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Quadrilaterals Test - 9

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Quadrilaterals Test - 9
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  • Question 1
    1 / -0
    In a rectangle ABCD if diagonal AC = 12 cm, then the diagonal BD is
    Solution

    In a rectangle, diagonals are equal to each other
    AC = BD = 12 cm
    AC = BD = 12 cm
    Diagonal BD = 12 cm
  • Question 2
    1 / -0
    In parallelogram ABCD, DAB = 300, BC = 20 cm and AB = 20 cm. The area of parallelogram is
    Solution
    According to the data given, the diagram is

    Find - Area of parallelogram = ?
    Sol. In AFD
    Sin 30o =
    20 = DF = DF = 10 cm
    Area of parallelogram = base height
    = 20 DF = 20 10
    = 200
    Area of parallelogram ABCD is 200 cm2
  • Question 3
    1 / -0
    In the given figure, ABCD is a parallelogram. The respective measures of angles x and y are

    Solution


    CAB = ACD = 50o [Alternate interior opposite angles]
    Since in parallelogram, opposite angels are congruent;
    x + y = 50o + 30o
    50o + y = 80o
    y = 30o
    Hence, x = 50o and y = 30o
  • Question 4
    1 / -0
    A quadrilateral in which diagonals are equal and bisect each other perpendicularly is
    Solution
    A quadrilateral in which diagonals are equal and bisect each other perpendicularly is a square.
  • Question 5
    1 / -0
    The length of a side of a rhombus is 10 m and one of its diagonals is 12 m. The length of the other diagonal is
    Solution


    As one side of rhombus is 10 m
    All sides would also be 10 m because all sides of a rhombus are equal.
    And also one of its diagonals is 12 m.
    As the diagonals of rhombus bisect each other at right angle
    6 m + 6 m = 12 m
    Now, In ABE
    (10)2 = (6)2 + (AE)2
    100 - 36 = (AE)2
    64 = (AE)2
    AE = 8 m
    length of the other diagonal = 2 x AE = 2 x 8 = 16 m
  • Question 6
    1 / -0
    In the parallelogram PQRS, the respective values of SQP and QSP are

    Solution
    SPQ + RQP = 180° (Sum of adjacent angles of a parallelogram is 180°.)
    RQP = 180° - SPQ = 180° - 75° = 105°
    SQP = 105° - 70° = 35°
    SPQ + PQS + QSP = 180°
    QSP = 180° - (75° + 35°) = 70°
  • Question 7
    1 / -0
    A quadrilateral is a rhombus but not a square if
    Solution
    A quadrilateral is a rhombus but not a square if the lengths of the diagonals are not equal.
  • Question 8
    1 / -0
    ABCD is a parallelogram and BD is a diagonal. If BAD = 70° and BDC = 65°, then DBC is equal to
    Solution
    BAD = 70°

    BDC = 65°

    DBA = BDC = 65° (Alternating angles as DC || AB and DB is transversal)

    DAB + ABC = 180°

    ABC = 180° - 70° = 110°

    DBC = 110° - 65° = 45°
  • Question 9
    1 / -0
    In the parallelogram ABCD, AB = 12 cm. The altitudes corresponding to the sides AB and AD are 9 cm and 8 cm respectively. Find length of AD.

    Solution
    Area of parallelogram = Base x altitude = 9cm x 12 cm = AD x 8cm
    Thus, AD = 13.5 cm
    Hence, answer option 1 is correct
  • Question 10
    1 / -0
    A quadrilateral is a rectangle but not a square when
    Solution
    A quadrilateral is a rectangle but not a square when its diagonals are not perpendicular.
  • Question 11
    1 / -0
    The angles of a quadrilateral, taken in order, measure x, x + 20, x + 30 and x + 50 degrees. This quadrilateral is a
    Solution
    Angles of a quadrilateral add up to 360o.
    x + (x + 20) + (x + 30) + (x + 50) = 360o
    4x + 100 = 360
    4x = 260
    x = 65o
    1st angle = x = 65o
    2nd angle = x + 20 = 65 + 20 = 85o
    3rd angle = x + 30 = 65 + 30 = 95o
    4th angle = x + 50 = 65 + 50 = 115o
    Out of the given options, options (1), (2) and (4) cannot be possible because all the angles must be equal in a square.
    In a rhombus and a parallelogram, opposite angles must be equal.
    It must be a trapezium.
  • Question 12
    1 / -0
    ABCD is a quadrilateral. If AC and BD bisect each other, then ABCD is a
    Solution
    ABCD is a quadrilateral. AB = BC = CD = DA and A = B = C = D = 90°
    Then ABCD can be called A SQUARE.
  • Question 13
    1 / -0
    In a parallelogram ABCD, if A = 60°, then B, C and D are respectively
    Solution
    In a parallelogram ABCD, if A = 60°, then C = A = 60
    Hence option 2 is correct
  • Question 14
    1 / -0
    In the given figure, ABCD is a square and ABX is equilateral. In this case, DCX is equal to

  • Question 15
    1 / -0
    ABCD is a parallelogram. The angle bisectors of A and D meet at O. The measure of AOD is
    Solution
    A + B + C + D = 360°
    We know, A = C and B = D
    This implies, 2(A + D) = 360°
    A + D = 180°
    In triangle AOD,
    (A + D) + AOD = 180°
    AOD = 90°
  • Question 16
    1 / -0
    In a parallelogram ABCD, AB = 4 cm and BC = 8 cm. Each of its diagonals is less than
    Solution
    In a parallelogram ABCD, AB = 4 cm and BC = 8 cm. Each of its diagonals is less than 12 cm because if we apply pythagorus theorem on a rectangular parallelogram, the diagonal will comes out to be 9 cm which is less than 12 cm.
  • Question 17
    1 / -0
    There is a plane quadrilateral enclosure ABCD. Diagonal AC = 8.2 cm and perpendiculars from B and D to AC are 3.4 cm and 2.6 cm respectively. If 1 cm represents 5 m, then the area enclosed is
    Solution

    Area of quadrilateral, Area of 2 triangles
    = Area of ABC + area of ADC
    = x 8.2 x 3.4 + x 2.6 x 8.2
    = 24.6 cm2
    As 1 cm represents 5 m
    1 cm2 represents 25 m2
    24.6 cm2 represents 24.6 x 25 m2
    = 615 m2
  • Question 18
    1 / -0
    ABCD is a quadrilateral. AB = BC = CD = DA and A = B = C = D = 90°. Then ABCD is
  • Question 19
    1 / -0
    In a square ABCD, its diagonals bisect at O. Then the triangle AOB is
  • Question 20
    1 / -0
    Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is
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