Exercise 11 Page 385

Let's modify the given figure a bit, keeping in mind that the question asks for the length of a segment. We focus on the length measures given on the figure and ignore the angle measures.

• We know that the incenter of a triangle is equidistant from the sides, so let's mark that FH, DH, and GH are congruent.
• Let's also highlight the segment BD — the length of which we are asked to find — and shade the right triangle △ BDH with this segment as one of the legs.
• Finally, let's shade the right triangle △ AFH, where the length of two sides are given.

We can find BD in three steps.

Finding FH.

We can use the Pythagorean Theorem in the green triangle to find FH.

a^2+b^2=c^2

SubstituteExpressions

Substitute expressions

AF^2+FH^2=HA^2

SubstituteII

AF= 24, HA= 25

24^2+FH^2= 25^2

▼

Solve for FH

CalcPow

Calculate power

576+FH^2=625

SubEqn

LHS-576=RHS-576

FH^2=49

CalcRoot

Calculate root

FH=7

Finding DH.

We already noticed that FH is congruent to DH, so their measure is the same. DH=FH=7

Finding BD.

The red right triangle HB=11 is given on the diagram, and we now know the length of DH. We can use the Pythagorean Theorem, this time in the red triangle, to find BD.

a^2+b^2=c^2

SubstituteExpressions

Substitute expressions

BD^2+DH^2=HB^2

SubstituteII

DH= 7, HB= 11

BD^2+ 7^2= 11^2

▼

Solve for BD

CalcPow

Calculate power

BD^2+49=121

SubEqn

LHS-49=RHS-49

BD^2=72

CalcRoot

Calculate root

BD=sqrt(72)

RoundDec

Round to 1 decimal place(s)

BD≈ 8.5

The length of BD is BD=sqrt(72), or approximately 8.5.