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Exercise - 3.1

Question-1 :-  quadrilateral
Classify each of them on the basis of the following.
(a) Simple curve  (b) Simple closed curve  (c) Polygon  (d) Convex polygon  (e) Concave polygon

Solution :-
(a) Simple curve: 1, 2, 5, 6, 7 
(b) Simple closed curve: 1, 2, 5, 6, 7 
(c) Polygon: 1, 2, 4 
(d) Convex polygon: 1 
(e) Concave polygon: 1, 4 
    

Question-2 :-  How many diagonals does each of the following have?
(a) A convex quadrilateral  (b) A regular hexagon  (c) A triangle

Solution :-
(a) A convex quadrilateral : Two Diagonals.
(b) A regular hexagon : Nine Diagonals.
(c) A triangle : No Diagonals.
    

Question-3 :- What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Solution :-
  Let ABCD is a convex quadrilateral, then we draw a diagonal AC which divides the quadrilateral in two triangles.
  ∠A + ∠B + ∠C + ∠D = ∠1 + ∠6 + ∠5 + ∠4 + ∠3 + ∠2
= ( ∠1 + ∠2 + ∠3) + ( ∠4 + ∠5 + ∠6)
= 180° + 180°	[By Angle sum property of triangle]
= 360°
  Hence, the sum of measures of the triangles of a convex quadrilateral is 360°.
  Yes, if quadrilateral is not convex then, this property will also be applied.
  Let ABCD is a non-convex quadrilateral and join BD, which also divides the quadrilateral in two triangles.
  Using angle sum property of triangle,
  In  ∆ABD,	∠1 +  ∠2 +  ∠3 = 180°.......(i) 
  In  ∆BDC,	∠4 +  ∠5 +  ∠6 = 180°.......(ii) 
  Adding eq. (i) and (ii),
  ∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 =  360°
  ∠1 + ∠2 + ( ∠3 + ∠4) + ∠5 + ∠6 = 360°
  ∠A + ∠B + ∠C + ∠D = 360°
  Hence proved.
    

Question-4 :-  Examine the table. (Each figure is divided into triangles and the sum of the angles deduced from that.) table
What can you say about the angle sum of a convex polygon with number of sides?
(a) 7    (b) 8    (c) 10    (d) n

Solution :-
(a) When  n = 7, then Angle sum of a polygon = (n - 2) x 180° = (7 - 2) x 180° = 5 x 180° = 900°
(b) When  n = 8, then Angle sum of a polygon = (n - 2) x 180° = (8 - 2) x 180° = 6 x 180° = 1080°
(c) When  n = 10, then Angle sum of a polygon = (n - 2) x 180° = (10 - 2) x 180° = 8 x 180° = 1440°
(d) When  n =  n,  then Angle sum of a polygon = (n - 2) x 180°
    

Question-5 :-  What is a regular polygon? State the name of a regular polygon of
(i) 3 sides    (ii) 4 sides    (iii) 6 sides

Solution :-
   A regular polygon: A polygon having all sides of equal length and the interior angles of equal size is known as regular polygon.
(i) 3 sides : Polygon having three sides is called a Triangle. 
(ii) 4 sides : Polygon having four sides is called a Quadrilateral.
(iii) 6 sides : Polygon having six sides is called a Hexagon.
    

Question-6 :- Find the angle measure x in the following figures. angle

Solution :-
(i) By the angle sum property of quadrilaeral
    50° + 130° + 120° + x = 360°
    300° + x = 360°
    x = 360° - 300°
    x = 60°
 
(ii) By the angle sum property of quadrilaeral
    90° + 60° + 70° + x = 360°
    220° + x = 360°
    x = 360° - 220°
    x = 140°
        
(iii) Linear pair angle = 180°
    First interior angle of base = 180° - 70° = 110°
    Second interior angle of base = 180° - 60° = 120°
    There are five sides, so n = 5
    Angle sum of polygon = (n-2) x 180° = (5-2) x 180° = 3 x 180° = 540°
    Now, 
    30° + x + 110° + 120° + x = 540°
    2x + 260° = 540°
    2x = 540° - 260°
    2x = 280°
     x = 280°/2 
     x = 140°
(iv) No. of angles = 5
    Angle sum of polygon = (n-2) x 180° = (5-2) x 180° = 3 x 180° = 540°
    Now, 
    x + x + x + x + x = 540°
    5x = 540°
     x = 540°/5 
     x = 108°
    

Question-7 :-  (a) Find x + y + z   (b) Find x + y + z + w question

Solution :-
(i) Sum of linear pair angle = 180°
    90° + x = 180°
         x = 180° + 90° 
         x = 90° and
    z + 30° = 180°
          z = 180° - 30°
          z = 150°
    Also, y = 90° + 30° = 120° [By Exeterior angle property]
    Hence, x + y + z = 90° + 120° + 150° = 360°
(ii) By using of angle sum property of quadrilateral,
    60° + 80° + 120° + n = 360°
    260° + n = 360°
           n = 360° - 260°
           n = 100°

    Now, sum of linear pair angles = 180°
    w + 100° = 180°
           w = 180° - 100°
           w = 80°

    x + 120° = 180°
           x = 180° - 120°
           x = 60°

    y + 80° = 180°
          y = 180° - 80°
          y = 100°
       
    z + 60° = 180°
          z = 180° - 60°
          z = 120°    
    According to question,
    x + y + z + w = 60° + 100° + 120° + 80° = 360°
    
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