Constructions

  • Question 1
    CBSEENMA9002835

    Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.



    Solution

    (i) ΔPDC and quadrilateral ABCD lie on the same base DC and between the same parallels DC and AB.
    (iii) ΔTRQ and parallelogram SRQP lie on the same base RQ and between the same parallels RQ and SP.

    (v) Quadrilaterals APCD and ABQD lie on the same base AD and between the same parallels AD and BQ.

    Question 2
    CBSEENMA9002836

    In figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. 


    Solution

    ar(parallelogram ABCD) = AB x AE = 16 x 8 cm2
    = 128 cm2    ...(1)
    ar(parallelogram ABCD) = AD x CF
    = AD x 10 cm2    ...(2)
    From (1) and (2), we get
    AD X 10 = 128
    rightwards double arrow space space AD equals 128 over 10
    rightwards double arrow  ad = 12.8 cm

    Question 3
    CBSEENMA9002837

    If E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar(EFGH) = 1 half ar(ABCD).

    Solution
    Given: E, F, G and H are respectively the mid-points of the sides of a parallelogram ABCD.

    To Prove: ar(EFGH) = 1 half

    ar(ABCD).
    Construction: Join OF, OG, OH and OE. Also, join AC and BD.
    Proof: In ΔBCD,
    ∵    F and G are the mid-points of BC and DC respectively.
    ∴ FG || BD    ...(1)
    In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side
    In ΔBAD,
    ∵ E and H are the mid-points of AB and AD respectively.
    EH || BD    ...(2)
    In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side
    From (1) and (2),
    EH || FG    ...(3)
    Similarly, we can prove that
    EF || HG    ...(4)
    From (3) and (4),
    Quadrilateral EFGH is a parallelogram
    | A quadrilateral is a parallelogram if its opposite sides are equal
    ∵    F is the mid-point of CB and O is the mid-point of CA
    FO || BA
    In a triangle, the line segment joining the mid-points of any two sides is parallel to the third side and is half of it ⇒    FO || CG    ...(5)
    V BA || CD (opposite sides of a parallelogram are parallel)
    ∴ BA || CG
    and    FO equals 1 half BA
space space space space equals space 1 half CD space space

    ∵ Opposite sides of a parallelogram are equal = CG    ...(6)
    | ∵ G is the mid-point of CD In view of (5) and (6),
    Quadrilateral OFCG is a parallelogram
    ∵ A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of
    equal length
    ∵ OP = PC
    | ∵ Diagonals of a || gm bisect each other
    ∵ ΔOPF and ∵CPF have equal bases
    (∵ OP = PC) and have a common vertex F Their altitudes are also the same ar(ΔOPF) = ar(ΔCPF)
    Similarly, ar(ΔOQF) = ar(ΔBQF)
    Adding, we get
    ar(ΔOPF) + ar(ΔOQF) = ar(ΔCPF) + ar(ΔBQF) ⇒ ar(|| gm OQFP) = ar(ΔCPF) + ar(ΔBQF)                                                               ....(7)
    Similarly,
    ar(|| gm OPGS) = ar(⇒GPC) + ar(⇒DSG)                 ...(8)
    ar(|| gm OSHR) = ar(ΔDSH) + ar(ΔHAR)
    ar(|| gm OREQ) = ar(ΔARE) + ar(ΔEQB)                 ...(10)
    Adding the corresponding sides of (7), (8), (9) and (10), we get
    ar(|| gm EFGH) = {ar(ACPF} + ar(ΔGPC)}
    + {ar(ΔDSG) + ar(ΔDSH)} + {ar(ΔHAR) + ar(ΔARE)} + (ar(ΔBQF) + ar(ΔEQB)}
    = ar(ΔFCG) + ar(ΔGDH) + ar(ΔHAE) + ar(ΔEBF)

    Question 4
    CBSEENMA9002838

    P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that ar(ΔAPB) = ar{ΔBQC).

    Solution
    Given: P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD.

    To Prove: ar(ΔAPB) = ar(ΔBQC).
    Proof: ∵ ΔAPB and || gm ABCD are on the same base AB and between the same parallels AB and DC.
    therefore space space ar left parenthesis increment APB right parenthesis equals 1 half ar left parenthesis parallel to space gm space ABCD right parenthesis space space space space space space space space space space space space space space space.... left parenthesis 1 right parenthesis
    ∵ ΔBQC and || gm ABCD are on the same base BC and between the same parallels BC and AD.
    therefore space space ar left parenthesis increment BQC right parenthesis equals 1 half ar left parenthesis space parallel to space gm space ABCD right parenthesis space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
    From (1) and (2),
    ar(ΔAPB) = ar(ΔBQC).

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