Work out with area

Geometry Level pending

The above shows a quadrilateral A B C D ABCD , with O O denotes the intersection point between the diagonals A C AC and B D BD .

Which of the following represents the area of this quadrilateral?

1 2 × sin ⁡ ( ∠ A O D ) ) × A C × B D \frac12 \times \sin (\angle AOD)) \times AC \times BD 1 2 × cos ⁡ ( ∠ A O D ) ) × A O × B D \frac12 \times \cos (\angle AOD)) \times AO \times BD 1 2 × cos ⁡ ( ∠ A O D ) ) × A C × C D \frac12 \times \cos (\angle AOD)) \times AC \times CD 1 2 × cos ⁡ ( ∠ A O D ) ) × A C × B D \frac12 \times \cos (\angle AOD)) \times AC \times BD

Arnab Kumar Debnath by Arnab Kumar Debnath , 21, Bangladesh

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1 solution

The area of this picture will be
area of triangle AOD+area of triangle DOC+area of triangle BOC+ area of triangle AOB
= ( 1 / 2 ) ∗ ( A O ∗ D O ∗ s i n ( A O D ) + D O ∗ C O ∗ s i n ( 180 − A O D ) + C O ∗ B O ∗ s i n ( B O C ) + B O ∗ A O ∗ s i n ( 180 − A O C ) ) =(1/2)*(AO*DO*sin(AOD)+DO*CO*sin(180-AOD)+CO*BO*sin(BOC)+BO*AO*sin(180-AOC))
= ( 1 / 2 ) ∗ ( s i n ( A O D ) ( D O ( A O + C O ) + ( B O ( A O + C O ) ) = (1/2)*(sin(AOD)(DO(AO+CO)+(BO(AO+CO)) [As angleAOD=angleBOC]
= ( 1 / 2 ) ∗ s i n ( A O D ) ∗ A C ∗ B D =(1/2)*sin(AOD)*AC*BD

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