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प्रश्न
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ADF) = ar (ABFC)
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उत्तर
Given in the question, A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F.
Prove that ar (ADF) = ar (ABFC)
Proof: ABCD is a parallelogram and AC divides it into two triangle of equal area.
ar (ΔADC) = ar (ΔABC) ...(I)
So, DC || AB and CF || AB
As we know that triangle on the same base and between the same parallels are equal in area.
So, ar (ΔACF) = ar (ΔBCF) ...(II)
Adding equation (I) and (II), we get
ar (ΔADC) + ar (ACF) = ar (ΔABC) + ar (ΔBCF)
ar (ΔADF) = ar (ABFC)
Hence proved.
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संबंधित प्रश्न
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(i) ar (DOC) = ar (AOB)
(ii) ar (DCB) = ar (ACB)
(iii) DA || CB or ABCD is a parallelogram.
[Hint: From D and B, draw perpendiculars to AC.]
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[Hint: Join CX.]
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In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-
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(ii) ar (BYXD) = 2 ar(MBC)
(iii) ar (BYXD) = ar(ABMN)
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