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Understanding Quadrilaterals Test - 17

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Understanding Quadrilaterals Test - 17
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  • Question 1
    1 / -0
    If in quadrilateral ABCDABCD, ABCDAB \parallel CD, then ABCDABCD is necessarily a
    Solution
    If in quadrilateral ABCD,ABCD,ABCD, AB \parallel CD, 
    condition
    1)BCDA1) BC \parallel DA then it becomes Parallelogram.
                     2)BC2) BC is not parallel to DADA then it is trapezium. 
    Hence, None of These is the correct answer.
  • Question 2
    1 / -0
    Complete the following statement .
    Number of measurements required to construct a rectangle are_______.
    Solution
    If we know the length and breadth of the rectangle, we can construct it since a rectangle has opposite sides equal in length and each of the four angles is equal to 900 {90}^{0}
    Hence, number of measurements required to construct a rectangle is 22.
  • Question 3
    1 / -0
    Two adjacent angles of a parallelogram are in the ratio 1 : 2. The angles are
    Solution
    The sum of 2 adjacent angles of a parallelogram is 180 degrees.

    Let the angles are xx and 2x2x.
    \thereforexx ++ 2x2x == 180o180^{o}
    \therefore3x=1803x = 180
    \thereforex=60ox = 60^{o}
    Hence angles are x=60ox = 60^{o} and 2x=120o2x = 120^{o}.
  • Question 4
    1 / -0
    How many sides does a polygon have whose sum of the interior angles is 1980o{1980}^{o}?
    Solution
    Sum of the interior angles of a polygon of n n sides is 1800(n2) {180}^{0} (n - 2)

    Given,
    sum of whose interior angles is 1980o {1980}^{o}
    =>1800(n2)=1980o => {180}^{0} (n - 2) = {1980}^{o}
    =>n2=11 => n - 2 = 11
    =>n=13 => n = 13

    Hence, the polygon has 13 13 sides.

  • Question 5
    1 / -0
    Under what conditions must PQRSPQRS be a parallelogram?

    Solution
    Opposite angles of a parallelogram are equal
    \Rightarrow S=Q\angle S=\angle Q and P=R\angle P=\angle R
    \Rightarrow 3x+42o=4x=3o3x+{42}^{o}=4x={3}^{o} and 6y=5y+16o6y=5y+{16}^{o}
    \Rightarrow x=45ox={45}^{o} and y=16oy={16}^{o}
  • Question 6
    1 / -0
    ABCD is a parallelogram. The angle bisectors of \angle A and \angleD meet at O. The measure of \angleAOD is:
    Solution
    Let A \angle A be $$ 2x

    $$
    Let D \angle D be 2y 2y



    Since ABDC AB \parallel DC , sum of angles on same side of the transversal AD AD will be 180o {180}^{o}
    =>2x+2y=180o => 2x + 2y = {180}^{o}
    x+y=90o x + y = {90}^{o}



    After bisection of angles A and D, we get 1 =x \angle 1  = x and 2=y \angle 2 = y as per the given figure.

    Now in AOD \triangle AOD , we have

    x+y+AOD=180o x + y + \angle AOD = {180}^{o}
     90o+AOD=180o {90}^{o} + \angle AOD = {180}^{o}
    =>AOD=90o => \angle AOD = {90}^{o}

  • Question 7
    1 / -0
    Which of the following properties need not be true for a parallelogram?
    Solution
    A) The diagonals of a parallelogram need not be equal. 
    Option AA need not be true.
    B)In a parallelogram, diagonals need not be perpendicular to each other.
    Option BB need not be true.
    C) In a parallelogram, diagonals need not divide it into four congruent triangles. 
    Option CC need not be true.
    So  the correct answer is option DD
  • Question 8
    1 / -0
    If ABCDABCD is a parallelogram with diagonals intersecting at OO, then the number of distinct pairs of congruent triangles formed is:
  • Question 9
    1 / -0
    Find the angles of a parallelogram if one angle is three times another.
    Solution
    Opposite angles of a parallel are equal to each other.

    So, if two angles are xo {x}^{o} each, then the other two opposite angles are 3xo {3x}^{o} , since it is given that one angle is three times the other.

    Sum of all four angles of a parallelogram =3600 = {360}^{0}
    =>xo+3xo+xo+3xo=3600 => {x}^{o} + {3x}^{o}+ {x}^{o} + {3x}^{o} = {360}^{0}
    =>8xo=3600 => {8x}^{o} = {360}^{0}
    =>xo=450 => {x}^{o} = {45}^{0}

    So 3xo=1350 {3x}^{o} = {135}^{0}

    Hence, the four angles of the parallelogram are 450,1350,450,1350 {45}^{0}, {135}^{0} , {45}^{0}, {135}^{0}
  • Question 10
    1 / -0
    ABCDABCD is a parallelogram. IF PP be a point on CDCD such that AP=ADAP=AD, then the measure of PAB+BCD\angle PAB+\angle BCD is:
    Solution
    In ADP\triangle ADP,
    AD=APAD=AP \Rightarrow APD=ADP\angle APD=\angle ADP
    Also, ADP+APD+PAD=180o\angle ADP+\angle APD+\angle PAD={180}^{o}
    \Rightarrow 2ADP=180oPAD2\angle ADP={180}^{o}-\angle PAD (\therefore APD=ADP)......(i)\angle APD=\angle ADP)......(i)
    In quad ABCDABCD
    A+B+C+D=360o\angle A+\angle B+\angle C+\angle D={360}^{o}
    PAD+PAB+2D+C=360o\angle PAD+\angle PAB+2\angle D+\angle C={360}^{o}(\because B=D\angle B=\angle D)
    PAD+PAB+180oPAD+C=360o\angle PAD+\angle PAB+{180}^{o}-\angle PAD+\angle C={360}^{o} (From (i)(i))
    \Rightarrow PAB+BCD=360o180o=180o\angle PAB+\angle BCD={360}^{o}-{180}^{o}={180}^{o}
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