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Question

In the given figure, E is mid-point of AB and DE meets diagonal AC at point F.
If ABCD is a parallelogram and area of ADF is 60cm2, then area of ADB is?
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Solution

Given, parallelogram ABCD E as a mid-point of AB and DE meets diagonal AC at point F.
In AEF and DFC
CAE=ACD ( Alternate angles, ABCD and AC intersects it. )
AFE=CFD ( Vertically opposite angle )
AED=EDC ( Alternate angles, AECD and ED intersects it. )
AEFDFC ( AAA Similarity property)
Using Similarity Property,
EFFD=AECD
or,EFFD=AEAB (CD=AB, As ABCD is a parallelogram )
or, EFFD=12 ( As E is mid point of AB , AB=2AE)
EF+FD=ED
FD=23 ED (1)
Drawing an altitude AG from A to meet DE at G.
Area of ADE=12×base×altitude=12×ED×AG
Area of ADF=12×base×altitude=12×DF×AG
=12×23×ED×AG ( FD=23 ED, From 1)
AreaofADEAreaofADF=12×ED×AG12×23ED×AG=32
or, Area of ADE=32× Area of ADF
=32× 60 cm2
= 90 cm2
Now, In ADB , E is mid point. So, DE is median.
Median divides the triangle into two triangles of equal areas.
Area of ADE= Area of BDE=12× Area of ADB
or, Area of ADB= 2 × Area of ADE
= 2 × 90 cm2
= 180 cm2

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