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Areas of Parallelograms and Triangles Test - 2

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Areas of Parallelograms and Triangles Test - 2
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  • Question 1
    1 / -0
    If ABC XYZ, which of the following options is true?
    Solution
    The congruent triangles have equal areas as they are of same shape and size.
    ar(ABC) = ar(XYZ)
  • Question 2
    1 / -0


    If KLMN is a parallelogram and KOL is a triangle with area x m2, then the area of KLMN is
    Solution
    The parallelogram KLMN and the triangle KOL are on the same base KL and between the same parallels KL and MN.
    So, the area of parallelogram KLMN = 2 ar (KOL)
    Area of parallelogram KLMN = 2x m2 [ Area of KOL = x m2]
  • Question 3
    1 / -0
    In the given figure, ABSP and ABQR are parallelograms.



    If PS = 6 units and SD = 5 units, the area of parallelogram ABQR is equal to _____.
    Solution
    ABSP is a parallelogram.



    AB = PS [Opposite sides of a parallelogram]
    AB = 6 units
    and SD = 5 units
    Area of parallelogram ABSP = Base Height
    = AB SD
    = 6 5 = 30 sq. units
    Now, parallelograms ABSP and ABQR are on the same base and between the same parallels. So, their areas are equal.
    i.e. Area of parallelogram ABQR = Area of parallelogram ABSP
    = 30 sq. units
  • Question 4
    1 / -0
    In the given figure, ABSP and ABQR are two parallelograms.



    If the area of the parallelogram ABQR is 20 cm2 and SD = 4 cm, then the length of AB is
    Solution
    Area of parallelogram ABQR = Base Height



    20 cm2 = AB 4 cm
    cm = AB
    AB = 5 cm
  • Question 5
    1 / -0
    MNO ABC. If the altitude and base of MNO are x and y, respectively, ar(ABC) is equal to _____.
    Solution
    Given, MNO ABC
    arMNO = ar ABC…... (1) [Congruent triangles are equal in area.]
    Also, area of MNO =
    = xy
    And from (1), area of ABC = xy
  • Question 6
    1 / -0
    In the given figure, ABCD is a parallelogram and ABP is a triangle. If ar(APB) = 16 sq. units, ar(ADP) + ar(BCP) will be _____.

    Solution
    The area of a triangle is half the area of a parallelogram when both are on the same base and between the same parallels.
    So, the area of ABCD = 2 the area of triangle ABP
    = 2 (16) sq. units
    = 32 sq. units
    Also, area of parallelogram ABCD = area of ABP + area of ADP + area of BPC
    32 = 16 + area of ADP + area of BCP
    32 – 16 = ar (ADP) + ar (BCP)
    ar (ADP) + ar (BCP) = 16 sq. units
  • Question 7
    1 / -0
    ABC and PQR are two congruent right triangles, right-angled at B and Q, respectively. In ABC, AB = 4 units and AC = 5 units. The area of PQR is equal to
    Solution
    In right-angledABC,



    AC2 = AB2 + BC2
    52 = 42 + BC2
    52 – 42 = BC2
    BC2 = 25 – 16
    BC2 = 9
    BC = 3 cm
    Now, ar(ABC) =
    = = 6 sq. units
    And ABC PQR
    ar(ABC) = ar(PQR) [ Congruent triangles are equal in area.]
    ar(PQR) = 6 sq. units
  • Question 8
    1 / -0
    In the given figure, ABCD is a trapezium. If DP || BC, then which of the following relationships is true?

    Solution
    In the trapezium ABCD, AB || DC.
    So, PB || DC [Parts of parallel lines]
    And DP || BC (Given)
    PBCD is a parallelogram.
    Parallelogram PBCD and triangle PQD are on the same base DP and between the same parallels DP and BC.
    Note that if a triangle and a parallelogram are on the same base and between the same parallels, then the area of triangle is equal to half the area of the parallelogram.

    So, ar(PBCD) = 2ar(PQD)
  • Question 9
    1 / -0
    ABCD and PQRS are two parallelograms with equal area and sides, i.e. AB = PQ. Which of the following relations is true?
    Solution
    Area of parallelogram ABCD = Area of parallelogram PQRS



    Base1 Height1 = Base2 Height2
    AB Height1 = PQ Height2
    Height1 = Height2 [ AB = PQ]
    Perpendicular distance between AB and CD = Perpendicular distance between PQ and RS
  • Question 10
    1 / -0
    In the given figure, DAB = ABC.



