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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

5. Volumes of Pyramids and Cones

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Exercise 37 Page 846

Use the Pythagorean Theorem.

Cornelio is correct. See solution.

Practice makes perfect

Let's analyze the given cone.

We are asked to check if Alexandra and Cornelio are correct. They are calculating the volume of the given cone. First, let's find this volume ourselves. Then let's analyze their solutions.

Volume of Cone

First, let's find the height of the cone h. To do this we will use Pythagorean Theorem for the right triangle.

5^2+h^2=13^2

▼

Solve for h

Calculate power

25+h^2=169

LHS-25=RHS-25

h^2=144

sqrt(LHS)=sqrt(RHS)

sqrt(h^2)=sqrt(144)

SqrtPowToNumber

sqrt(a^2)=a

h=sqrt(144)

Calculate root

h=12

Now, let's use the formula for the volume of a cone.

V=1/3π r^2h

▼

Substitute values and evaluate

r= 5, h= 12

V=1/3π( 5)^2( 12)

Calculate power and product

V=1/3* 300π

MoveRightFacToNumOne

1/b* a = a/b

V=300π/3

Calculate quotient

V=100π

π ≈ 3.142

V≈ 100( 3.142)

Multiply

V≈ 314.2

Therefore, the volume of the given cone is about 314.2 cubic centimeters.

Alexandra's Solution

Alexandra got that the volume of the cone is about 340.3 cubic centimeters. Therefore, her solution is incorrect. Let's find the mistake she made.

Alexandra V &=1/3Bh & =1/3π (5^2)(13) & = 340.3 cm^3

As we can see, Alexandra substituted 13 for h, which is not correct.

Cornelio's Solution

Cornelio's solution is exactly the same as ours. Therefore, he is correct.

Subchapter links

Exercises p.844-847