The base of a right pyramid is an equilateral triangle of side 4cm and slant edge 5cm. Find volume of pyramid?
The perpendicular drawn from the apex of the pyramid to the triangular will hit it right at the center of the perpendicular drawn from any vertex of the triangle to the opposite side.[math]\text{Area of equilateral triangle} = \dfrac{\sqrt{3}}{4}\times 4^2 = 4\sqrt{3}[/math][math]\dfrac{1}{2}\times 4\times x = 4\sqrt{3}[/math][math]\dfrac{1}{2}x=\sqrt{3}[/math][math]h=\sqrt{5^2-(\sqrt{3})^2}=\sqrt{22}[/math][math]\text{Volume} = \dfrac{1}{3}\times \text{base area}\times\text{height}[/math]This actually reminds of the area of a tetrahedron, which is what the figure shows.[math]V=\dfrac{1}{3}\times 4\sqrt{3}\times \sqrt{22}=\dfrac{4\sqrt{66}}{3}[/math]
Find the angle between plane/edge and base?
AD = 2 (5m)cos(53.13 deg) = 6.00 meters; apparently triangles ADE and BCF are congruent. That in itself doesn't prove that sides AB, CD, and EF are parallel; I'm simply going to assume that they are. Call the midpoint of AD by the name O. The distance from E to O is (5m)sin(53.13) = 4 meters. Call the midpoint of EF by the name P. If EF is parallel to AB and CD, then it is in fact horizontal, and at a uniform elevation above plane ABCD. The horizontal component of the distance OP is 8 meters. Note that PE = 6 meters, so the horizontal component of EO is 2 meters, even thought the actual distance EO is 4 meters. Since 2/4 = cos(60 degrees), the 60 degrees is the angle between plane ADE and plane ABCD. The vertical component of EO is sqrt(16-4) = 2 sqrt(3) meters; this is the uniform vertical elevation of EF. The angle between AE and the vertical is arccos(2 sqrt(3)/5) = 46.15 degrees; then the angle between AE and the horizontal plane ABCD is 90-46.15 = 43.85 degrees.
In a regular square pyramid,the lateral edge is inclined at an angle 60 dgree to the plane of the base.?
Hi, The dihedral angle between Base and face is same as the inclination of lateral edge to the plane of the base of regular pyramid and let that be Then cos(P) = - a^2/4s^2, refer http://mathforum.org/dr.math/faq/formula... P = arccos[- cos^2(B)] = arccos[ - cos^2(60)] = arccos[ - 0.5^2] = 180 - arccos(0.25) = 180 - 75.52 = 104.48 degrees Good luck
How do I calculate the height of a triangular pyramid if I have only the length of the three sides of the base of the pyramid?
The only way this problem has a solution is if you mean a regular tetrahedron — a triangular pyramid with equilateral triangles for the base and all three sides — since a pyramid with an arbitrary triangle for a base could have any height.To determine the height of a tetrahedron, we can imagine dropping a vertical line through the peak to the base, connecting that point to a corner, and back up that edge to the peak, and solving for the legs of that triangle. (We could also create a triangle from peak down to base to midpoint of a side.)In the lefthand diagram below, I used the characteristics of a 30°-60°-90° right triangle to determine the distance from the center — circumcenter, but irrelevant for an equilateral triangle — to a vertex, if edge length is s, as [math]\frac{s}{\sqrt{3}}[/math] or [math]\frac{s\sqrt{3}}{3}[/math].In the righthand diagram, I used the Pythagorean theorem; solving for h yields:[math]h=\sqrt\frac{2}{3}s=\frac{\sqrt{6}}{3}s[/math]… which matches the Google results for “height of a tetrahedron”.
A block of granite is in the form of the frustum of a regular sure pyramid whose upper and lower base edges a?
The frustrum is everything but the tip of the pyramid right? Flat bottom and flat top? I wish I could draw on here but I think what your saying is the lower base is 7x7 and the and the upper base is 3x3? The faces are 31 feet from lower to upper at a 62 degree angle. So we need to find the heigt from bottom to top. Draw a triangle. Since all of the angles in a triangle = 180 degrees we can figure out what it will look like since we already have on degree measurement. One angle is 62 180-62 =118 We have 118 degrees left to use Another angle is 90 degrees 118-90=28 So this tells us the last angle is 28 degrees So draw a horizontal line on paper On the right end point of the line draw a line straight up at a 90 degree angle Then from the left end point connect the two lines. the lower left angle is "62" degrees and the top angle is "28" degrees The line traveling at 62 degrees is the 31 ft line that connects the lower and the upper base, this is also the hypoteneuse. We now have three angles and a side. So we can use the AAS (Angle-Angle-Side) technique to figure out the other two sides. Label your triangle like this Angle A=62 degrees Angle B= 28 Degrees Angle C= 90 Degrees Use lowercase for your side labels Side a= the vertical side Side b= base Side c= hypoteneuse (31 ft side) We use the law of sines: a/SinA = b/sinB = c/sinC We only need the height to solve the problem so we will solve for side "a" Since we know that Side c = 31 ft and Angle C= 90 and angle A=62 degrees we set it up like this a/Sin 62 = 31/Sin 90 31/Sin 90= 31 So now we have a/Sin 62 = 31 Solve for "a" Sin 62 x 31 = a = 27.37 So the distance between the lower and upper base is 27.37 ft (height) To find the volume of a frustrum use these variables H=Height (27.37) A1= Area of lower base (49 or 7x7) A2= Area lower base (9 or 3x3) The volume formula is (H/3) x (A1+A2+ sq root of (A1xA2) So plug in the numbers (27.37 / 3) x (49+9+21)= 720.743 ft cubed There it is 720.743 ft cubed... Sorry this took so long, if I could draw it would have went a lot faster.