    If 1 = 2, then which of the following relationships is true?
    Solution
    In ADB andBCA,
    1 =2 [Given]
    AB = BA [Common]
    DAB = ABC [Given]
    ADB ABC
    Hence, ar(ADB) = ar(ABC) [Congruent triangles are equal in area.]
  • Question 11
    1 / -0
    Which of the following options includes figures on the same base and between the same parallels?
    Solution
    In the figure given in option (3), the triangle ADB and rectangle ABCD are on the same base AB and between the same parallels AB and CD.
  • Question 12
    1 / -0
    In the given figure, if ar(PQR) = 17 sq. units, ar (XYZ) is _____.

    Solution
    PQR XYZ [ Corresponding parts are equal.]
    Area of PQR = Area of XYZ
    ar (XYZ) = 17 sq units.
  • Question 13
    1 / -0
    Which of the following options includes figures on the same base?
    Solution
    Clearly, parallelograms ABCD and ABQP are on the same base AB.
  • Question 14
    1 / -0
    If two triangles are congruent, it can be said that
    Solution
    Congruent triangles are equal in area and their corresponding sides are also equal. Hence, their perimeters and corresponding angles are also equal.
  • Question 15
    1 / -0
    ABCD and ABPQ are two parallelograms on the same base AB and between the same parallels. Which of the following relationships is true?

    Solution
    Area of ABCD = Area of ABPQ [Parallelograms on the same base and between the same parallels are equal in area.]



    ar(ABPD) + ar(BPC) = ar(ABPD) + ar(ADQ)
    ar(BPC) = ar(ADQ)
  • Question 16
    1 / -0
    MNO and ABC are two congruent isosceles triangles with base measuring 7 cm and altitude measuring 6 cm. Find the areas of MNO and ABC.
    Solution
    MNO and ABC are two congruent triangles. So, their areas are equal.
    So, area of MNO = area of ABC =



    =
    = cm2
  • Question 17
    1 / -0
    In the given figure, if || m and AB = PQ, then which of the following statements is true?

    Solution

    Parallelogram ABQP andABP are on the same base AB and between the same parallels l and m.
    Area of ABQP = 2ar(ABP)….. (1) [ The area of a parallelogram is double the area of a , which are on the same base and between the same parallels]
    Also, area of ABQP = 2ar(PBQ)….. (2) [ The area of a parallelogram is double the area of a , which are on the same base and between the same parallels]
    From (1) and (2), we get

    2ar(ABP) = 2ar(PBQ)
    ar(ABP) = ar(PBQ)
  • Question 18
    1 / -0
    DEF and PQR are two congruent equilateral triangles. The side PQ of PQR is 4 cm. The area of DEF is equal to _____.
    Solution
    Area of equilateral triangle = , where 'a' is the side of a triangle.
    Area (PQR) =
    = 4 cm2
    And PQR DEF
    ar(PQR ) = ar(DEF) [Congruent triangles are equal in area.]
    ar (DEF) = 4 cm2
  • Question 19
    1 / -0
    Which of the following options represents different figures on the same base and between the same parallels?
    Solution
    ADC and rectangle ABCD are on the same base DC and between the same parallels DC and AB.
  • Question 20
    1 / -0
    In the given figure, ABED is a rectangle and ABCF is a parallelogram.



    If the area of ABED = y cm2, then the area of ABCF is
    Solution
    A rectangle is also a parallelogram.
    So, ABED is a parallelogram.
    Area of ABED = Area of ABCF [ Parallelograms on the same base and between the same parallels are equal in area.]
    Area of ABCF = y cm2
  • Question 21
    1 / -0
    In the given figure, ABCD and PQCD are two parallelograms.



    Which of the following relationships is true?
    Solution



    CD = AB …. (1) [ ABCD is a parallelogram.]
    CD = PQ …..(2) [ PQCD is a parallelogram.]
    From (1) and (2), AB = PQ
    Subtracting PB from both the sides, we get
    AB – PB = PQ – PB
    AP = BQ
  • Question 22
    1 / -0
    In the given figure, ABCD and PQCD are two parallelograms.



