In right triangle ABC, the measure of angle ACB is 90^\circ. Altitude CD is drawn from ver… — GRE General Test

In right triangle $ABC$ , the measure of angle $ACB$ is $90^\circ$ . Altitude $CD$ is drawn from vertex $C$ to hypotenuse $AB$ , with point $D$ lying on line segment $AB$ . If $AD = 9$ and $DB = 16$ , what is the perimeter of triangle $ABC$ ?

Answer: 60

Answer

The perimeter of triangle

ABC

is 60.

By the Geometric Mean Theorem for right triangles, the altitude

CD

to hypotenuse

AB

satisfies

CD^2 = AD \cdot DB = 9 \cdot 16 = 144

, giving

CD = 12

. Applying the Pythagorean Theorem to the smaller right triangles

\triangle ADC

and

\triangle BDC

yields

AC = \sqrt{9^2 + 12^2} = 15

and

BC = \sqrt{16^2 + 12^2} = 20

. The hypotenuse

AB = 9 + 16 = 25

. Summing the side lengths gives the perimeter:

15 + 20 + 25 = 60

Step-by-Step Solution

Calculate the length of altitude

CD

using the Geometric Mean Theorem.

CD = \sqrt{AD \cdot DB} = \sqrt{9 \cdot 16} = \sqrt{144} = 12

In a right triangle, the altitude to the hypotenuse divides the hypotenuse into two segments such that the altitude is the geometric mean of the two segment lengths.

Calculate leg

AC

using the Pythagorean Theorem in right triangle

ADC

AC = \sqrt{AD^2 + CD^2} = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15

Triangle

ADC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

3

Calculate leg

BC

using the Pythagorean Theorem in right triangle

BDC

BC = \sqrt{BD^2 + CD^2} = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20

Triangle

BDC

is a right triangle with right angle at

D

(

3-4-5

triple scaled by

4

Calculate the total perimeter of triangle

ABC

Perimeter =

AC + BC + AB = 15 + 20 + (9 + 16) = 15 + 20 + 25 = 60

The perimeter is the sum of the three outer sides of triangle

ABC

Key Concept

Right Triangle Altitude Relationships and Pythagorean Triples

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