Volume of a Pyramid with base area 36 and faces are equilateral triangle?
Thankfully, the base is a square (area = 36 --> base edges = 6). As difficult as pyramids can get, square pyramids are the best ones. The volume of a pyramid = Base area * height / 3. So we just need the height. Not as easy as it sounds, but not too bad. The slant height of the pyramid (basically the altitude of one of those equilateral triangles) is 3√3. This is found with a simple 30-60-90 relationship in the triangle. Imagine the height dropping straight down from the top vertex to the center of the square. Then the center of the square meet the bottom of that slant height, forming a right triangle floating inside the pyramid. The hypotenuse of this right triangle is 3√3, the leg that lies on the square base is 3 (half the square's edge), so by using Pthagorean theorem, you find the pyramid's height is (3√3)² - 3² = h² ==> h = √18 = 3√2. So the volume is: V = 36 * 3√2 / 3 = 36√2 units³
Given the base of a pyramid is a square and its lateral faces form 45 degree angles with the base, how do I prove the pyramid is a regular pyramid?
Let ABCD is the square base of side A.Let P, Q, R & S are mid points of sides AB, BC, CD, DA.Let E the apex of the pyramid. Let F is perpendicular drop of E on base.Consider vertical triangles ∆EFP & ∆EFR.Angle EFP= angle EFR= 90°Angle EPF= angle ERF=45°EF is side common to both.∆EPF & ∆EFR are congruent.PF=FR=a/2In similar way, QF=FS (∆EFQ & ∆EFS congruent)Also, QF=FR (∆EFQ & EFR congruent)Finally FP=FQ=FR=FS=r(say).P,Q,R,S lie on circle of radius r.PQ=QR=RS=SP=a/√2=sides of rhombus PQRS (in fact square). PR and QS are mutually orthogonal.Now consider ∆ FPS & ∆FRS.PF=FR (already proved)PS=√(PA.PA+AS.AS)=√(RD.RD+SD.SD)=RSFS is common side.Triangles are congruent.Angle PFS=angle SFR= angle RFQ=angle QFP=360/4=90°.F is the intersection of PR and QS.F is the centre of square base, to which EF is vertical.The pyramid is a right pyramid with base of regular polygon. That is the pyramid is regular.
If a regular tetrahedron has edges of length 8 centimeters, what is the tetrahedron's volume?
A regular tetrahedron is made up of four congruent equilateral triangles. It can be thought as as a pyramid with a base that is an equilateral triangle.The volume of a pyramid is [math]\frac{1}{3}Ah[/math], where [math]A[/math] is the area of the base and [math]h[/math] is the height of the pyramid perpendicular to the base.The area of an equilateral triangle is [math]\frac{\sqrt{3}}{4}s^2[/math], where [math]s[/math] is the length of a side or edge.The height of a regular tetrahedron is [math]\sqrt{\frac{2}{3}}s[/math].Therefore, the volume of the regular tetrahedron is:[math]V = \frac{1}{3}·\frac{\sqrt{3}}{4}s^2·\sqrt{\frac{2}{3}}s[/math][math] =\frac{\sqrt{2}}{12}s^3[/math].Fos s = 8 cm, the volume is [math]\frac{128\sqrt{2}}{3}\text{ cm³}[/math], which is approximately 60.34 cm³.
A prism has an angle of 60 degrees and its incident angle is 48 degrees. What is the least deviation angle?
Least angle of deviation (or angle of minimum deviation) is 36 degrees.Angle i denotes angle of incidence.Angle e denotes angle of emergence.i + e = Angle of prism + Angle of deviationHere, in case of minimum deviation, where the prism angle is 60 degrees, i and e would be equal to each other.So, 2i = A + Angle of deviation2 (48) = 60 + Angle of deviation96 - 60 = Angle of deviation36 degrees is equal to the angle of deviation. Below is the proof for the formula and the above calculations, in case it is necessary.Note- MN and TN are normal rays (perpendicular to AB and AC respectively), shown with dashed lines. As for the quadrilateral QNRA, its interior angles sum up to 360 degrees and that quadrilateral angle sum property is used here.