    Which of the following relationships is true?
    Solution
    ar(ABCD) = ar(PQCD)



    [Parallelograms ABCD and PQCD are on the same base CD and between the same parallels AQ and CD, so their areas are equal.]
  • Question 23
    1 / -0
    In the given figure, ABCD and ABPQ are two parallelograms on the same base AB. If ar(ABCD) = x sq. units and ar(ABPQ) = y sq. units, then which of the following relationships is true?

    Solution
    Parallelograms ABCD and ABPQ are on the same base AB and between the same parallels AB and DP.
    ar(ABCD) = ar(ABPQ) [ Parallelograms on the same base and between the same parallels are equal in area.]
    x = y
  • Question 24
    1 / -0
    In the given figure, || m, AD || BC and AE || BD.



    If the area of parallelogram ABCD is 46 sq. units, then the area of AED is equal to
    Solution
    Area of parallelogram ABCD = area of parallelogram ABDE



    [ These parallelograms (ABCD and ABDE) are on the same base AB and between the same parallels and m.]

    ar(ABD) + ar(BDC) = ar(AED) + ar(ABD)
    ar(BDC) = ar(AED) … (1)

    Now, ar(ABCD) = 2 ar(BDC) [The area of a parallelogram is double the area of a triangle when both are on the same base and between the same parallels.]

    46 sq. units = 2 ar(BDC)
    sq. units = ar(BDC)
    ar(BDC) = 23 sq. units

    From (1), ar(AED) = 23 sq. units
  • Question 25
    1 / -0
    Directions: In the given figure, line || m, AD || BC and AE || BD.



    If ar(ABDE) = 50 sq. units, the ar(ABCD) = ______.
    Solution
    Parallelograms on the same base and between the same parallels are equal in area.
    Parallelogram ABCD and parallelogram ABDE are on the same base AB and between the same parallel lines and m.
    Therefore, ar (ABDE) = ar (ABCD)
    ar (ABCD) = 50 sq. units
  • Question 26
    1 / -0
    In the given figure, PQRS is a parallelogram and PQM is a triangle. If ar(PQM) = 'a' sq. units and ar(PQRS) = 'b' sq. units, then which of the following relationships is true?

    Solution
    The area of a parallelogram is double the area of a triangle when both are on the same base and between the same parallels.



    ar(PQRS) = 2ar(PQM) [ These are on the same base PQ and between the same parallels PQ and RS.]
    b = 2a
    2a – b = 0
  • Question 27
    1 / -0
    In the given figure, if || m, then which of the following relationships is true?

    Solution
    ar(ABC) = ar(ABE) + ar(ACD) – ar(ADE)
    RHS = ar(ABE) + ar(ACD) – ar(ADE)
    = ar(ABD) + ar(ADE) + ar(AEC) + ar(ADE) – ar(ADE)
    = ar(ABD) + ar(ADE) + ar(AEC)
    = ar(ABC) = LHS
  • Question 28
    1 / -0
    In the given figure, ABDC and KLCD are two parallelograms.



    If the area of trapezium CLBD is 150 cm2 and that of DBK is 50 cm2, then what is the area of ABDC?
    Solution
    Area of ACDB = area of CLBD + area of CLA
    Since area of △DBK = area of △CLA;
    Area of ACDB = 150 cm2 + 50 cm2
    = 200 cm2
  • Question 29
    1 / -0
    In the given figure, ABCD is a rectangle and ABEF is a parallelogram. If AB = 4 cm and BC = 3 cm, then the area of ABEF is


    Solution
    Area of rectangle ABCD = b
    = 3 4 = 12 cm2
    A rectangle is also a parallelogram; and parallelograms on the same base and between the same parallels are equal in area.
    ar(ABCD) = ar(ABEF)
    ar(ABEF) = 12 cm2
  • Question 30
    1 / -0


    If ABCD and ABDE are two parallelograms and ar(ABCD) – ar(ABDE) = a cm2, then a is
    Solution
    Parallelograms on the same base and between the same parallels are equal in area.

