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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 $$ABCD$$, $$AB \parallel CD$$, then $$ABCD$$ is necessarily a
    Solution
    If in quadrilateral $$ABCD, AB \parallel CD,$$ 
    condition     $$1) BC \parallel DA$$ then it becomes Parallelogram.
                     $$2) BC$$ is not parallel to $$DA$$ 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 $$ {90}^{0} $$
    Hence, number of measurements required to construct a rectangle is $$2$$.
  • 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 $$x$$ and $$2x$$.
    $$\therefore$$$$x$$ $$+$$ $$2x$$ $$= $$$$180^{o}$$
    $$\therefore$$$$3x = 180$$
    $$\therefore$$$$x = 60^{o}$$
    Hence angles are $$x = 60^{o}$$ and $$2x = 120^{o}$$.
  • Question 4
    1 / -0
    How many sides does a polygon have whose sum of the interior angles is $${1980}^{o}$$?
    Solution
    Sum of the interior angles of a polygon of $$ n $$ sides is $$ {180}^{0} (n - 2) $$

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

    Hence, the polygon has $$ 13 $$ sides.

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

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

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



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



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

    Now in $$ \triangle AOD $$, we have

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

  • 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 $$A$$ need not be true.
    B)In a parallelogram, diagonals need not be perpendicular to each other.
    Option $$B$$ need not be true.
    C) In a parallelogram, diagonals need not divide it into four congruent triangles. 
    Option $$C$$ need not be true.
    So  the correct answer is option $$D$$. 
  • Question 8
    1 / -0
    If $$ABCD$$ is a parallelogram with diagonals intersecting at $$O$$, 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 $$ {x}^{o} $$ each, then the other two opposite angles are $$ {3x}^{o} $$, since it is given that one angle is three times the other.

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

    So $$ {3x}^{o} = {135}^{0} $$

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