Exercise 5 Page 754

We are asked to find x. Let's start by paying close attention to the diagram. Since BC is tangent to the circle, the angle it makes with the radius at the point of tangency is a right angle. Therefore, the triangle formed is a right triangle.

We can see that the lengths of the legs are 30 and x. The length of the hypotenuse is x+18. If we substitute these values in the equation of the Pythagorean Theorem, we will be able to find the value of x. Let's do it!

a^2+b^2=c^2

SubstituteValues

Substitute values

30^2+ x^2=( x+18)^2

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Solve for x

ExpandPosPerfectSquare

(a+b)^2=a^2+2ab+b^2

30^2+x^2=x^2+2x(18)+18^2

CalcPowProd

Calculate power and product

900+x^2=x^2+36x+324

SubEqn

LHS-x^2=RHS-x^2

900=36x+324

SubEqn

LHS-324=RHS-324

576=36x

DivEqn

.LHS /36.=.RHS /36.

16=x

RearrangeEqn

Rearrange equation

x=16

We found that the value of x is 16.