GEOMETRIC MEAN THEOREM

Solve for the missing variable.

Problem 1 :

Solution:

ABC is a right triangle and AD is the altitude drawn from the right angle A.

So, AD is the geometric mean between BD and DC.

AD = √(BD ⋅ DC)

x =  √(12 ⋅ 8)

x =  √96

x = 9.79

Problem 2 :

Solution:

ABC is a right triangle and AD is the altitude drawn from the right angle A.

So, AD is the geometric mean between BD and DC.

AD = √(BD ⋅ DC)

x =  √(9 ⋅ 3)

x =  √27

x =  3√3

Problem 3 :

Solution:

AC = AD + DC

AC = 4 + 6

AC = 10

ABC is a right triangle and BD is the altitude drawn from the right angle B.

So, BA is the geometric mean between AD and AC.

BA = √(AD ⋅ AC)

x =  √(4 ⋅ 10)

x =  √40

x = 6.32

So, BC is the geometric mean between CD and AC.

BC = √(CD ⋅ AC)

y =  √(6 ⋅ 10)

y =  √60

y = 7.74

Problem 4 :

Solution:

ABC is a right triangle and AD is the altitude drawn from the right angle A.

So, AD is the geometric mean between BD and DC.

AD = √(BD ⋅ DC)

10 = √(x ⋅ 25)

10² = 25x 

100 = 25x

x = 4

In right triangle ABD,

y² = x² + 10²

y² = 4² + 10²

y² = 16 + 100

y² = 116

y = √116

y = 10.77

Problem 5 :

Solution:

ABC is a right triangle and AD is the altitude drawn from the right angle A.

So, AD is the geometric mean between BD and DC.

AD = √(BD ⋅ DC)

x = √(7 ⋅ 28)

x = √196

x = 14

In right triangle ABD,

y² = x² + 7²

y² = 14² + 7²

y² = 196 + 49

y² = 245

y = √245

y = 15.65

Problem 6 :

Solution:

AC = AD + DC

AC = 5 + 15

AC = 20

ABC is a right triangle and BD is the altitude drawn from the right angle B.

So, AB is the geometric mean between AD and AC.

AB = √(AD ⋅ AC)

x = √(5 ⋅ 20)

x = √100

x = 10

Problem 7 :

Solution:

ABC is a right triangle and BD is the altitude drawn from the right angle B.

So, BD is the geometric mean between AD and DC.

BD = √(AD ⋅ DC)

12 = √(x ⋅ 16)

12² = 16x 

144 = 16x

x = 9

In right triangle ABD,

y² = x² + 12²

y² = 9² + 12²

y² = 81 + 144

y² = 225

y = √225

y = 15

Problem 8 :

Solution:

BC = BD + DC

12 = BD + 3

BD = 9

ABC is a right triangle and AD is the altitude drawn from the right angle A.

AC is the geometric mean between DC and BC.

AC = √(DC ⋅ BC)

x = √(3 ⋅ 12)

x = √36

x = 6

AB is the geometric mean between BD and BC.

AB = √(BD ⋅ BC)

y = √(9 ⋅ 12)

y = √108

x = 10.39