Home / IB DP Maths / Analysis and Approach HL / MAA HL Flashcards / SL 3.1 Distance volume and Surface Area

# IB Mathematics AA HL Flashcards- SL 3.1 Distance volume and Surface Area

[qdeck ” ]

[h] SL 3.1 Distance volume and Surface Area

[q] Distance between two points.

[a] This is found by creating a right-angled traingle and using Pythagorean Theorem to find the distance (hypotenuse).
2D: d = $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}$$
3D: d = $$\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}}$$

[q] Cylinder ” Circular Prism”

[a] Volume = $$\Pi(r)^2h$$
Curved Surface Area = $$2\Pi(rh)$$
Total Surface Area = $$2\Pi(rh)+2\Pi(r^2)$$

[q] Cone

[a] Volume = $$\frac{1}{3}\Pi(r)^2h$$
Curved Surface Area = $$\Pi(rL)$$
Total Surface Area = $$\Pi(rL)+2\Pi(r^2)$$

[q] Sphere

[a] Volume = $$\frac{4}{3}\Pi(r)^3$$
Curved Surface Area = Total Surface Area = $$4\Pi(r^2)$$

[q] Pyramid

[a] Volume = $$\frac{1}{3}×Base Area×Height$$
Curved Surface Area = Total Surface Area = Sum of areas of all faces

[x] Exit text

(enter text or “Add Media”; select text to format)

[/qdeck]

## IB Mathematics AA HL Flashcards- SL 3.1 Distance volume and Surface Area

IB Mathematics AA HL Flashcards- All Topics

Scroll to Top