Posted by **sunil** on Wednesday, December 12, 2012 at 2:48am.

ABCD IS A PARALLELOGRAM, G IS THE POINT ON AB SUCH THAT AG = 2GB, E IS A POINT OF CD SUCH THAT CE = 2DE AND AND F IS A POINT OF BC SUCH THAT BF = 2FC. PROVE THAT

1) ar(EGB) = 1/6 ar(ABCD)

2)ar(EFC) = 1/2 ar(EBF)

3) ar(EBG) = ar(EFC)

4) FIND WHAT PORTION OF THE ARE OF PARALLELOGRAM IS THE AREA OF EFG.

- maths --plse help me.. -
**Steve**, Wednesday, December 12, 2012 at 12:41pm
1)

height h of ABCD and EGB is the same

base of EGB = 1/3 AB

area(ABCD) = h*AB

area(EGB) = 1/2 h*(1/3 AB) = 1/6 h*AB

2)

base of EFC = 1/2 base of EBF

so area is 1/2 as well

3)

if g = altitude from AD to BC

area(EFC) = 1/2 * (2/3 g) * (1/3 BC) = 1/9 area(ABCD)

So, I get area(EFC) = 2/3 area(EBG)

Is there a typo here?

4)

area(EFG) = area(EGB)+area(EBF)+area(EFC)

= 1/6(ABCD)+2area(EFC)+area(EFC)

= 1/6(ABCD) + 3(1/9 ABCD)

= 1/2(ABCD)

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