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In the Following Figure, Ce is Drawn Parallel to Diagonals Db of the Quadrilateral Abcd Which Meets Ab Produced at Point E. Prove that δAde and Quadrilateral Abcd Are Equal in Area - Mathematics

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प्रश्न

In the following figure, CE is drawn parallel to diagonals DB of the quadrilateral ABCD which meets AB produced at point E.
Prove that ΔADE and quadrilateral ABCD are equal in area.

योग
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उत्तर

Since ΔDCB and ΔDEB are on the same base DB and between the same parallels i.e. DB // CE, therefore we get

Ar. ( ΔDCB) = Ar. ( ΔDEB )
Ar. ( ΔDCB + ΔADB ) = AR. (ΔDEB + ΔADB )
Ar. ( ABCD ) = Ar. ( ΔADE )
Hence proved.

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Figures Between the Same Parallels
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 16: Area Theorems [Proof and Use] - Exercise 16 (A) [पृष्ठ १९६]

APPEARS IN

सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 16 Area Theorems [Proof and Use]
Exercise 16 (A) | Q 6 | पृष्ठ १९६

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