The quadrangle ABCD in a figure is a parallelogram. QS and BC are parallel and RT and CD are also parallel.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
The point P which is an intersection of QS and RT is on the diagonal line BD.
(The length of BP) : (length of PD) = 2 : 1.
Find the area ratio between the sum of the area of the triangle AQT and CSR and the area of parallelogram ABCD.
Answer
Each area of parallelograms AQPT and PRCS is set to be 2 × 1 = 2.
Then, the sum of the area of △AQT and △CSR can also be set to 2.
The area of parallelogram ABCD can be set to (2+1) × (2+1) = 9.
Therefore, the area ratio is 2 : 9.
2 : 9
Solution
In a figure, since BP : PD = 2 : 1, AT : TD = BR : RC = BQ : QA = CS : SD = 2 : 1. Each area of parallelograms AQPT and PRCS is set to be 2 × 1 = 2.
Then, the sum of the area of △AQT and △CSR can also be set to 2.
The area of parallelogram ABCD can be set to (2+1) × (2+1) = 9.
Therefore, the area ratio is 2 : 9.
