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Quadrilaterals Test - 6

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Quadrilaterals Test - 6
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  • Question 1
    1 / -0
    Which of the following quadrilaterals is formed by joining the mid-points of the consecutive sides of a quadrilateral?
    Solution


    The quadrilateral formed by joining the mid-points of the consecutive sides of a quadrilateral is a parallelogram.
  • Question 2
    1 / -0
    Which of the following quadrilateral has no pair of parallel sides?
    Solution
    As can be seen from the diagram given below, there are no pair of parallel sides.
  • Question 3
    1 / -0
    The opposite angles of a quadrilateral are equal and the adjacent angles are in ratio 1 : 3. What will be the measure of all the angles in ascending order?
    Solution
    Let the angles be x, x, 3x and 3x.
    According to angle sum property,
    x + x + 3x + 3x = 360°
    8x = 360°
    x = 45°
    So, the angles are 45°, 45°, 135° and 135°.
  • Question 4
    1 / -0
    In the given figure, LMPN is a trapezium with LM || NP, MPO = 30° and MO ⊥ NP. What is the measure of LNO if LMP = 2LNO?

    Solution


    Given: MPO = 30° and MO⊥NP
    In triangle MPO,
    MOP + OPM + PMO = 180° (Sum of interior angles of a triangle)
    90° + 30° + PMO = 180°
    120° + PMO = 180°
    PMO = 60°
    Now, LMP = 90° + 60°
    LMP = 150°
    Given: LMP = 2LNO
    2LNO = 150°
    LNO = 75°
  • Question 5
    1 / -0
    In the given figure, ABCD is a rectangle in which ∠OAB = y°, ∠OBC = x° and ∠COD = 110°, find the values of x and y, respectively.

    Solution
    We know that the diagonals of a rectangle are equal and they bisect each other.
    So, in △AOB,
    AO = OB
    ∠OAB = ∠OBA (Base angles are equal.)
    i.e. ∠OBA = y°
    (∠OAB = y°, given)
    ∠AOB = 180° - y° - y° = 180° - 2y° and ∠DOC = 110° = ∠AOB = 110° (Vertically opposite angles)
    ∠AOB = 180° - 2y°
    110° = 180° - 2y°
    2y° = 70°
    y° = 35° and ∠OBA = 35° .... (i)
    Now, consider the right triangle, △ABC, right angled at B.
    So, ∠ABC = 90° (As ABCD is a rectangle)
    Now, consider △OBC.
    So, ∠OBC = x° = ∠ABC - ∠OBA = 90° - 35° = 55° (From (i))
    x° = 55° and y° = 35°
  • Question 6
    1 / -0
    If two consecutive angles of a parallelogram are (x + 80)° and (5x + 40)°, what is the name of the quadrilateral?
    Solution
    We know that consecutive interior angles of a parallelogram are supplementary.
    (x + 80)° + (5x + 40)° = 180°
    6x° + 120° = 180°
    6x° = 60°
    x° = 10°
    Thus, two consecutive angles are (10 + 80)° and (5 × 10 + 40)°, i.e. 90° and 90°. Hence, the name of the given parallelogram is rectangle.
  • Question 7
    1 / -0
    In a quadrilateral ABCD, if ∠A = 90° and ∠B : ∠C : ∠D = 3 : 4 : 8, then what will be the measures of the other angles?
    Solution
    Let ∠B = 3x, ∠C = 4x and ∠D = 8x
    Now, ∠A + ∠B + ∠C + ∠D = 360°
    90° + 3x + 4x + 8x = 360°
    15x = 270°
    x = 18°
    So, ∠B = 3 × 18, ∠C = 4 × 18 and ∠D = 8 × 18
    ∠B = 54°, ∠C = 72° and ∠D = 144°
  • Question 8
    1 / -0
    A rhombus is made inside a circle. One diagonal BD of the rhombus passes through the centre of the circle. If ∠ACB = 50°, find∠ADB.
    Solution
    ∠ACB = 50° = ∠CAD or ∠OAD (Alternate angles)
    ∠AOD = 90° (Diagonals of a rhombus intersect at right angles.)
    In △AOD,
    ∠OAD + ∠AOD + ∠ADO = 180° (Angle sum property of a triangle)
    ∠ADB = 40°
  • Question 9
    1 / -0
    In the given quadrilateral, what is the sum of A and C?

