Answer:
The value is
![A = 39315 \ m^2](https://img.qammunity.org/2021/formulas/physics/college/psgyjdtyiccnijw05n18yjf2qgfk5zkype.png)
Step-by-step explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is
![\rho = 1.2 \ kg/m^3](https://img.qammunity.org/2021/formulas/physics/college/f19ln8i0stugtyjegje5ys54iidmm2f5zk.png)
The acceleration due to gravity in Mars is
![g_m = 3.689 \ m/s^2](https://img.qammunity.org/2021/formulas/physics/college/emgmp7g48yy095kja6mx0h6fgccypvj70s.png)
Generally the Mars atmosphere density is mathematically represented as
![\rho_m = 0.71 * \rho](https://img.qammunity.org/2021/formulas/physics/college/7vlh901tijjqtxxo53mx0pre76myi0hw21.png)
=>
![\rho_m = 0.71 * 1.2](https://img.qammunity.org/2021/formulas/physics/college/h1csqzmy7jm0yf2fu3usfbhn27u104ftwn.png)
=>
![\rho_m = 0.852 \ kg/m^3](https://img.qammunity.org/2021/formulas/physics/college/2nuzaupc1zif0srhzjpbdkmx4k6q6libt5.png)
Generally the drag force on the rover and the parachute is mathematically represented as
![F__(D)} = m * g_(m)](https://img.qammunity.org/2021/formulas/physics/college/pacnailg6xykfwu8ohv8l1e28hde8x2xp1.png)
=>
=>
Gnerally this drag force is mathematically represented as
![F__(D)} = C__(D)} * A * (\rho_m * v^2 )/(2)](https://img.qammunity.org/2021/formulas/physics/college/v4a45enuu3ed4cjmpbwx3j8te1brienonq.png)
Here A is the frontal area
So
![A = (2 * F__D )/( C__D) * \rho_m * v^2 }](https://img.qammunity.org/2021/formulas/physics/college/7ki214zpkjbh8x6fcx20jnrug1mq2u88ww.png)
=>
![A = (2 * 8374 )/( 0.5 * 0.852 * 1 ^2 )](https://img.qammunity.org/2021/formulas/physics/college/dwguwedggvalzy1b1mla4bszfqe9m6owxa.png)
=>
![A = 39315 \ m^2](https://img.qammunity.org/2021/formulas/physics/college/psgyjdtyiccnijw05n18yjf2qgfk5zkype.png)