solution of it........fstt
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area (APB)
= area (ACB) [ same base AB and same perpendicular height from AB ]
= area (BQC) [ same base BC and same perpendicular height from BC ]
SHAHAN186:
what is acb
what is it
ACB is the triangle with the vertices A, C and B. Just like APB is the triangle with vertices A, P and B. Draw in the line AC if it helps.
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since, triangle BQC and IIgm ABCD are on same base BC and between same parallels
AD & BC
therefore ar(BQC)=1/2 ar(ABCD) .......(1)
since triangle APB and IIgm ABCD are on same
base AV and between same parallels AB&CD
therefore ar(APB)=ar(ABCD).........(2)
from (1) & (2),
ar(BQC)=ar(APB)
PROOVED
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AD & BC
therefore ar(BQC)=1/2 ar(ABCD) .......(1)
since triangle APB and IIgm ABCD are on same
base AV and between same parallels AB&CD
therefore ar(APB)=ar(ABCD).........(2)
from (1) & (2),
ar(BQC)=ar(APB)
PROOVED
HOPE This WILL HELP U..
MARK IT AS BRAINEST
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