# Finding the Volume and Surface Area of a Pyramid (Video & Practice)

Hello, and welcome to this video about pyramids ! In this video recording, we will explore different types of pyramids and how to calculate their volume and come on area. Let ’ s learn about pyramids ! We can not talk about pyramids without mentioning the most celebrated pyramids in the world, located in Egypt, and built as a grave for egyptian kings. Although pyramids are not quite common, their form is so contact that it seems it is used to make a instruction, possibly that is the reason the egyptian kings used it as their grave. The Louvre Museum in Paris, which is the largest art museum in the global, is besides shaped like a pyramid. As you can see from the images of the egyptian Pyramids, a pyramid is a three-dimensional figure with triangular sides that meet at the edges and at the top to form an apex and it has a polygon as its base. The polygonal base determines the type of pyramid. here you see the pyramid with a triangle as its base is a trilateral pyramid, the pyramid with a rectangle as its base is a rectangular pyramid, and the pyramid with a pentagon as its base is a pentangular pyramid. We besides have hexangular pyramids and heptagonal pyramids, and so on. Let ’ s take a moment and recall what the bulk and surface sphere of a three-dimensional figure means.

The volume of a three-dimensional figure is the measure of how much it can hold, and it is measured in cubic units. The surface area of a three-dimensional calculate is the measurement of the total area that the coat of the figure covers and is measured in square units. Before we can calculate the book and surface sphere of a pyramid, we must know the dispute between the acme and pitch height. The height of a pyramid is the perpendicular duration from the vertex to the base, and the slant height is the distance from the vertex to the center of the bottomland edge of one of the triangular faces. here are the formulas for the volume and surface area of any pyramid. \ ( V=\frac { 1 } { 3 } Bh\ )
\ ( SA=\frac { 1 } { 2 } B+ps\ ) To calculate the volume and surface area of any pyramid we need B, which represents the area of the root, and phosphorus, which represents the perimeter of the base. It is crucial to note, since the nucleotide of a pyramid can be any polygon, we will be using our prior cognition of finding the area and perimeter of different polygons to calculate the volume and open area of a pyramid. \ ( V=\frac { 1 } { 3 } Bh\ )
\ ( SA=\frac { 1 } { 2 } B+ps\ ) \ ( B\ ) = area of base
\ ( h\ ) = acme of pyramid
\ ( p\ ) = perimeter of base
\ ( s\ ) = slant acme Let ’ s look at an exemplar. A trilateral pyramid has an equilateral triangle as its base with side lengths 6 in and a height of 8 in. What is the volume and surface sphere of the pyramid ? To find the open area of a pyramid, we use the formula \ ( SA=B+\frac { 1 } { 2 } ps\ ), where \ ( B\ ) is the area of the infrastructure, \ ( p\ ) is the perimeter of the base, and \ ( s\ ) is the lean acme. Since the base is a triangle, we will use the convention for the area of a triangle to find \ ( B\ ). \ ( \text { Area of a triangle } =\frac { 1 } { 2 } bh\ ), where the \ ( b\ ) and \ ( h\ ) are the base and height of the trilateral base. We will draw a perpendicular line to the base, which is the altitude of the trilateral base and it divides the base of the triangle into 2 equal parts. Since the triangle is now turned into 2 right triangles, we can use the Pythagorean theorem to find the acme. So the Pythagorean theorem is : \ ( c^ { 2 } =a^ { 2 } +b^ { 2 } \ ) And then we ’ re gon na look at our triangle, which says we have : \ ( 6^ { 2 } =h^ { 2 } +3^ { 2 } \ )
\ ( 36=h^ { 2 } +9\ )
We ’ ll subtract 9 from both sides. This gives us : \ ( 27=h^ { 2 } \ ) And then we ’ ll squarely rout both sides. Which gives us : \ ( h=3\sqrt { 3 } \ )
Or : \ ( h\approx 5.2\ ) consequently, the area of the base can be found by doing : \ ( A=\frac { 1 } { 2 } bh\ )
\ ( A=\frac { 1 } { 2 } ( 6 ) ( 5.2 ) \ )
\ ( A\approx 15.6 \text { in } ^ { 2 } \ ) The perimeter of the base is peer to all the sides added together, so : \ ( p=6+6+6=18\text { inches } \ ) nowadays to solve for surface area, all we have to do is plug these values in for our variables. so our surface area is equal to :