    Solution
    In ΔADC,
    DAC + DCA = CDR (Exterior angle of a triangle is equal to the sum of interior opposite angles.)
    So, DAC + DCA = 40° --- (1)
    In ΔABC,
    BAC + BCA = ABT (Exterior angle of a triangle is equal to the sum of interior opposite angles.)
    So, BAC + BCA = 60° --- (2)
    Adding equations 1 and 2:
    DAC + DCA + BAC + BCA = 40° + 60° --- (3)
    BAC + DAC = A
    BCA + DCA = C
    So, from equation 3,
    A + C = 100°
  • Question 10
    1 / -0
    If a quadrilateral has equal diagonals but unequal and non-parallel adjacent sides, then which of the following quadrilaterals can it be?
    Solution
    The quadrilateral which has equal diagonals but unequal and non-parallel adjacent sides can be an isosceles trapezium.

    ABCD is an isosceles trapezium, with AD = BC and AB parallel to DC. Diagonals AC and BD are equal.
  • Question 11
    1 / -0
    Find the largest angle of a parallelogram, if one angle of a parallelogram is 4° less than one-third of its adjacent angle.
    Solution
    Let the angle be x.

    Other angle =x – 4
    x +x – 4 = 180° (Sum of adjacent angles)



    4x - 12 = 540

    4x = 552°

    Largest angle = 138°
  • Question 12
    1 / -0
    Consider the following information to identify the shape(s).

    (1) The quadrilateral has two pairs of opposite sides parallel.
    (2) The diagonals bisect the angles and are perpendicular bisectors of each other.
    (3) Consecutive angles are supplementary but not equal.
    Solution
    All the mentioned properties are of a rhombus.
    In point 3, it is mentioned that consecutive angles are not equal, whereas a square has equal measure of consecutive angles.In point 2, it is mentioned that the diagonals are perpendicular bisectors, whereas not all the parallelograms have diagonals which are perpendicular bisectors.
  • Question 13
    1 / -0
    A parallelogram PQRS is inscribed in a circle. PR and QS are chords of the circle that intersect at point O. Find SQP.

    Solution
    PQRS is a parallelogram.
    PQ ∥ SR
    Now, POS is an exterior angle of △OPQ.
    QPR + SQP = POS (Exterior angle = Sum of two interior opposite angles)
    SQP = 75° - 34° = 41°
    SQP = 41°
  • Question 14
    1 / -0
    In a rectangle MNOP, Q, R, S, and T are the mid-points of MN, NO, OP and PM, respectively. The perimeter of the rectangle is 32 cm and the area is 60 cm2. Find the length of TQ, given the largest side is MN.
    Solution
    According to the given information, the figure is as given below.



    Given: Perimeter = 32 cm; Area = 60 cm2
    Let the width of the rectangle MN be x cm and the length NO be y cm.
    Perimeter = 2(x + y)
    32 = 2(x + y)
    x + y = 16
    x = 16 - y .. (i)
    Area = x × y
    x × y = 60
    Using (i), we have
    (16 - y) × (y) = 60
    16y - y2 = 60
    y2 - 16y = -60
    y2 - 16y + 60 = 0
    y2 - 10y - 6y + 60 = 0
    y(y - 10) - 6(y - 10) = 0
    (y - 10)(y - 6) = 0
    y = 10 or 6
    Since MN is the largest side;
    y = 6 cm and x = 10 cm
    Now, in △MTQ,
    MT = 3 cm (half of length)
    MQ = 5 cm (half of width)
    Using Pythagoras theorem,
    (Hypotenuse)2 = (Base)2 + (Perpendicular)2
    (QT)2 = (MT)2 + (MQ)2
    (QT)2 = 52 + 32
    (QT)2 = 25 + 9
    (QT)2 = 34
    QT =
  • Question 15
    1 / -0
    In a parallelogram ABCD, DE is the angle bisector of ADC, EG is produced such that EDG = EGD and EF is the bisector of DEG. If EG ∥ BC, and points F and G are on line DC, then
    Solution
    According to the question:

