## 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

Solution
- math_hunter
**Posts:**22**Joined:**Mon Sep 24, 2018 10:33 pm-
**Contact:**

### 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?

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

$Solution:$ 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, $(PQRS) = 1/2$

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

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

Adding those 2 equations,

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

Implies, $(PQRS) = 1/2$

Oops, I forgot