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### BdMO National Junior 2019/6

Posted: **Mon Mar 04, 2019 8:55 am**

by **samiul_samin**

$ABCD$ 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 $PQRS$.

### Re: BdMO National Junior 2019/6

Posted: **Wed Mar 13, 2019 10:16 am**

by **samiul_samin**

Diagram

- Screenshot_2019-03-12-22-38-58-1.png (38.11 KiB) Viewed 6770 times

Solution

### Re: BdMO National Junior 2019/6

Posted: **Fri Mar 15, 2019 7:45 pm**

by **math_hunter**

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?

### Re: BdMO National Junior 2019/6

Posted: **Sun Jan 19, 2020 10:21 pm**

by **Arifa**

$Solution:$ Let x be the area of $BOC$ ; $(OQR) = (SOR)$ & $(POS) = (PQO)$ then,

$2(OQR) = 2(1010 - (2020+x)/4 ) = 2020 -1010 - x/2$

$2(POS) = 2 ( (2019+x)/4 - 2019/2) = x/2 + 2019/2$

Adding those 2 equations,

$2(OQR) + 2(POS)= 2020-1010- (2019/2) = 1/2$

Implies, $(PQRS) = 1/2$

### Re: BdMO National Junior 2019/6

Posted: **Thu Jan 23, 2020 2:49 pm**

by **Arifa**

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-

$2(POS) = x/2 - 2019/2$