1. A cone was formed by rolling a thin sheet of metal in the form of a sector of a circle 72cm in diameter with a central angle of 210deg. What is the volume of the cone?

2. Through one vertex of an equilateral triangle with sides a draw a line l perpendicular to the altitude upon the opposite side. Find the volume and area of the figure developed by revolving the triangular about the line l.

3. The base of a right pyramid is a square of side 12 cm and the dihedral angle between its base and a lateral face is 60 degrees. Find its height and volume.

1. Dihedral angleof three vectors,definedas an exterior sphericalangle. The longer and shorter black segments are arcs of the great circles passing through and and through and , respectively. In geometry, adihedralor torsionangleis theanglebetween two planes.2. Right Pyramid3. Apyramidthat has itsapexaligned directly above the center of thebase.Solid view: right pyramid with asquarebaseFrame view: right pyramid with asquarebase

Area of an Octagon FormulaArea of an Octagon is the amount of space occupied by the octagon. An octagon is a 8 sided polygon. A regular octagon has equal sides and internal angles. The internal angle in any vertex of a regular polygon is 135o.

TheArea of an Octagon Formulais,

Solid GeometryA pyramid whose altitude is 5 feet weighs 800lbs . At what distance from its vertex must it be cut by a plane parallel to its base so that the two solids of equal weight will be formed?

i have noideahow to solve this1. I assume that the material of the pyramid is homogenous. Then the weight correspond directly to the volume:

where k is a real constant.

2.denotes the base area of the complete pyramide, b the base area of the smaller pyramide. Both areas are similar. Then

with

3. h denotes the height of the smaller pyramid. Then the volume is calculated by:

with

4. Sinceandare similar you can use the proportion:

5.Plug inthis term into the equation at #3:

Solve for h.

Best Answer:Weight of cutoff portion = 1/2 wt of full solid. =. its (ht)^3 = (ht of original)^3 /2 = 128.ht of cutoff =(128) = 5.04 ft (approx).Ravindra P 5 years ago Just plug in x for the height of a pyramid with the same properties whose weight is 450 pounds. (1/2 of 900) then x will be your answer. You can use a volume equation if you know the weight/cu ft of material and sub the volume equivalent for the weight. for instance if the weight of the material is 5 pounds per cu ft then 90 cu ft will be the answer for the volume.

19.a pyramid whose altitude is 4 ft weighs 600lb. At what distance from its vertex must it be cut by a plane parallel to its base so that two solids of equal weight will be formed?answer:3.1748 ft####7 ==(p370) 9. Find the volume and area of a figure formed by revolving an equilateral triangle about an altitude, the sides of the triangle beings.10. Find the area and volume of the figure developed by an equilateral triangle with sidesiif it is revolved about one of its sides.

An equilateral triangle of side 14 centimeters is revolved about an altitude to form a cone. What is the number of cubic centimeters in the volume of the cone? Express your answer to the nearest whole number, without units.

I just want the answer. Thanks! :) ADVANCED AMC8 GEOMETRY QUESTION/MATH -Jai, Saturday, August 10, 2013 at 2:02amNote that an equilateral triangle is a triangle where all lengths of the sides are equal.If you revolve the triangle about its altitude (or the height), you'll generate a cone (just try to imagine or draw the figure). Therefore, the height of the cone is equal to the height of triangle, and its radius is equal to half of one side of the equilateral triangle, which is 14/2 = 7 cm (radius).To get the height, use pythagorean theorem:c^2 = a^2 + b^2wherec = hypotenuse (in this case, 14 cm)a = height of triangle (which is also cone height)b = base (which is also the cone radius)Substituting,14^2 = a^2 + 7^2a^2 = 14^2 - 7^2a^2 = 147a = 7*sqrt(3)Finally, we get the volume of cone. Recall that the volume of cone is justV = (1/3)*(pi)*(r^2)*hSubstituting,V = (1/3)(3.14)(7^2)(7*sqrt(3))

Now solve for V. Units in cubic centimeters.1 pt. = .125 galPlane Solid Mensuration Problem.?Situation: A vessel is in form of an inverted regular square pyramid of altitude 9.87 inches and base edge 6.27 inches The depth of the water it contains is 6 inches.Question:a.) How much will the surface rise when 1 pt of water is added? (1 gallon = 231 cubic inches)b.) Find the wetted surface when the depth of the water is 9.23 in.

How do I do this step by step if the answer is:a.) 1.6108 inchesb.) 113.57 square inchesBest Answer:a) This part is done by the difference in volume between two pyramids. You can get the width of the water surface by using similar triangles. From this the original volume of the water in cubic inches. Now work out the ratio of new to original volumes. The cube root of this is the scale factor by which both height and width have increased (although it is only the height you are asked for).

b) Again, use similar triangles to find the width of the square that forms the water surface. Then use Pythagoras rule to find the length of the line up the middle of a side. From these you can work out the area of a side and multiply by 4.