    In parallelogram ABCD,
    AB = DC and AD = BC ...(i) (Opposite sides of a parallelogram are parallel and equal.)
    Since EG ∥ BC;
    EB = GC and EG = BC ... (ii)

    In ΔADE and ΔDEG,
    AE = DG (From (i) and (ii))
    AD = EG (From (i) and (ii))
    ED (Common)
    Therefore, using SSS congruency rule, ΔADE ≌ ΔGED
    Now, in ΔDEG, we know that EF is the bisector of DEG, so it cuts ΔDEG into half triangles ΔEFD and ΔEFG.
    Therefore, ar(ΔADE) = ar(ΔEFG)
  • Question 16
    1 / -0
    In the following figure, PQRS is a quadrilateral, SR is produced to N and PQ is produced to M such that QMNR is a parallelogram. C and D are the mid-points of QM and RN, respectively. Also, SC and PD intersect at O and bisects RQ. If OD = OC, then which of the following options is correct?
    Solution


    In triangles ORD and OQC,
    RD = QC (because QMNR is a parallelogram, and C and D are the mid-points of QM and RN, respectively.
    So, RN = QM
    Therefore, RN = QM)
    OD = OC (Given)
    OR = OQ (As O bisects RQ)
    Therefore, △ORD is congruent to △OQC.
    ∠ORD = ∠OQC ... (i)
    Also, ∠ORD = ∠OQP (Alternate angles) ... (ii)
    From (i) and (ii),
    ∠OQC = ∠OQP
  • Question 17
    1 / -0
    In the given rectangle ABDC, AE is the bisector of CAB and AE ∥ HI. If GFD = 48°, then GJH equals

    Solution


    In the given rectangle ABCD, CAB = 90° (Interior angles of a rectangle are 90°.)
    We know that AE is the bisector of CAB.
    So, CAE = 45°
    Now, in △ACE, CAE = 45°
    ACE = 90° (Interior angles of a rectangle are 90°.)
    Therefore, CEA = 180 - (CAE + ACE) (Sum of interior angles of a triangle = 180°)
    CEA = 180° - (45°+90°)
    = 180° - 135° = 45°
    Also, GFD = 48° (Given)
    In ΔKFE , KFE = 48° (Same as GFD)
    FEK = 45° (Same as CEA)
    Therefore, EKF = 180° - (KFE + FEK) (Sum of interior angles of a triangle = 180°)
    EKF = 180° - (48° + 45°)
    = 180° - 93° = 87°
    Now, we know that AE ∥ HI.
    So, FKE = GJH (Exterior alternate angles on parallel lines are same.)
    GJH = 87°
  • Question 18
    1 / -0
    In the following figure, X is the mid-point of PH. What would be the shape of figure XUZH, if PUIH is a parallelogram and Z is the mid-point of IU?
    Solution
    X is the mid-point of PH. (Given)
    So, XP = HX = PH
    Also, Z is the mid-point of IU. (Given)
    So, IZ = ZU = IU
    But PH = IU (Opposite sides of parallelogram)
    So, HX = ZU
    Again, PH is parallel to IU.
    Therefore, HX is parallel to ZU.
    Now, HX is parallel to ZU and HX = ZU.
    Thus, XUZH is a parallelogram.
  • Question 19
    1 / -0
    In the given figure, if PQ || NO and QR || FG, find the value of QF.

