## BdMO National Junior 2019/6

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### BdMO National Junior 2019/6

\$ABCD\$ is a trapezium and \$AB\$ parallel \$CD\$.\$P,Q,R,S\$ are the midpoint of \$AB,BD,CD,AC\$. \$AC\$ & \$BD\$ intersect at \$O\$.The area of \$\triangle AOB=2019\$ & \$\triangle COD=2020\$. Find the area of \$PQRS\$.

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BdMO National Junior 2019/6

Diagram
Screenshot_2019-03-12-22-38-58-1.png (38.11 KiB) Viewed 6760 times
Solution

math_hunter
Posts: 22
Joined: Mon Sep 24, 2018 10:33 pm
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### Re: BdMO National Junior 2019/6

I don't understand the solution.
How the area of APQD is (3/4)ADB, PBCS is (3/4)ABC, CSR is (1/4)ACD, DQR is (1/4)DBC???
Can you please narrate the solution?

Arifa
Posts: 7
Joined: Wed Nov 22, 2017 8:09 pm

### Re: BdMO National Junior 2019/6

\$Solution:\$ Let x be the area of \$BOC\$ ; \$(OQR) = (SOR)\$ & \$(POS) = (PQO)\$ then,
\$2(OQR) = 2(1010 - (2020+x)/4 ) = 2020 -1010 - x/2\$
\$2(POS) = 2 ( (2019+x)/4 - 2019/2) = x/2 + 2019/2\$
\$2(OQR) + 2(POS)= 2020-1010- (2019/2) = 1/2\$
Implies, \$(PQRS) = 1/2\$
Oops, I forgot

Arifa
Posts: 7
Joined: Wed Nov 22, 2017 8:09 pm

### Re: BdMO National Junior 2019/6

Arifa wrote:
Sun Jan 19, 2020 10:21 pm
\$2(POS) = 2 ( (2019+x)/4 - 2019/2) = x/2 + 2019/2\$
There is a typo , \$sorry\$ the \$2nd\$ equation should be-
\$2(POS) = x/2 - 2019/2\$
Oops, I forgot