EDIT. I also get an answer of about 1.55 inches for the first part.

Since several calculations need to be done on pyramids of various sizes you might consider appealing to the properties of similar solids. You will know that if solids are similar with similarity ratio r, then corresponding distances are in the ratio r, areas in the ratio r and volumes in the ratio r. If we calculate the area/volume of the vessel we can then use similarity to do efficient calculations on other pyramids .

viz

For the vessel V = x 6.27 x 9.87 = 129.3394

Volume V when depth is 6 is given by V/129.3394 = (6/9.87) V = 29.0558

New volume when pint is added is 29.0558 + 231/8 = 57.9308

New altitude h comes from (h/6) = 57.9308/29.0558 h = 7.5517

First, we find the s/a of the vessel for which we need the altitude of a lateral .

By Pythagoras this is (9.87+3.135) = 10.3559

Lateral surface area of vessel = 4 x x 6.27 x 10.3559 = 129.8630

The lateral s/a S for h=9.23 is given by S/129.8630 = (9.23/9.87) S = 113.56Indica 5 years ago0Thumbs up0Thumbs downCommentReport Abuse Knowing the height and base edge, s, of a square pyramid, it's volume is given by:

V = (1/3)hs

Proportionally, when the water is 6 inches deep inside the inverted pyramid, the base edge of the water, s, is given by:

s/6.27 = 6/9.87

s = 6(6.27/9.87) = 3.81155 inches

From this, the current volume of water is:

V = (1/3)hs = (1/3)(6)(3.81155) = 29.056 in

One pint of water is 1/8 of a gallon, i.e., 231/8 = 28.875 in.

From above, the length of the base edge, s, of the new water level is given by:

s = h(6.27/9.87)

where "h" is the height of the water. Thus, the equation for the volume becomes:

V = (1/3)hs = (1/3)h[h(6.27/9.87)] = (1/3)(6.27/9.87)h

Therefore, given a new volume of 29.056 + 28.875 = 57.931 in, the height of the water, h, becomes:

V = (1/3)(6.27/9.87)h

h = 3V(9.87/6.27)

h = [3V(9.87/6.27)] = [3(57.931)(9.87/6.27)] = 7.552 in

Therefore, the surface rise is 7.552 - 6 = 1.552 inches. (Obviously, I'm not getting 1.6108 inches; I believe my calculations are correct.)

b) (I'm assuming the "wetted surface" only includes the four surfaces of the square pyramid, and not the surface of the water.) Given a water depth of 9.23 inches, the base edge, s, is:

s = (9.23)(6.27/9.87) = 5.8634 inches

Using the Pythagorean theorem, the slant length of the pyramid is simply:

l = [(h) + (s/2)] = [(9.23) + (5.8634/2)] = 9.6844 inches

Thus, the total area of the four wetted surfaces is:

A = 4(1/2)(5.8634)(9.6844) = 113.57 in2

Best Answer:height to base ratio: h/b = 9.87/6.27b = 6.27h/9.87

volume of pyramidv = bhv = (6.27h/9.87)hwhen h = 6in; v = (6.27h/9.87)h = 29.0558in

after adding a pint: 1 pt = 28.875inv = 29.0558 + 28.875 = 57.9308in

v = (6.27/9.87)hlet rise = Hv = (6.27/9.87)(6 + H)57.9308 = (6.27/9.87)(6 + H)H = 1.55168H = 1.55 in----------------- (rises from 6in to 7.55in level)

A vessel is in the form of an inverted regular square pyramid of altitude 9.87 in. and a base edge 6.27 in. The depth of the water it contains is 6 in. How much will the surface rise when 1 pint of water is added? (One gallon = 231 cubic inches) Find the wetted surface when the depth of the water is 9.23 in.

Solution:

To illustrate the problem, let's draw the figure and label as follows

Photo by Math Principles in Everyday Life

The volume of the empty vessel is calculated as follows

When the depth of the water is 6 in, the volume is calculated by ratio and proportion as follows

When 1 pint of water is added, the volume in cubic inches is

Therefore, after the addition of 1 pint of water, the height of water can be calculated by ratio and proportion as follows

The difference of the height of water after the addition of 1 pint of water is

When the depth of the water is 9.23 in, the length of a square base is calculated by ratio and proportion as follows

By Pythagorean Theorem, the slant height will be equal to

Photo by Math Principles in Everyday Life

Therefore, the wetted surface of a vessel is