ABCD is a parallelogram in which the side BC is produced to E such that CE = BC.AE intersects CD at F.If area triangle DFB is equal to 3 cm², find the area of parallelogram ABCD.
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u can use mid pt theorem here as BC=CE
C is the mid pt and CF is parallel to AB
this implies F is the mid pt of AE
This implies that BF is the median for the triangle BDC
we know that median divides triangle into two triangles with equal area
ar of DFB = ar of BFC = 3cm^2
ar of BDC = 6 cm^2
diagonal of parallelogram divides it into two triangles with equal areas
this implies ar of ABCD =2( ar of BDC)= 12cm^2
hope this helps you!
C is the mid pt and CF is parallel to AB
this implies F is the mid pt of AE
This implies that BF is the median for the triangle BDC
we know that median divides triangle into two triangles with equal area
ar of DFB = ar of BFC = 3cm^2
ar of BDC = 6 cm^2
diagonal of parallelogram divides it into two triangles with equal areas
this implies ar of ABCD =2( ar of BDC)= 12cm^2
hope this helps you!
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hey mate here is your answer
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