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Core Connections Integrated II, 2015

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Core Connections Integrated II, 2015 View details

Chapter Closure

Exercise 91 Page 635

The surface area of the pyramid is the sum of the base area and its lateral surfaces. The lateral sides are all congruent isosceles triangles.

Total Surface Area: About 161.28 units^2
Volume: 120units^3

Let's draw the described pyramid, including the side length and height in our drawing.

Let's use this to find the requested information.

Surface Area

The surface area of the pyramid is the sum of the areas of its base and lateral surfaces. As we can see, the base is a square, so its area is its side squared.

6^2=36 units^2 The lateral surfaces make 4 congruent isosceles triangles. To determine their areas we recognize that the pyramid's slant height is the height of these isosceles triangles. The slant height is also the hypotenuse of a right triangle where the height and half the side of the base make up its legs.

Using the Pythagorean Theorem, we can calculate the slant height.

a^2+b^2=c^2

a= 3, b= 10

3^2+ 10^2=c^2

▼

Solve for c

Calculate power

9+100=c^2

Add terms

109=c^2

Rearrange equation

c^2=109

sqrt(LHS)=sqrt(RHS)

c=± sqrt(109)

c > 0

c=sqrt(109)

When we know the height of the triangles we can determine their combined area by calculating the area of one of them and multiplying by the number of lateral sides, which is 4. 4(1/2(sqrt(109))(6))=12sqrt(109) units^2 Finally, we will add the base area and lateral surfaces to determine the pyramid's surface area. 36+12sqrt(109) =161.28 units^2

Volume

We also need to calculate the volume of the drawn pyramid. In order to do so, let's first recall the formula for it. V = 1/3 Bh In the formula above B is the area of the base and h is the height of the pyramid. We already know that B is 36 square units and h is 10 units. We only need to substitute that into the formula above and simplify.

V = 1/3 Bh

B= 36, h= 10

V = 1/3( 36)( 10)

Multiply

V = 1/3(360)

MoveRightFacToNumOne

1/b* a = a/b

V = 360/3

Calculate quotient

V = 120

The volume of the pyramid is 120units^3.

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