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If a square based pyramind has sides of 20 what is the the height.

If a square based pyramind has sides of 20 what is the the height. All sides are 20 and the triangles are obviously, equilatteral. Thanks :)

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A pyramid has a square base with sides 8" and a slant height ...

A pyramid has a square base with sides 8" and a ... A pyramid has a square base with sides 8" and a slant height of 5". total area = a ... 20 la suta din rest sunt ...
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A pyramid has a square base with side s. The height of the ...

A pyramid has a square base with side s. The height of the pyramid is 2 ... height of the pyramid is 2/3 that of its side. ... this is a square and the height is ...
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A square pyramid each side of the base has a length of 60 m ...

A square pyramid each side of the base has a length of 60 m and the height of each triangle is 20 m. ... Each side of the base has 60m and the height is 20 m.
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Square Pyramid -- from Wolfram MathWorld

A square pyramid is a pyramid with a square ... slant height s of a right square pyramid of side length a and height h are e ... be tangent to the sides, so ...
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Spinning Square Pyramid - Math Is Fun

Square Pyramid Facts. ... The 4 Side Faces are Triangles; The Base is a Square ; It has 5 Vertices (corner points) It has 8 ... Volume = 1 3 × [Base Area] × Height ...
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[11.04] A square pyramid has a base side of 5 in.... - OpenStudy

The lateral area of the pyramid is _____ in.². 20 40 ... [11.04] A square pyramid has a base side of 5 in. and a slant height ... A square pyramid has a base side of ...
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Pyramid A is a square pyramid with a base side le... - OpenStudy

Pyramid A is a square pyramid with a base side length of 7 inches and a height of 6 inches. Pyramid B has a ... it has a square base. and all the slant sides go and ...
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Suggested Questions And Answer :

If a square based pyramind has sides of 20 what is the the height.

Take one of the equilateral triangles and drop a perpendicular from the top to the base. The perpendicular bisects the base and forms two back-to-back right-angled triangles. The length of the perpendicular is sqrt(20^2-10^2) by Pythagoras. That's sqrt(300)=10sqrt(3). Now view the pyramid side on and drop a perpendicular from the apex where the four equilateral triangles meet on to the square base. The perpendicular meets the base at the centre of the square. Take just one of the triangular sides and view it joining the perpendicular from the apex. We have a right-angled triangle where the hypotenuse has length 10sqrt(3) and the base is half the side of the square so its length is 10. The third side of this internal triangle is the height of the pyramid=sqrt(300-10^2) where 300 is (10sqrt(3))^2. So the height is sqrt(200)=10sqrt(2)=14.14 approx.
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how do you determine height of a triangle

The side of the square is 10 inches. The area of the square is thus = side * side = 10 * 10 = 100 inches^2 The square and triangle have the same area, so the area of the triangle is also 100 inches^2. The formula for the area of a triangle is 1/2 * base * height. Since the base of the triangle is 10 inches and its area is 100 inches^2, we have: 100 = 1/2 * 10 * height 100 = 5 * height height = 100 / 5 height = 20 inches Hence, the height of the triangle is 20 inches.
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Open rectangle box (no top) with square base and rectangular sides: total area of 5 faces = 975 sq.m. Find max volume

We can call the side of the square base B and the height H. The total surface area of the open box is 4HB+B^2=975, so H=(975-B^2)/4B or B^2+4HB+4H^2-4H^2=975, (B+2H)^2=975+4H^2 and B=sqrt(975+4H^2)-2H. The volume V=H(sqrt(975+4H^2)-2H)^2. Let X=sqrt(975+4H^2), X^2=975+4H^2 and 2XdX/dH=8H, so dX/dH=4H/X. V=H(X-2H)^2. If we differentiate V with respect to H we get: dV/dH=(X-2H)^2+2H(X-2H)(dX/dH-2)= (X-2H)^2+4H(X-2H)(2H/X-1)= (X-2H)^2-4H(X-2H)^2/X=(X-2H)^2(1-4H/X)/X. When dV/dH=0 V is at a maximum or minimum: so X=2H or 4H. (X=2H is not valid, because 4H^2=975+4H^2, resulting in 0=975, which is never true.) So 16H^2=975+4H^2 and 12H^2=975 so H=sqrt(975/12)=sqrt(325/4)=5sqrt(13)/2=9.014m approx. From this value of H, B=sqrt(975+4H^2)-2H=sqrt(975+325)-5sqrt(13)=10sqrt(13)-5sqrt(13)=5sqrt(13)=18.03m approx. V=5sqrt(13)/2*325=2929.51 sq m. To prove this is a maximum, we need to take H=9 (close to 9.014) making B=sqrt(975+324)-18=18.04m. V=2929.50 sq m. Now take H=9.1, so B=17.94 and V=2929.41 sq m. So maxV=2929.51 sq m, height=9.014m and base side is 18.028m. The height is half the base side length and the maximum volume is B^(3/2)/2=325^(3/2)/2.  
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how do you find the area of a prism???

