and only one line can pass, so (c) is the correct choice. 3. How many lines can be drawn through four collinear points? (a) One (b) Four (c) Six (d) No line can be drawn. Sol: When all the points are collinear, then they fall on the same straight line. So, only one line can be drawn. Hence, option (a) is the right choice.

4. The number of points of intersection of two lines is (a) Two (b) Three (c) One (d) Zero Sol: We know that two lines intersect in point only, so option (c) is the right choice. 5. The number of point(s) of intersection of four parallel lines is (a) Two (b) Four (c) One (d) Zero Sol: We know that parallel lines never intersect among themselves, so their point of intersection is zero. Hence option (d) is the right choice. 6. The meeting of two walls of a room is an example of: (a) Line (b) Ray (c) Line segment (d) None of them Sol: Walls of a room meet in a (part of) line, called line segment. So (c) is the right choice.

7. The point(s) of intersection of four concurrent lines is/are (a) 6 (b) 1 (c) 4 (d) 0 Sol: We know that concurrent lines meet in one point only, so option (b) is the right choice. 8. The number of rays drawn through a given point is: (a) No ray can be drawn through a point (b) 1 (c) Infinitely many (d) 2 Sol: We know that through one point infinitely many rays can be drawn, so option (c) is the right choice. 9. The adjacent sides of a rectangle intersect in a (a) Ray (b) Line (c) Line segment (d) Point Sol: The adjacent sides of rectangle are line segments. The intersection of two line segments is a point. So option (d) is the right choice. 10. ‘II” is the symbol of

(a) Concurrent lines (b) Intersecting lines (c) Perpendicular lines (d) Parallel lines Sol: It is a symbol of parallel lines. So option (d) is the right choice. 11. The unit of measurement of a point is (a) Kilogram (b) Metre (c) Litre (d) None of the above Sol: We know that a point is a mark of position. It has no length, no breadth and no height. It has position and location but no magnitude. Since it has no magnitude, so it has no units of its measurement. So, Option (d) is the right choice. 12. A line segment has (a) Two end points (b) One end point (c) No end point (d) None of the above Sol: A line segment has two end points. So option (a) is the right choice. 13. The end points of the same side of the polygon are known as its:

(a) Adjacent sides (b) Adjacent vertices (c) Opposite rays (d) All of them Sol: We know that the end points of the same side of a polygon are known as adjacent vertices. So option (b) is the right choice. 14. If we take a piece of paper and doodle it, then the pictures obtained as a result of doodling are called: (a) Lines (b) Rays (c) Line segments (d) Curves Sol: A doodled paper may form closed or open curves or line segments. All these are covered by a curve as a line segment is a curve. So option (d) is the right choice. 15. If the ends of a curve are joined, it is called a/an: (a) Open curve (b) Closed curve (c) Opposite rays (d) None of the above Sol: We know that a curve is said to be closed if its ends are joined, so optin (b0 is the right choice.

suur suur suur 16. Look at figure 1-41 and answer whether lines AB, QD, RF ,

(a) Intersect an another at A, O and E (b) are concurrent at A, O and E (c) are parallel to each other (d) all of them Sol: They are parallel to each other, so option (c) is the right choice. 17. There are 8 points on the page of a notebook such that no three of them are collinear. The number of lines one can get by joining them in pairs is: (a) 28 (b) 56 (c) 14 (d) 35 Sol: We know that when no three points are collinear, the number of lines one can m(m − 1) . get by joining them in a pairs is given by the formula 2 Here m = 8 , substituting the value of m = 8 in the above formula, we get =

m(m − 1) 8(8 − 1) 8 × 7 56 = = = = 28 lines 2 2 2 2

Hence, (a) is the right choice. 18. The number of diagonals of a heptagon (a closed figure bounded by seven sides) is: (a) 14 (b) 7 (c) 21 (d) 28 Sol: We know that the number of diagonals of a polygon is given by the n(n − 1) − n , where n is the number of its sides. formula 2 A heptagon has 7 sides, so n = 7 here. Substituting the value of n = 7 in the above formula, we get =

n(n − 1) 7(7 − 1) 7×6 −n = −7 = − 7 = 21 − 7 = 14 diagonals 2 2 2

Hence (a) is the right choice. 19. There are 9 points on a notebook such that 5 of them are collinear. The number of lines one can get by joining them in pairs is (a) 26 (b) 27 (c) 28 (d) 14 Sol: To obtain the number of lines passing through ' m ' points when ' n ' points of it m(m − 1) n(n − 1) are collinear, the formula used is − +1 2 2

Here m = 9 and n = 5 . Substituting these values of ' m ' and ' n ' in the above formula, we get =

m( m − 1) n(n − 1) − +1 2 2

=

9(9 − 1) 5(5 − 1) − +1 2 2

=

9 ×8 5× 4 − +1 2 2

= 36 − 10 + 1 = 27

Hence, option (b) is the right choice. 20. A polygon has 20 diagonals. The number of its sides is (a) 12 (b) 7 (c) 10 (d) 8 Sol: The number of diagonals of a polygon is given by the formula where ' n ' is the number of its sides. Since it has 20 diagonals Therefore

n(n − 1) − n = 20 2

n2 − n − n = 20 2

n(n − 1) −n, 2

n 2 − n − 2n = 20 2 n 2 − 3n = 20 2 n 2 − 3n = 40 n 2 − 3n − 40 = 0 n 2 − 8n + 5n − 40 = 0 n(n − 8) + 5(n − 8) = 0 (n − 8)(n + 5) = 0 n = 8, n = −5 Since ' n ' is the number of sides of the polygon, it cannot be negative Hence n=8 Thus, the number of sides of the polygon is 8. Now, we conclude that it is an octagon as it has 8 sides. So, option (d) is the right choice. 21. A polygon has 9 diagonals. It is a (a) Pentagon (b) Quadrilateral (c) Hexagon (d) None of the above Sol: To find the number of diagonals of a polygon, formula used is Here number of diagonals =9

n(n − 1) −n. 2

Now

n(n − 1) −n =9 2

Multiplying the whole equation by 2, we get

⎡ n(n − 1) ⎤ − n⎥ = 9 × 2 2⎢ ⎣ 2 ⎦ 2n(n − 1) − 2n = 18 2 n(n − 1) − 2n = 18 n 2 − n − 2n = 18 n 2 − 3n − 18 = 0 n 2 − 6n + 3n − 18 = 0 n(n − 6) + 3(n − 6) = 0 (n − 6)(n + 3) = 0 n = 6, n = −3 Since ' n ' is the number of sides of a polygon, it cannot be negative. So, n = 6 . Hence, it is a hexagon, so option (c) is the right choice.

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