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Quadrilateral Test 241
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Quadrilateral Test 241

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  • Question 1/10
    1 / -0.25
    In the given figure, MNOP is a parallelogram. PM is extended to Z. OZ intersects MN and PN at Y and X respectively. If OX = 27cm and XY = 18cm, then what is the length (in cm) of YZ?

    Solutions
    ∠YNX = ∠OPX (similar triangle)
    =
    = ,
    YN =
    Now, ∠ZMY and ∠ZPO are similar
    =
    =
    = ,
    =

  • Question 2/10
    1 / -0.25
    ABCD is a rhombus. A straight line through C cuts AD produced at P and AB produced at Q. If   ;  then the ratio of the lengths of BQ and AB is
    Solutions

    angle P is common in triangle DPC and APQ. Apart from that DC || AB hence other two angles are also same. triangle DPC and APQ are similar.
    Let AB=BC=CD=DA=2 and PD=1 to satisfy the question condition.
    Now using similarity
    PD/PA = DC/AQ
    1/3 = 2/AQ
    AQ = 6
    BQ = AQ - AB = 6-2 =4
    BQ:AB = 4:2 = 2:1
  • Question 3/10
    1 / -0.25
    In quadrilateral PQRS, RM ⊥ QS, PN ⊥ QS and QS = 6 cm. If RM = 3 cm and PN = 2 cm, then the area of PQRS is:
    Solutions

    Area (quad PQRS) = Area (triangle PQS) + Area (triangle SRQ)

    = (1/2) × QS × PN + (1/2) × QS × NR

    = (1/2) × 6 × 2 + (1/2) × 6 × 3

    = 15 cm2

  • Question 4/10
    1 / -0.25
    The areas, of a square and a rectangle with equal perimeter are denoted by S and R, respectively. Which one of the following is correct?
    Solutions
    4 × side = 2(l + b)
    Side of square = (l + b)/2
    Area of rectangle = lb = R
    Area of square= (side)2=(l + b)2/4 = (l2 + b2 + 2lb)/4
    Now l2 + b2 > 2lb (holds always)
    So Area of square > 4lb/4
    S > R
  • Question 5/10
    1 / -0.25
    A quadrilateral ABCD circumscribes a circle and AB = 6 cm. CD = 5 cm and AD = 7 cm. The length of side BC is
    Solutions
    Let ABCD is a quadrilateral in which a circle is inscribed
    Given AB = 6cm; CD = 5cm

    By the property of circle, sum of opposite sides are equal
    AB + CD = BC + AD
    6 + 5 = BC + 7
    ⇒ BC = 11 - 7 = 4cm
  • Question 6/10
    1 / -0.25
    The perimeter of a rhombus in 20 cm and one of the diagonals is 8 cm. What is the area (in cm2) of the rhombus?
    Solutions
    Perimeter of rhombus= 20
    So side of rhombus=20/4=5cm
    One diagonal is 8 cm, let the other one be 2d
    Then 52 =(8/2)2 +d2
    d2=25-16=9
    d=3
    2d=6
    Hence diagonal of rhombus =6
    Area of rhombus =
  • Question 7/10
    1 / -0.25
    In the given figure, area of isosceles triangle ABE is 72 cm2. If BE = AB, AB = 2AD and AE||DC, then what is the area (in cm) of the trapezium ABCD ?

    Solutions
    Given: AB = BE
    And area(∆ABE) = 72 cm2
    1/2 (AB × BE) = 72
    ⇒AB2 = 144
    ⇒AB = 12 cm = BE
    It is also given that, AB = 2AD
    ⇒AD = AB/2
    = 6 cm
    Now, the area of the trapezium = 1/2 × AB × (AD + BC)
    = 1/2 × 12 × (6 + 12 + 6)
    = 144 cm2
  • Question 8/10
    1 / -0.25

    Q is a point in the interior of a rectangle ABCD. If QA = 3 cm, QB = 4 cm and QC = 5 cm, then the length of QD in centimetre is

    Solutions

    Hint: Use pythagorous theore in all 4 small rectangles to derive this formula.
  • Question 9/10
    1 / -0.25
    MNOP is parallelogram such that angle PMN = 600. If the angle bisector of angles M & N meet at the point X which lies on the line OP then find the value of OX = ?
    Solutions
    angle N = 1200 ........(180-60=120, Sum of straight angle)
     
    In the above figure, 
    angle MNX = angle OXN= 600  ............(alternate angles)
    also, angle MNX = angle ONX ........(angle bisector)
    therefore ΔOXN is an isosceles triangle
    so, OX = ON

    also,
    angle XMN = angle MXP= 300  ............(alternate angles)
    also, angle XMN = angle PMX ........(angle bisector)
    therefore ΔPMX is an isosceles triangle
    so, PX= PM

    As we know that, PM = ON since it is a parallelogram
    therefore,
    PX= PM= ON = OX
    we can write,
    OX = PX
  • Question 10/10
    1 / -0.25
    The diagonals are not perpendicular in a __________.
    Solutions
    The diagonals of a parallelogram are not perpendicular.
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