All prisms take a two-dimensional shape and use it as the top and bottom faces in parallel. The top and bottom are then joined to make a solid, so that all the sides are parallelograms. These parallelograms can be rectangles or even squares. The area of the prism is the surface area of the solid, usually including the top and bottom surfaces. The top and bottom faces can each be the same triangle, quadrilateral (including squares, rectangles, trapezoids, etc.), or polygon (star, pentagon, hexagon, etc.). If the top and bottom have n sides, there will be n parallelograms forming the length of the prism. The area of each parallelogram is base times vertical height, where the base length is the length of the particular side of the base figure on which the parallelogram stands. The volume of the prism is the area of the base times the vertical height. For example, if the top and bottom faces are the same regular hexagon (like a pencil), there will be 6 rectangles forming the sides of the prism, each with the same area, A. The surface area of the prism is 6A plus twice the area of the base, or bottom face.
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equilateral math equation help

The side of the square is the square root of its area=sqrt(16)=4. The area of the triangle is half the base times the height. So we need the height of the equilateral triangle. The perpendicular from vertex E bisects AD and the angle AED. The equilateral triangle now consists of two back to back right-angled congruent triangles. Angle EAD=60 (same as  the other two angles of triangle AED) so half the angle is 30 degrees. The common side of the two right-angled triangles is the height of the equilateral triangle and has length=sqrt(4^2-2^2), by Pythagoras, because the hypotenuse is length 4 (same as the side of the square) and the shortest side is 2, half of AD. The height is sqrt(12)=sqrt(4*3)=2sqrt(3). The area of the triangle is 1/2*4*2sqrt(3)=4sqrt(3)=6.93 sq units approx. Another way of finding the area is to take the two right-angled triangles and join them together along their common hypotenuse forming a rectangle of sides 2sqrt(3) and 2. The hypotenuse is a diagonal of the rectangle. The area of the rectangle (2sqrt(3)*2=4sqrt(3)=6.93) must be the same size as the area of the equilateral triangle, because the rectangle and the equilateral triangle are made up of the same two right-angled triangles.
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What is the radius of a right circular cone with a height of 10 cm and surface area of 125 square cm

me take it, yu meen the area av the side av the kone=125 & yu dont overlap the sides (maebee tape edjes tugether) side area av kone=triangel with base=rim-leng av serkel triangel base=2pi*radius triangel side area=(1/2)base*slant-hite=pi*radius*slant-hite slant hite av kone=sqrt(rad*rad + hite*hite) slant hite=sqrt(100 + rad*rad) area=125=sqrt(100+rad*rad)*pi*rad area/pi=125/pi=rad*sqrt(100+rad*rad)=29.7887382 bi tri & tri agin (iterate) me get rad=2.863758
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candle fundraiser to brighten lives

I have this as a math question now where are the answers..
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what can you say about the triangles formed by the corners of the square outside the octagon?

There are only four triangles each occupying a corner of the square encompassing the octagon. The internal angle of each of the eight isosceles triangles inside the octagon is 360/8=45 degrees. So the other two angles are (180-45)/2=67.5 degrees. Each of the four external triangles is a right-angled isosceles triangle, so the other angles are 45 degrees and the hypotenuse is the length of the side of the octagon. The four hypotenuses are BC, DE, FG and HA. If each octagon triangle has a base length of 10 (side length of the octagon) then the hypotenuse of the corner triangles is 10 and the other two sides are given by x in the equation 2x^2=100 (Pythagoras), so x=sqrt(50)=7.071 approx., and the area of each corner triangle is 1/2x^2 (1/2 base times height), which is 50/2=25. The length of the side of the enclosing square is 2*7.071+10=24.142. If the radial sides of the octagon triangles are 10, the side of the octagon is 7.6537 (20sin(22.5 degrees)) and the side of the triangle is about 5.412 (sqrt(7.6537^2/2)). The side of the square is the side of the octagon plus 2*the side of the corner triangle, 7.6537+2*5.412=18.478. The question doesn't specify what the measurement 10 is and doesn't specify what property of the triangle x is.
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the height of a squere based pyramid is 12cm

Let v1 and v2 be the volumes of the larger and smaller pyramids, and let A1 and A2 be their base areas. v1=A1h1 and v2=A2h2 where h represent the heights. The masses are m1 and m2. We have the following values: h1=12,cm, h2=7.2cm, m1=37.5g. Density=mass/volume, and since the pyramids are similar their density is the same, so d=m1/v1=m2/v2, and m1/m2=v1/v2=A1h1/(A2h2). So 37.5/m2=12A1/(7.2A2). However, the pyramids are geometrically similar. This means that the ratio of the heights of the two pyramids is also the ratio of the sides of their square bases. Thus, if a1 is the base length of the larger pyramid then the base area is a1^2=A1 and A2=a2^2, and h1/h2=a1/a2. The ratio of the areas=A1/A2=a1^2/a^2. But h1/h2=12/7.2=1/0.6=5/3, so A1/A2=(h1/h2)^2=25/9. Therefore, 37.5/m2=12/7.2*25/9=125/27 or (5/3)^3. So m2=37.5*27/125=8.1g.
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How do you find the volume if you have the height and base of a pyramid

B= 4 square root of 3 m squared. V=1/3 Bh =1/3 (4 square root of 3) (9) V=62.35 cubic meter  
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