    Solution
    In MNO, MP = PN = 4
    P is the mid-point of MN.
    As PQ || NO, Q is the mid-point of MO. (By converse of mid-point theorem, i.e. the line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side)
    MQ = QO = 5
    In QRO, OG = GR = 3
    And QR || FG
    Also, G is the mid-point of OR. (OG = GR)
    F is the mid-point of QO. (By converse of mid-point theorem)

    QF = FO = QO

    QF = FO = = 2.5
  • Question 20
    1 / -0
    In the given figure, PQRS is a square, and M and N are the points on SR and RQ, respectively. What is the ratio of the length of MN to that of SQ, if MN is 6 cm shorter than SQ?

    Solution
    The given square has a side of cm.
    ∠SRQ = 90°
    In triangle SRQ, by Pythagoras theorem,
    SQ² = SR² + RQ²
    SQ² =
    SQ = = 8 cm
    SQ = 8 cm
    MN = 8 – 6 = 2 cm (Given)
    Ratio of the length of MN to that of SQ = =
    Therefore, required ratio = 1 : 4
  • Question 21
    1 / -0
    Read the statements carefully and find out the option that holds.

    Statement 1: Opposite angles of a rhombus have equal measure.
    Statement 2: Arhombus is an ortho-diagonalquadrilateral. Its diagonals bisectopposite angles.
    Solution
    Statement 1 is true as a rhombus has two diagonals connecting pairs of opposite vertices, and two pairs of parallel sides. Using congruent triangles, one can prove that the rhombus is symmetric across each of these diagonals. This theorem states that opposite angles in a rhombus are equal.

    If we draw a line joining the two points B and D, two triangles are created - DAB and BCD.

    Now, in these triangles, AB = CD, AD = BC, and side DB is common. Therefore, according to the side-side-side property of triangles, these are congruent.
    Therefore, corresponding angles would be equal.
    The same can be proved for the other two angles as well.
    Thus, the opposite angles in a rhombus are equal.
    Statement 2 is true. The proof is as follows:



    Given: Rhombus ABCD with diagonals BD and AC
    First we will use only diagonal AC.



    Segment DA is congruent to segment AB. Segment CD is congruent to segment CB. Segment CA is congruent to segment CA. Triangle DAC is congruent to triangle BAC. Angle CAD is congruent to angle BAC. Angle DCA is congruent to angle ACB.
    Segment AC bisects angles BAD and DCB.
    Then we will use only diagonal BD.



    Segment DA is congruent to segment DC. Segment AB is congruent to segment BC. Segment BD is congruent to segment BD. Triangle DAB is congruent to triangle DCB. Angle ADB is congruent to angle BDC. Angle DBA is congruent to angle CBD.
    Segment BD bisects angles ADC and CBA.
    So, a rhombus is an ortho-diagonal quadrilateral. Its diagonals bisect opposite angles.
  • Question 22
    1 / -0
    Read the statements carefully and mark them as true (T) or false (F).

    1. Every rhombus is a parallelogram.
    2. Two pairs of equal-length sides that are adjacent to each other is a kite.
    3. Rhombus with one angle as 90° is a square.
    4. Every parallelogram is a rectangle in which two opposite sides are equal.
    Solution
    1. Every rhombus is a parallelogram. TRUE



    It has two pairs of parallel sides, which makes it fit for the condition.
    2. A quadrilateral whose opposite sides are not equal is a kite. TRUE
    Only kite satisfies the situation as the diagonals of a kite meet at a right angle. Also, a kite has exactly one pair of opposite angles that are congruent.
    3. Rhombus with one angle as 90° is a square. TRUE
    All the sides of the rhombus are equal and opposite angles of rhombus are equal. So, if one angle measures 90° then the opposite angle will also measure 90°. The sum of other two angles is 180° so each angle would measure 90°. This means that it will be a square as all the sides are same and all the angles measure 90°.
    Hence, the given statement is true.
    4. Every parallelogram in which two opposite sides are equal is not a rectangle. As each pair of co-interior angles is supplementary because two right angles add to a straight angle, the opposite sides of a rectangle are parallel. This means that a rectangle is a parallelogram, so its opposite sides are equal and parallel. Its diagonals bisect each other. Hence, the given statement is false.
  • Question 23
    1 / -0
    Fill in the blanks:

