## Friday, October 28, 2016

### Geometry Problem 1279 Parallelogram, Midpoint, Triangle, Area, 3/10

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1279. 1. https://goo.gl/photos/L9wsdrtPF7pDebWf7
Let DE extended meet BC at P
Since EB= ½. CD => PB=BC=AD
Triangle GAD similar to GFP => GA/GF=AD/PF= 2/3=> GA/AF=2/5 => S(DGF)/S(ADF)=3/5
S(DGF)=3/5.S(ADF)= 3/5 x ½ S(ABCD)= 3/10. S(ABCD)

2. Let M be the mid point of AF.

EM = AD/4 so AD = 4/5AM

Now S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
Hence S(AED) = 4/5 X S/16 = S/20

S1 = S -S(ABF) - S(ADE) - S(CDF) + S(ADE)

Hence S1 = S -S/4 - S/4 - S/4 + S/20 = 3S/10

Sumith Peiris
Moratuwa
Sri Lanka

1. In the 2nd line it should read as

AG = 4/5AM not AD = 4/5AM

3. S(AED) = S/20 ??

1. S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
Hence S(AED) = 4/5 X S/16 = S/20

2. Sumith
Per observation, S(AED) is not equal S/20.
Peter Tran

3. Area of triangle AED is 1/4 the area of ABCD, because triangle has same base as ABCD and height 1/2 of height of parallelogram ABCD

4. Extremely sorry Omid and Peter. My mistake.

I should have said S(AEG) = S/20 which is easily seen since S(AEM) = S/16 and AG = 4/5AM

4. Extend AF to meet DC at P
Since FC is half of AD, from mid-point theorem we have
CP = CD & FP = AF = l (say) and AG = x (say)
Also the triangles DGP and EGA are similar
=> DP/EA = PG/AG
=> 4 = 2l-x/x
=> x = AG = (2/5)l and GF = (3/5)l

Since the area of triangle AFD = 1/2(S) => Area of triangle GFD (S1) = 3/5*1/2*S (S being the are of //gm ABCD)

Therefore S1 = 3/10*S

5. Draw CG, G midpoint of AD, H on ED. Tr GHD ~ tr AEG => S(AEG)=1/20 (S)
AGD + AEG + EGFB + FCD + S1 = S => 1/5 + 1/20 + 1/5 + 1/4 + S1 = S
S1 = 7/10 S