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    Parallelograms on the Same Base and Between the Same Parallels (Theorem and Examples)

    Parallelograms on the same base and between the same parallels theorem and its proof is explained here in detail. Visit BYJU’S to learn the theorem and an example with a complete explanation.

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    Parallelograms on the Same Base and Between the Same Parallels

    Parallelograms on the same base and between the same parallels are explained here in this article with the help of the theorem.

    Theorem:

    A parallelogram on the same base and between the same parallels are equal in area.

    Proof:

    Consider two parallelograms ABCD and EFCD on the same base DC and the same parallels AF and DC, as shown in the figure.

    To prove: ar (ABCD) = ar (EFCD)

    It means that we need to prove the area of parallelogram ABCD is equal to the area of a parallelogram EFCD.

    Consider the triangles ADE and BCF,

    ∠ DAE = ∠ CBF [Corresponding angles from AD || BC and transversal AF] …(1)

    ∠ AED = ∠ BFC [Corresponding angles from ED || FC and transversal AF] …(2)

    By using the angle sum property of a triangle, we can write:

    ∠ ADE = ∠ BCF ….(3)

    Also, from the opposite sides of a parallelogram ABCD, we can write:

    AD = BC ….(4)

    Using the equations (1), (3), and (4) and by Angle-Side-Angle (ASA) rule,

    ∆ ADE ≅ ∆ BCF

    Therefore, the area of triangle ADE is equal to the area of a triangle BCF. (As congruent figures have equal area)

    (i.e) ar (ADE) = ar (BCF) …(5)

    Thus, the ar (ABCD) = ar (ADE) + ar (EDCB)

    ar (ABCD) = ar (BCF) + ar (EDCB) (using equation (5))

    ar (ABCD) = ar (EFCD)

    Thus, the area of parallelogram ABCD is equal to the area of parallelogram EFCD.

    Hence, a parallelogram on the same base and between the same parallels are equal in area, is proved.

    Example

    Now, let us solve the below problem with the help of the theorem.

    Question:

    Consider a parallelogram ABCD and EFCD is a rectangle. Also, AL is perpendicular to DC.

    Prove that:

    ar (ABCD) = ar (EFCD)

    ar (ABCD) = DC × AL

    Solution:

    Given that ABCD is a parallelogram, and EFCD is a rectangle.

    To prove : ar (ABCD) = ar (EFCD)

    A rectangle is also a parallelogram, and by using the theorem “parallelogram on the same base and between the same parallels are equal in the area”, we can write

    ar (ABCD) = ar (EFCD)

    Hence proved.

    To prove: ar (ABCD) = DC × AL

    By using ar (ABCD) = ar (EFCD)

    ar (ABCD) = DC × FC (As, the area of a rectangle = Length × Breadth) …(1)

    Also given that AL ⊥ DC, we can say AFCL is also a rectangle.

    Therefore, AL = FC …(2)

    From (1) and (2), we can get

    ar (ABCD) = DC × AL Hence, proved.

    Practice Problems

    Solve the following problems:

    The point P and Q lie on the sides DC and AD respectively of a parallelogram ABCD. Prove that ar (APB) = ar (BQC)

    In the figure given below, PQRS and ABRS are parallelograms. “X” is the point on the side BR.

    Prove that:

    (1) ar (PQRS) = ar (ABRS)

    (2) ar (AX S) = ½ ar (PQRS)

    Stay tuned with BYJU’S – The Learning App and learn all Maths-related topics easily by exploring more videos.

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    Prove the theorem : Parallelogram on the same base and between the same parallels are equal in area.

    Click here👆to get an answer to your question ✍️ Prove the theorem : Parallelogram on the same base and between the same parallels are equal in area.

    Question

    Prove the theorem : Parallelogram on the same base and between the same parallels are equal in area.

    Medium Open in App Solution Verified by Toppr

    Given two parallelogram ABCD and EFCD that have the same base CD and lie between same parallel AF and CD.

    We have prove that ar(ABCD)=ar(EFCD)

    Since opposite sides of ∥gm are parallel AB∥CD and ED∥FC with transversal AB

    ⇒∠DAB=∠CBF  [ Corresponding angles ]

    with transversal EF

    ⇒∠DEA=∠CFE  [ Corresponding angles ]

    ⇒AD=BC   [ Opposite sides of parallelogram are equal ]

    In △AED ξ △BFC ⇒∠DEA=∠CFE ∠DAB=∠CBF ∴AD=BC

    ⇒△AED≅△BFC  [ AAS congruency ]

    Hence, ar(△AED)=ar(△BFC)

    ( Areas of congruent figures are equal )

    ⇒ar(ABCD)=ar(△ADE)+ar(EBCD)

    =ar(△BFC)+ar(EBCD) =ar(EBCD) ∴ar(ABCD)=ar(EBCD)

    Hence, the answer is proved.

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    Theorem 9.1

    Theorem 9.1 Parallelograms on the same base and between the same parallels are equal in area. Given : Two parallelograms ABCD & EFCD, that have the same base CD & lie between same parallels AF & CD. To Prove : r (ABCD) = r (EFCD) Proof : Since opposite sides of pa

    Check sibling questions

    Theorem 9.1 - Chapter 9 Class 9 - Areas of Parallelograms and Triangles

    Last updated at Aug. 25, 2021 by Teachoo

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    Transcript

    Theorem 9.1 Parallelograms on the same base and between the same parallels are equal in area. Given : Two parallelograms ABCD & EFCD, that have the same base CD & lie between same parallels AF & CD. To Prove : r (ABCD) = r (EFCD) Proof : Since opposite sides of parallelogram are parallel Also, AD = BC In AED and BFC DAB = CBF DEA = CFE AD = BC AED BFC AED BFC Hence, r ( AED) = r ( BFC) Now, r (ABCD) = r ( ADE) + r(EBCD) = r ( BFC) + r (EBCD) = r ( EFCD) Hence, proved

    Next: Theorem 9.2 Important →

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    Davneet Singh

    Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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