\ ( SA=B+\frac { 1 } { 2 } ps\ )
\ ( SA= ( 15.6 ) +\frac { 1 } { 2 } ( 18 ) ( 7 ) \ )
\ ( SA\approx 78.6 \text { in } ^ { 2 } \ ) And all this added together is approximately equal to 78.6 square inches. To find the volume of the trilateral pyramid, we need the area of the base, \ ( B\ ), and the stature of the pyramid, which is 8 inches. so permit ’ s plug this in. So : \ ( V=\frac { 1 } { 3 } Bh\ )
\ ( V=\frac { 1 } { 3 } ( 15.6 ) ( 8 ) \ ) now all we have to do is calculate this out, which is equal to approximately 41.6 cubic inches. \ ( V=41.6\text { in } ^ { 3 } \ ) Let ’ s look at another case. here is a pyramid with a public square root, with side lengths of 5 centimeters. The height of the pyramid is 11 centimeters. What is the volume of the pyramid ? so our volume formula is : \ ( V=\frac { 1 } { 3 } Bh\ ) Before we can calculate the volume of the pyramid, we need to find the area of the base. Since the nucleotide is a square, we use the formula for the area of a feather, which is \ ( s^ { 2 } \ ). so to find our sphere, we ’ rhenium gon na use \ ( s^ { 2 } \ ) and our side length is 5. \ ( A=s^ { 2 } \ )
\ ( A= ( 5 ) ^ { 2 } \ )
\ ( A=25\text { curium } ^ { 2 } \ ) now we can plug this value into our recipe. \ ( V=\frac { 1 } { 3 } Bh\ )
\ ( V=\frac { 1 } { 3 } ( 25 ) ( 11 ) \ )
\ ( V\approx 91.67\text { centimeter } ^ { 3 } \ ) then if we plug this into a calculator, we ’ ll catch that it ’ randomness approximately peer to 91.67 cubic centimeters. But when would we ever need this in real number life ? Well, I ’ molarity glad you asked ! Take a spirit at this future exercise and try it on your own. The roof of a wooden bungalow is shaped like a square-based pyramid. The length of the sides of the square basis are 22 feet and the acme of the trilateral face is 14 feet. Peter wants to paint the roof of his wooden bungalow and needs to determine how much paint he needs to buy. We will assume 1 pint of paint covers 400 feather feet. How much paint does Peter need to buy ? The roof of the bungalow does not include the free-base of the pyramid. consequently, we only need to find the area of the 4 triangular faces. This is called the lateral area. So the lateral area is equal to the surface sphere minus the sphere of the base. So all we need is : \ ( LA=\frac { 1 } { 2 } ps\ ) The circumference of our hearty is equal to 4 times the side length. \ ( LA=\frac { 1 } { 2 } ( 4\times 22 ) ( 14 ) \ ) Which we can plug into a calculator to get : \ ( LA=616 \text { foot } ^ { 2 } \ ) So we will need 616 square feet to be covered. If 1 pint of paint covers 400 square feet, we need to divide 616 by 400 to figure out how much paint we need. so : \ ( 616\div 400=1.54\ )

indeed, Peter will need to buy 2 pints of paint to cover the roof of the bungalow because 1 pint won ’ metric ton be enough and you can ’ t get 0.54 of a pint. then 2 pints of key will cover the whole roof. I hope this video recording on finding the bulk and open area of a pyramid was helpful ! Thanks for watch, and felicitous perusal !

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