Guys even this is difficult for me
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Anonymous:
nails Kato
I didn't ask this Mr.
But i m telling this u
tomorrow is my exams and u don't irritate me by the way do you know this answer
tomorrow is ur xam but dont be serious.. bcz....single piece of paper cannot decide ur futurs
nice quote bro but even a half mark can rank u down right?
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Given: 1) ABCD is a trapezium.
2) AB // DC
3) AO / OC = BO / OD = 1 / 2
To find: DC
Solution: 1) In triangle AOB and triangle DOC,
AO / OC = BO / OD = 1 / 2 ... (Given)
2) Angle AOB = angle DOC ... (Vertically Opposite Angles)
3) Therefore,triangle AOB is similar to triangle DOC ... (SAS Similarity Test)
4) Thus, AB / DC =AO / OC = BO / OD = 1 / 2
Therefore, AB / DC = 1 / 2 i.e. 3 / DC = 1 / 2
Hence, DC = 6
2) AB // DC
3) AO / OC = BO / OD = 1 / 2
To find: DC
Solution: 1) In triangle AOB and triangle DOC,
AO / OC = BO / OD = 1 / 2 ... (Given)
2) Angle AOB = angle DOC ... (Vertically Opposite Angles)
3) Therefore,triangle AOB is similar to triangle DOC ... (SAS Similarity Test)
4) Thus, AB / DC =AO / OC = BO / OD = 1 / 2
Therefore, AB / DC = 1 / 2 i.e. 3 / DC = 1 / 2
Hence, DC = 6
tq u saved me dude
Wlcm
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