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McGraw Hill Integrated II, 2012 View details

1. Circles and Circumference

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Exercise 9 Page 719

Find the length of a hypotenuse of the inscribed triangle, which is also the diameter of the circle.

4πsqrt(13) cm

Practice makes perfect

To find the circumference of the circle, we first need to find its diameter. Since one of the sides of the inscribed triangle passes through the center, it is a diameter of the circle. Therefore, the triangle is a right triangle and this side is its hypotenuse.

To find the length of the hypotenuse, we will substitute 12 for a and 8 for b into the Pythagorean Theorem.

a^2+b^2=c^2

SubstituteII

a= 12, b= 8

12^2+ 8^2=c^2

▼

Solve for c

CalcPow

Calculate power

144+64=c^2

AddTerms

Add terms

208=c^2

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(208)=c

SplitIntoFactors

Split into factors

sqrt(16* 13)=c

SqrtProd

sqrt(a* b)=sqrt(a)*sqrt(b)

sqrt(16)* sqrt(13)=c

CalcRoot

Calculate root

4sqrt(13)=c

RearrangeEqn

Rearrange equation

c=4sqrt(13)

Note that since c is the length of the hypotenuse of a right triangle — and the diameter of the circle — we know that it is a positive number. This is why we keep the principal root when solving the equation.

Now that we know that the diameter of the circle is 4sqrt(13) centimeters, we can calculate its circumference.

C=π d

Substitute

d= 4sqrt(13)

C=π ( 4sqrt(13))

Multiply

C=4πsqrt(13)

The circumference of the circle is 4πsqrt(13) centimeters.

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Exercise 9 Page 719

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