Solve ques. 7 and 8 ...I need correct and..
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See the diagram for both Qn 7 and Qn 8.
DG and EF are perpendiculars from points D and E respectively onto the base BC.
Qn 7: part (i)
Given DE || BC.
=> DG = EF
=> 1/2 * BC * DG = 1/2 * BC * EF
=> ar(ΔBCD) = ar(ΔBCE) ------(1)
ar(ΔACD) = ar(ΔABC) - ar(ΔBCD)
= ar(ΔABC) - ar(ΔBCE) using (1)
= ar(ΔABE)
Hence proved.
Qn7 part (ii):
From part (i), ar(ΔBCD) = ar(ΔBCE)
=> ar(ΔBCO) + ar(ΔOBD) = ar(ΔBCO) + ar(ΔOCE)
=> ar(ΔOBD) = ar(ΔOCE)
Proved.
==================
Qn 8.
See the diagram.
Given
ar(ΔBCE) = ar(ΔBCD)
=> 1/2 * BC * DG = 1/2 * BC * EF
=> DG = EF
As the perpendiculars DG and EF onto BC from D and E, are equal :
DE || BC.
Proved.
DG and EF are perpendiculars from points D and E respectively onto the base BC.
Qn 7: part (i)
Given DE || BC.
=> DG = EF
=> 1/2 * BC * DG = 1/2 * BC * EF
=> ar(ΔBCD) = ar(ΔBCE) ------(1)
ar(ΔACD) = ar(ΔABC) - ar(ΔBCD)
= ar(ΔABC) - ar(ΔBCE) using (1)
= ar(ΔABE)
Hence proved.
Qn7 part (ii):
From part (i), ar(ΔBCD) = ar(ΔBCE)
=> ar(ΔBCO) + ar(ΔOBD) = ar(ΔBCO) + ar(ΔOCE)
=> ar(ΔOBD) = ar(ΔOCE)
Proved.
==================
Qn 8.
See the diagram.
Given
ar(ΔBCE) = ar(ΔBCD)
=> 1/2 * BC * DG = 1/2 * BC * EF
=> DG = EF
As the perpendiculars DG and EF onto BC from D and E, are equal :
DE || BC.
Proved.
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kvnmurty:
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Answered by
2
Answer done here.........
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Please mark the answer as brain list answer
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Please mark the answer as brain list answer
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