In \triangle ABC, the length of side AB is 10 and the length of side BC is 17. Point D lie… — GRE General Test

In $\triangle ABC$ , the length of side $AB$ is $10$ and the length of side $BC$ is $17$ . Point $D$ lies on the line containing segment $AC$ such that segment $BD$ is perpendicular to line $AC$ . If the length of altitude $BD$ and the length of side $AC$ are both integers, and the area of $\triangle ABC$ is strictly greater than $36$ , what is the perimeter of $\triangle ABC$ ?

A
$36$
B
$42$
$48$ Answer
D
$52$
E
$56$

Answer

The perimeter of

\triangle ABC

48

Applying the Pythagorean theorem to both right triangles formed by altitude

BD = h

gives

100 - m^2 = h^2

and

289 - n^2 = h^2

, where

m = AD

and

n = CD

. Subtracting these equations yields

n^2 - m^2 = 189

, which factors as

(n - m)(n + m) = 189

. Testing integer factor pairs of

189

while enforcing

m \le 10

isolates a single non-degenerate solution:

m = 6

n = 15

, and

h = 8

. When point

D

lies between

A

and

C

AC = 6 + 15 = 21

. This gives an area of

\frac{1}{2} \times 21 \times 8 = 84

(which is strictly greater than

36

) and a total perimeter of

10 + 17 + 21 = 48

Step-by-Step Solution

Set up Pythagorean relationships for the right triangles formed by altitude

BD

Let

BD = h

AD = m

, and

CD = n

. In right

\triangle ABD

m^2 + h^2 = 10^2 = 100

. In right

\triangle CBD

n^2 + h^2 = 17^2 = 289

Altitude

BD

divides the figure into two right triangles sharing leg

h

Subtract the two equations to eliminate

h^2

and factor the difference of squares.

n^2 - m^2 = 289 - 100 = 189 \implies (n - m)(n + m) = 189.

Since

h

and

AC

are integers,

m

and

n

must also be integers for

AC = n \pm m

to be an integer.

Analyze integer factor pairs

(u, v)

189

where

u = n - m

and

v = n + m

Factor pairs

(u, v)

with

u \cdot v = 189

:
-

(1, 189) \implies m = 94

(invalid,

m \le 10

)
-

(3, 63) \implies m = 30

(invalid,

m \le 10

)
-

(7, 27) \implies m = 10, h = 0

(degenerate triangle)
-

(9, 21) \implies n = 15, m = 6, h = \sqrt{100 - 36} = 8

The leg

m = AD

cannot exceed the hypotenuse

AB = 10

Evaluate side

AC

, area, and perimeter for valid geometric configurations.

Case 1:

D

lies on segment

AC \implies AC = n + m = 15 + 6 = 21

.
Area =

\frac{1}{2} \times 21 \times 8 = 84 > 36

.
Perimeter =

10 + 17 + 21 = 48

.
Case 2:

D

lies outside segment

AC \implies AC = n - m = 15 - 6 = 9

.
Area =

\frac{1}{2} \times 9 \times 8 = 36

, which does not satisfy area

> 36

The problem specifies that the area must be strictly greater than

36

Key Concept

Properties of triangles, Pythagorean theorem system solver, and geometric area constraints

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