Exercise 20 Page 482

Let's analyze the given right triangle so that we may find the value of x.

Let's analyze the given right triangle and recall the corollary that says that the length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

Let's compare the theorem's corollary to the expressions in our figure. In our case, x is the altitude of the triangle, 5 and 7 are the lengths of partial segments of the hypotenuse. AD/DC=DC/DB ⇔ 7/x=x/5 Now we can use the Cross Product Property to find the value of x.

7/x=x/5

CrossMult

Cross multiply

7 * 5 = x * x

▼

Solve for x

Multiply

Multiply

35 = x^2

RearrangeEqn

Rearrange equation

x^2 = 35

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(x^2) = sqrt(35)

SqrtToAbs

sqrt(a^2)=|a|

|x| = ±sqrt(35)

x ≥ 0

x = sqrt(35)