Surface Area and Volume in Three Dimensi

Surface Area and Volume in Three Dimensions : 

Surface Area and Volume in Three Dimensions

Slide 2: 

See and appreciate first the following pictures. Be amazed and have fun!

Slide 11: 

See how strange are they? That is how geometry is applied.

Geometric Solids in Three Dimensions : 

Geometric Solids in Three Dimensions Height Length Width

3D : 

3D Straight-Edged Objects Cones and Cylinders Sphere, Ellipsoid and Torus

Straight-Edged Objects : 

Straight-Edged Objects Solids with straight edges, flat faces (facets), each of which is a plane polygon. A.k.a. polyhedron Tetrahedron Rectangular Pyramid Cube Rectangular Prism Parallelepiped

Problem 1 : 

Problem 1 Suppose you want to paint the interior walls of a room in a house. The room is shaped like a rectangular prism. The ceiling is 3 m above the floor. The floor & the ceiling both measure 1.5 m high by 1 m wide. There is one doorway, the outer frame of which measures 2.5 m high by 1 m wide. One liter of paint can be expected to cover exactly 20 square meter of wall area. How much paint, in liters, will you need to completely do the job?

Cones and Cylinders : 

Cones and Cylinders A cone has a circular or elliptical base and an apex point. A cylinder has a circular or elliptical base, & a circular or elliptical top that is congruent to the base & that lies in a plane parallel to the base. Right circular cone Frustum of Right circular cone Slant circular cone Right circular cylinder Slant circular cylinder

Problem 2 : 

Problem 2 Suppose a cylindrical water tower is exactly 30 m high & exactly 10 m in radius. How many liters of water can it hold, assuming the entire interior can be filled with water? ( 1 liter = cubic decimeter, or the volume of a cube measuring 0.1 m on an edge.) Round the answer to the nearest 10 liters.

Sphere, Ellipsoid and Torus : 

Sphere, Ellipsoid and Torus These are geometric solids with curve spaces throughout. Sphere Ellipsoid Torus

Slide 19: 

Suppose a football field is to be covered by an inflatable dome that takes the shape of a half-sphere. If the radius of the dome is 100 m, what is the volume of air enclosed by the dome in cubic meters? Find the result to the nearest 1000 cubic meters. Problem 3

The Tetrahedron : 

The Tetrahedron Four-faced polyhedron Each face is a triangle Has four vertices Surface Area = √3 s2 Volume = Ah/3 h where A = Base Area

Rectangular Pyramid : 

Rectangular Pyramid Has rectangular base & 4 slanted faces. If the base is a square, & if the apex lies directly above a point at the center of the base, then it is a symmetrical square pyramid. S.A. = t2 + 2t √(s2 – t2/4) V = Ah/3 s = slant height t = edge of the base

The Cube : 

The Cube It is a regular hexahedron Has 12 edges & 6 faces S.A. = 6s2 V = s3

The Rectangular Prism : 

The Rectangular Prism A hexahedron that has 6 rectangular faces. Has 12 edges but not necessarily of the same length. S.A. = 2s1s2 + 2s1s3 + 2s2s3 V = s1s2 s3 S1 S2 S3

The Parallelepiped : 

The Parallelepiped A six-faced polyhedron Each face is a parallelogram Opposite pairs of faces are congruent Has 12 edges S.A. = 2s1s2 sin x + 2s1s3 sin y + 2s2s3 sin z V = hs1s3 sin y h s1 s2 s3 x y z

The Right Circular Cone : 

The Right Circular Cone Has circular base Has an apex point that lies on a line perpendicular to the plane of the base. S.A. = Πrs, where s = √(r2 + h2) = Πr √(r2 + h2) S.A. = Πr2 + Πrs = Πr2 + Πr√(r2 + h2) V = Πr2 h/3 S LATERAL AREA SURFACE AREA OF THE CONE, INCLUDING THE BASE

Frustum of Right Circular Cone : 

Frustum of Right Circular Cone It is when a right circular cone was truncated by a plane parallel to the base. S.A. = Π(r2 + r2) √[h2 + (r2 – r1)2] + Π(r12 + r22), where s = √[h2 + (r2 – r1)2], then = Πs(r2 + r2) + Π (r12 + r22) S.A. = Π(r1 + r2) √[h2 + (r2 – r1)2] where s = √[h2 + (r2 – r1)2], then = Πs(r1 + r2) V = Πh( r12 +r1r2 + r22)/3 S.A. INCLUDING THE TOP & THE BASE S.A. EXCLUDING THE TOP & THE BASE

The Slant Circular Cone : 

The Slant Circular Cone Has circular base Has an apex point that does not pass through the center of the base V = Πr2 h/3 r h

The Right Circular Cylinder : 

The Right Circular Cylinder Has circular base & circular top. The base & the top lie in a parallel planes S.A. = 2Πrh + 2Πr2 or = 2Πr (h + r) S.A. = 2Πrh V = Πr2 h S.A. INCLUDING THE BASE S.A. EXCLUDING THE BASE

The Slant Circular Cylinder : 

The Slant Circular Cylinder It has circular base and circular top The base & the top lie in parallel planes V = Πr2 h r h

The Sphere : 

The Sphere Surface Area: A = 4Πr2 Volume: V = 4Πr3 /3 r

The Ellipsoid : 

The Ellipsoid Volume: V = 4Πr1r2r3 /3 r1 r2 r3

The Torus : 

The Torus Surface Area: B = Π2 (r2 + r1)(r2 – r1) Volume: V = Π2 (r2 + r1)(r2 – r1)2 /4 r1 r2

THANKS FOR LISTENING! : 

THANKS FOR LISTENING! PRESENTED BY: CYNDI M. QUINONES AND JEFERSON D. KARAGDAG