In the given figure- ABCD is a quadrilateral in which diagonals AC and BD intersect at a point E then-A- Area - BEC - Area - CDE - Area - AED - Area - ABEB- Area - AED - Area - BEC - Area - ABE - Area - CDEC- Area - ABE - Area - BEC - Area - AED - Area - CDED- None of these

The correct option is B Area Δ AED × Area Δ BEC = Area Δ ABE × Area Δ CDEDraw AM ⊥ BD and CN ⊥ BD Area Δ AED × area Δ BEC = ( 1/2ED × AM )× ( 1/2BE × CN ) =1/ ...

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Byju's Answer

Standard IX

Mathematics

Area of a Triangle

In the given ...

Question

In the given figure, ABCD is a quadrilateral in which diagonals AC and BD intersect at a point E then,

A r e a Δ A E D \times A r e a Δ B E C = A r e a Δ A B E \times A r e a Δ C D E

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A r e a Δ A B E \times A r e a Δ B E C = A r e a Δ A E D \times A r e a Δ C D E

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A r e a Δ B E C \times A r e a Δ C D E = A r e a Δ A E D \times A r e a Δ A B E

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Solution

The correct option is A $A r e a Δ A E D \times A r e a Δ B E C = A r e a Δ A B E \times A r e a Δ C D E$
Draw $A M ⊥ B D a n d C N ⊥ B D$
$A r e a Δ A E D \times a r e a Δ B E C$
$= (\frac{1}{2} E D \times A M) \times (\frac{1}{2} B E \times C N)$
$= \frac{1}{4} E D \times A M \times B E \times C N$
$= (\frac{1}{2} B E \times A M) \times (\frac{1}{2} D E \times C N)$
$= a r e a Δ A B E \times a r e a Δ C D E$

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In the given figure, ABCD is a quadrilateral in which diagonals AC and BD intersect at a point E then,

The correct option is A Area ΔAED×Area ΔBEC=Area ΔABE×Area ΔCDEDraw AM⊥BD and CN⊥BD Area ΔAED×area ΔBEC =(12ED×AM)×(12BE×CN) =14ED×AM×BE×CN =(12BE×AM)×(12DE×CN) =area ΔABE×area ΔCDE