    1. The opposite sides of a _______ are parallel and equal.
    2. If each angle measures 90° and diagonals are not perpendicular bisectors of each other, then it is a __________.
    3. A parallelogram cannot be a _______.
    Solution
    1. The opposite sides of a parallelogram are parallel and equal.
    2. If each angle measures 90° and diagonals are perpendicular bisectors of each other, then it is a rectangle.


    3. A parallelogram cannot be a trapezoid because it does not have all sides parallel to each other.


  • Question 24
    1 / -0
    In the given rectangle ABCD, FEG is a triangle and DF = GC = . The perimeter of rectangle ABCD is 56 cm and that of rectangle IHCD is 42 cm. If J and K are the respective mid-points of EF and EG, find the area of the trapezium FJKG given that the length and width of the rectangle ABCD are in the ratio 4 : 3.

    Solution


    We know that length AB and width BC of rectangle ABCD are in the ratio 4 : 3.
    So, let the length of rectangle ABCD be 4x and the width be 3x.
    Perimeter of rectangle ABCD = 56 cm (GIven)
    Perimeter of rectangle ABCD = 2(length + width)
    56 = 2(4x + 3x)
    56 = 8x + 6x
    56 = 14x
    x = 56 ÷ 14
    x = 4
    Therefore, length AB of rectangle ABCD = 4x
    = 4(4)
    = 16 cm
    Also, DC = 16 cm
    In rectangle IHCD, length DC = 16 cm
    Let the width HC of rectangle IHCD be y cm.
    Perimeter of rectangle IHCD = 42 cm (Given)
    Perimeter of rectangle IHCD = 2(length + width)
    42 = 2(16 + y)
    42 = 32 + 2y
    2y = 42 - 32
    2y = 10
    y = 10 ÷ 2
    y = 5 cm
    Therefore, width HC of rectangle IHCD = 5 cm

    We know that DF = GC =
    FG = DF+GC
    DC = FG + DF + GC
    16 = DF + GC + DF + GC
    Since we know that DF + GC = FG;
    FG = 8 cm

    Now, in △EFG, J and K are the mid-points of EF and EG, respectively.

    So, JK =
    JK =

    JK = 4 cm

    In trapezium FJKG,
    FG = 8 cm
    JK = 4 cm
    And we know that the height of trapezium FJKG is same as HC.
    So, height of trapezium FJKG = 5 cm

    Area of trapezium FJKG =
    =

    = 6 × 5 = 30 cm2
  • Question 25
    1 / -0
    Match the columns:

    1. A rectangle in which two adjacent sides have equal length is a a. rhombus
    2. If the diagonals are perpendicular to each other and the opposite corner angles are equal to each other, then the parallelogram is a/an b. rectangle
    3. A parallelogram whose diagonals are congruent is a c. square
    4. The two non-parallel sides are equal and form equal angles at one of the bases, then it is called a/an d. isosceles trapezium
    Solution
    1. A rectangle in which two adjacent sides have equal length is a square.
    2. If the diagonals are perpendicular to each other, then the parallelogram is a rhombus.
    These are the properties of a rhombus.
    • 1. The diagonals are perpendicular to each other.
    • 2. Opposite angles are equal.
    3. A rectangle has the following properties:
    • All the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite sides are congruent, and diagonals bisect each other).
    • All angles are right angles by definition.
    • The diagonals are congruent.
    4. It's a property of an isosceles trapezium that the two non-parallel sides are equal and form equal angles at one of the bases.
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