 Loi sur les exemples de Questions partie 2 All Modules View all topics Close topics
Mathématiques de la Loi sur l'examen partie 2 ### Ce cours a été révisé!

Pour une expérience d'apprentissage plus agréable, nous vous recommandons d'étudier la version mobile relogée de ce cours.

Emmenez-moi au cours révisé.

- or -

Continue studying this course

# Loi sur les exemples de Questions partie 2

• Study Reminders

We'll email you at these times to remind you to study

You can set up to 7 reminders per week

#### You're all set

We'll email you at these times to remind you to study

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

• Notes d'étude
• Révisions du sujet  Petra S. 0 0 I never knew how to do ratio between a circle and a square. Helpful to now know the formula. One side of square is x and r is the radius of the circle. Formula to get r squared is x/2 squared plus x/2 squared. Figure out r. Square area equals length times width and circle equals Pi times r squared. Divide area of square by area of circle and the answer is 2/Pi.  Jorge Carlos V. 0 0 There is a simpler solution: The square is two triangles with base 2r and height r, that is 2*squared(r). The area of circle is pi*squared(r).The requested ratio is the square area divided by the circle area, that is: 2/pi  Khalid B. 1 0 5/2π  Khalid B. 0 0 3  Khalid B. 0 0 1  Khalid B. 0 0 0.25  Khalid B. 0 0 0.50  Linda E. 0 0 It was very helpful  Candance S. 0 0 I guess I was the only one who was lost by that whole equation  Hafid A. 0 0 G 