Exercise 13 Page 730

Let's take a look at the given square pyramid.

To calculate the volume of a pyramid, we can use a known formula where B is the area of the base and h is the height. V=1/3Bh The base of the pyramid is a square with the side length of 23mm. Let's calculate its area! B = s^2 ⇒ B = 23^2 B= 529mm^2 We are given that the slant height l is 24mm. Now let's focus on the right triangle created within the pyramid.

The slant height of the pyramid is the hypotenuse of the triangle. The legs of the triangle are the height of the pyramid h and half of the side length of the pyramid.

Notice that we can find the height of the pyramid using the Pythagorean Theorem.

(23/2)^2+ h^2= 24^2

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Solve for h

CalcQuot

Calculate quotient

11.5^2+h^2=24^2

CalcPow

Calculate power

132.25+h^2=576

SubEqn

LHS-132.25=RHS-132.25

h^2=443.75

SqrtEqn

sqrt(LHS)=sqrt(RHS)

h=sqrt(443.75)

Note that, when solving the above equation, we only needed to consider the principal root because h is a positive number. Finally let's substitute the base area and the height of the pyramid into the formula and calculate V.

V=1/3Bh

SubstituteII

B= 100, h= sqrt(119)

V=1/3( 529)( sqrt(443.75))

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Simplify right-hand side

Multiply

V=1/3(529sqrt(443.75))

MoveRightFacToNumOne

1/b* a = a/b

V=529sqrt(443.75)/3

UseCalc