Perimeter = area + b1 + b2 + c
P = a + b1 + b2 + c
A hexagonal prism is a polyhedron with two parallel bases bounded by congruent hexagons and with lateral faces bounded by parallelograms that connect corresponding sides of the bases. The height h of the prism is any perpendicular segment drawn from a point on the base to the plane containing the other base. If the segments that join corresponding vertices of the bases are perpendicular to the bases, then the prism is a right hexagonal prism. Otherwise, it is called oblique. The perimeter formula of a right regular hexagonal prism is P = 12s + 6h where s = side and h = height
A pentagonal prism * * * * * No. A pentagonal prism has 7 faces and 10 vertices! A rectangular pyramid has 5 faces and 5 vertices.
(1)Best formula to use is as follows -V = h/3(Areatop + √(Areatop*Areabottom) + Areabottom)(2)To find (h) using a tape measure -Areatop => bdAreabase => acLateral edge remaining => e (from top corner to base corner)k = 1 - √(bd/ac)H= √([e/k]² - [a/2]² - [c/2]²)h = HkV = H/3*(ac-bd+bd*k)(3)Lets say Top is a rectangle with sides b & dand bottom is a rectangle with sides a & c respectively.Let height be hin that case the volume of Truncated Pyramid with rectangular base will be -V = 1/3((a²c-b²d)/(a-b))hBUT BE CAREFUL - a,b,c,d are not all independent variables (one depends on the others) so this answer is misleading!!!Proof -Suppose the height of Full Pyramid is HFrom parallel line property(H-h)/H = b/aRearrangingH = ah/(a-b) --------------------(1)AlsoSince V=1/3 Base area X HeightVolume of full pyramid = 1/3 X ac X HVolume of removed Pyramid = 1/3 X bd X (H-h)So volume of truncated part V = 1/3(acH-bd(H-h))=1/3((ac-bd)H + bdh)From (1)V = 1/3((ac-bd)ah/(a-b) + bdh)reducing and rearranging we getV = 1/3((a²c-b²d)/(a-b))h(4)In case the truncated solid forms a prism instead, we have following formula -V = ( h/6)(ad + bc + 2ac + 2bd)Proof -Fig(1)Fig(2)Lets divide the fig(1) into four different shapes as shown in fig(2)VA = Volume of cuboid = bdhVB = Volume of prism after joining both Bs= ½ X base X height X width = ½ (a-b) (d)(h)VC = Volume of prism after joining both Cs = ½ X base X height X width = ½ (c-d) (b)(h)VD = Volume of rectangular pyramid after joining all Ds = 1/3 X base area X height =1/3 (a-b) (c-d) hThen V = VA + VB + VC + VDOr, V = bdh + 1/2(a-b)dh +1/2(c-d)bh + 1/3(a-b)(c-d)hArranging and simplifying we get -V = ( h/6)(ad + bc + 2ac + 2bd)
You must be thinking of a triangular prism. In that case, c is the length of the third side of the triangle at the end of the prism.
area of base x height area of base x height
I dont know:d
There must be a typo in this question, "Why does the formula for finding the surface area of arectangular prism is helpful?" What does that even mean?
its not i dont no why
Volume = 1/2*(a+b)*h*l where a and b are the lengths of the parallel sides of the trapezium, h is the height of the trapezium, and l is the length of the prism.
If you mean volume of a trapezoidal prism then it is: 0.5*(sum of parallel sides)*height*length
I am not sure that a rectangular prism is in any position to care!
the volume of a trapezoidal prism is equal to the height times the base area of the trapezoid. First you find the area of trapezoid h(a+b)/2 h is the height of the trapezoid, not the height of the prism a is the length of the top b is the length of the bottom Then you find the volume of the trapezoidal prism with this formula H*h(a+b)/2 H is the height of the prism. Multiply H by the area of the trapezoid that you found in step one.
Surface Area = 2 × Base Area + Base Perimeter × Length
A trapezoidal prism has 8 vertices:A trapezoid has 4 vertices.A trapezoidal prism is composed of 2 trapezoids. 2 X 4 = 8.
It is helpful because when you do the problem you know what to do.