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Quadrilaterals Test - 7

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Quadrilaterals Test - 7
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  • Question 1
    1 / -0

    In quadrilateral ABCD, if ∠A = 60 and ∠B : ∠C : ∠D = 2:3:7, then ∠D is :

    Solution

    In quadrilateral, the sum of the all four angles equal to 360. let angle B = 2x, angle C = 3x and angle D = 7x.

    angle A + angle B + angle C + angle D = 360

    60 + 2x + 3x + 7x = 360

    12x = 300

    x= 25

    So, angle D = 7x = 7 (25) = 175

  • Question 2
    1 / -0

    If APB and CQD are 2 parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form, square only if

    Solution

    The diagonals of a square bisect its angles. Opposite sides of a square are both parallel and equal in length. All four angles of a square are equal.

  • Question 3
    1 / -0

    The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if

    Solution

    A quadrilateral formed by joining the mid points of a square is a square. So, ABCD is a square. In Square, diagonals are equal and perpendicular.

  • Question 4
    1 / -0

    D and E are the mid-points of the sides AB and AC of △ABC and O is any point on the side BC, O is joined to A. If P and Q are the mid-points of OB and OC res, Then DEQP is

    Solution

    By mid point theorem, DE is parallel to BC. In triangle BOA,  DP parallel to OA and OA is parallel to QE in triangle AOC( mid point theorem) because D and P are mid points in triangle BOA and E and Q are mid points in triangle AOC.

    So, DP is parallel to EQ. In quadrilateral DPQE, both pair of opposite sides are parallel. So, it become parallelogram.

  • Question 5
    1 / -0

    The bisectors of the angles of a Parallelogram enclose a

    Solution

    Let ABCD be a parallelogram. angle A + angle D = 180 ( co-interior angles)

    1/2 of (angle A + angle D) = 90.

    Triangle formed by bisectors of angle A and angle D, have sum of two angles equals to 90 therefore, remaining angle is of 90. similarly we can prove that other angles formed are of 90 each by bisectors of angles of ABCD. The quadrilateral formed by angle bisectors of ABCD has all angles equal to 90 ( Vertically opposite angles). A quadrilateral with all right angles is a Rectangle.

  • Question 6
    1 / -0

    Rhombus is a quadrilateral

    Solution

    Let ABCD be a rhombus.

    Join BD which forms two triangles ABD and DCB. In triangle ABD, AB = AD.

    So, angle ADB = angle ABD ( angles opposite to equal sides are equal) ----------(1)

    But , angle ABD = angle BDC and angle ADB = angle CBD ( alternate angles)-------------(2)

    So, from (1) and (2)

    angle ADB = angle ABD = angle BDC  = angle CBD

    therefore, diagonal BD bisects the angle B and angle D.

  • Question 7
    1 / -0

    In Quadrilateral ABCD, ∠A= 110, ∠B = 75 and ∠C = 350. Find ∠D?

    Solution

    angle A + angle B + angle C + angle D = 360 ( angle sum property of quadrilateral)

    110 + 75 + 35 + angle D = 360

    angle D = 360-220 = 140

  • Question 8
    1 / -0

    In Quadrilateral ABCD, ∠A = (3x) , ∠B = (5x) , ∠C = (20x) , ∠D = (8x) . Find the value of x?

    Solution

    angle A + angle B + angle C + angle D = 360 ( angle sum property)

    3x + 5x + 20x + 8x = 360

    36x = 360

    x = 10

  • Question 9
    1 / -0

    In Quadrilateral ∠A = 38, ∠C = 3 ∠A, ∠D = 4∠A. Find the value of ∠B?

    Solution

    angle A + angle B + angle C + angle D = 360 ( angle sum property)

    angle C = 3 ( 38) = 114

    angle D = 4 ( 38) = 152

    So, 38 + angle B + 114 + 152 = 360

    angle B = 360 - 304 = 56

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