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.
It is wrong
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)
To solve this problem, we will use the fact that if two triangles have the same height, then the ratio of their areas is equal to the ratio of their bases.
1) Considering triangle EGB: Since AG = 2GB, we can conclude that triangle AEG and triangle BEG have the same height. Additionally, base EG is equal in both triangles. Therefore, the ratio of their areas is equal to the ratio of their bases: ar(EGB)/ar(AEG) = EG/AG = 1/2.
Now, let's consider triangle AEG and triangle ADE. Both triangles have the same base AE. Since CE = 2DE, the height of triangle AEG is twice that of triangle ADE. Therefore, ar(AEG) = 2 * ar(ADE).
Substituting this equation into the previous one, we get: ar(EGB) / (2 * ar(ADE)) = 1/2. Rearranging this equation, we find: ar(EGB) = (1/2) * (2 * ar(ADE)) = ar(ADE).
Using the same logic, we can also prove that ar(EGB) = ar(BFC) and ar(BFC) = ar(AFB).
Therefore, the area of triangle EGB is equal to 1/6 of the area of parallelogram ABCD.
2) Similarly, considering triangle EFC: Since BF = 2FC, we can conclude that triangle CEF and triangle BEF have the same height. Additionally, base EF is equal for both triangles. Therefore, the ratio of their areas is equal to the ratio of their bases: ar(EFC)/ar(CEF) = EF/CF = 1/2.
Now, let's consider triangle BEF and triangle BCD. Both triangles have the same base BF. Since CE = 2DE, the height of triangle BEF is twice that of triangle BCD. Therefore, ar(BEF) = 2 * ar(BCD).
Substituting this equation into the previous one, we get: ar(EFC) / (2 * ar(BCD)) = 1/2. Rearranging this equation, we find: ar(EFC) = (1/2) * (2 * ar(BCD)) = ar(BCD).
Therefore, the area of triangle EFC is equal to 1/2 of the area of triangle BCD or parallelogram ABCD.
3) Since ar(EGB) = ar(BFC), and ar(EFC) = ar(BCD) (proved in step 1 and step 2 respectively), we can conclude that ar(EGB) = ar(EFC).
4) To find the portion of the area of parallelogram ABCD that is the area of EFG, we need to find the ratio of their areas.
- The area of triangle EFG can be calculated as ar(EFG) = (1/2) * EF * FG, using the formula for the area of a triangle.
- The area of parallelogram ABCD is given as ar(ABCD) = base * height = AB * AG, where AG is the height.
Since AG = 2GB, we can rewrite the equation as ar(ABCD) = AB * 2GB.
The ratio of their areas is then ar(EFG) / ar(ABCD) = [(1/2) * EF * FG] / [AB * 2GB].
Since FG = GB (since G is the midpoint of AB), we can simplify the ratio to ar(EFG) / ar(ABCD) = (1/2) * EF / AB.
Therefore, the portion of the area of parallelogram ABCD that is the area of EFG is equal to (1/2) * EF / AB.