     ar(ABCD) = ar(ABDE) [ They are on the same base AB and between the same parallels AB and CD.]



    ar(ABCD) – ar(ABDE) = 0 cm2

    a = 0
  • Question 31
    1 / -0
    In the given figure, ABCD is a parallelogram and ABP is a triangle. If ar(ABCD) = 22 sq. units, ar(ABP) will be

    Solution
    The area of a triangle is half the area of a parallelogram when both are on the same base and between the same parallels.
    Moreover, ABP and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.



    ar (ABP) = ar (ABCD)
    ar (ABP) = = 11 sq. units
  • Question 32
    1 / -0
    In the given figure, CD || BQ, ABCD is a parallelogram and CDQP is a rectangle. If AB = x, then PQ is equal to

    Solution
    ABCD is a parallelogram.
    AB = CD … (1) [ Opposite sides of a parallelogram are equal.]
    In addition, CDPQ is a rectangle.
    PQ = CD … (2) [ Opposite sides of a rectangle are also equal.]
    From (1) and (2), AB = PQ
    PQ = x
  • Question 33
    1 / -0
    In the given figure, if AB || CD, AB = CD and CN is perpendicular to AB (produced), then which of the following relationships is true?

    Solution
    The area of triangle is half the area of parallelogram when both are on the same base and between the same parallels.



    ADB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
    ar(ADB) = ar(ABCD)
  • Question 34
    1 / -0


    In the given figure, if ar(PBC) = 20 sq. units, ar(ABPC) = _____.
    Solution
    In ABC and PBC,
    AB = BP [5 cm each]
    BC = BC [Common]
    AC = PC [4 cm each]
    ABC PBC [By SSS rule]
    ar (ABC) ar (PBC) [Areas of congruent triangles are equal.]
    And ar (ABPC) = ar (PBC) + ar (ABC)
    = 20 + 20
    = 40 sq units
  • Question 35
    1 / -0
    If CN AB and AD || BC, then which of the following relationships is true?

    Solution


    From the figure,
    ar(ABC) + ar(CBN) = ar(ACN)
  • Question 36
    1 / -0
    In the given figure, if AD || BC, AD = BC and CN is perpendicular to AB (produced), which of the following relationships is true?

    Solution
    If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.



    Now, ACB and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
    ar(ACB) = ar(ABCD)
  • Question 37
    1 / -0
    In the given figure, p || q and BC || AD.



    If ar(ABE) = 16 cm2, then ar(ABCD) is equal to
    Solution
    If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.



    Here, ABE and parallelogram ABCD are on the same base AB and between the same parallels p and q.
    ar(ABE) = ar(ABCD)
    ar(ABCD) = 2ar(ABE)
    = 2(16 cm2) = 32 cm2
  • Question 38
    1 / -0
    In the given figure, p || q and BC || AD.



    If ar(ABCD) = 50 cm2, then ar(ABE) is equal to
    Solution
    If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.



    Now, ABE and parallelogram ABCD are on the same base AB and between the same parallels p and q.
    ar(ABE) = ar(ABCD)
    ar(ABE) =
    = 25 cm2
  • Question 39
    1 / -0
    In ABC and PQR, AB = PQ, BC = QR and AC = PR. If ar(ABC) = 20 sq. units, ar(PQR) = ______.
    Solution
    In ABC and PQR,



    AB = PQ [Given]
    BC = QR [Given]
    AC = PR [Given]
    ABC PQR [By SSS rule]
    Also, congruent triangles have equal areas.
    ar (ABC) = ar (PQR)
    20 = ar (PQR)
    So, ar (PQR) = 20 sq units.
  • Question 40
    1 / -0


    In the given figure, ABCD is a parallelogram. Which of the following relationships is correct?
    Solution
    Parallelogram ABCD and ADB are on the same base AB and between the same parallels AB and CD.
    ar(ABCD) = 2ar(ADB) … (1)
    Also, parallelogram ABCD and DBC are on the same base CD and between the same parallels AB and CD.
    ar(ABCD) = 2ar(DBC) … (2)
    From (1) and (2), 2 x ar(ADB) = 2 x ar(DBC)
    ar(ADB) = ar(DBC)
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