In \triangle ABC, point D lies on segment BC such that segment AD is perpendicular to BC. … — GRE General Test

In $\triangle ABC$ , point $D$ lies on segment $BC$ such that segment $AD$ is perpendicular to $BC$ . The ratio of the area of $\triangle ABD$ to the area of $\triangle ADC$ is $5 : 16$ . If $AB = 13$ and the perimeter of $\triangle ABC$ is $54$ , what is the area of $\triangle ABC$ ?

A
96
B
104
126Answer
D
130
E
252

Answer

126

The correct answer is 126. Since triangles

\triangle ABD

and

\triangle ADC

share height

AD

, their areas are in proportion to their bases

BD:DC = 5:16

. Setting

BD = 5k

and

DC = 16k

, the Pythagorean theorem yields altitude

AD = \sqrt{169 - 25k^2}

and hypotenuse

AC = \sqrt{169 + 231k^2}

. Substituting these into the perimeter equation

13 + 21k + AC = 54

yields

k = 1

(after rejecting an extraneous root). Thus

BC = 21

and

AD = 12

, making the area

\frac{1}{2} \times 21 \times 12 = 126

Step-by-Step Solution

Relate the areas of the sub-triangles to their base lengths.

\frac{\text{Area}(\triangle ABD)}{\text{Area}(\triangle ADC)} = \frac{\frac{1}{2} \cdot BD \cdot AD}{\frac{1}{2} \cdot DC \cdot AD} = \frac{BD}{DC} = \frac{5}{16}

. Thus,

BD = 5k

and

DC = 16k

for some positive constant

k

, giving

BC = 21k

Triangles sharing the same altitude have areas proportional to their bases.

Express altitude

AD

and side

AC

in terms of

k

using the Pythagorean theorem.

In right

\triangle ABD

AD = \sqrt{AB^2 - BD^2} = \sqrt{13^2 - (5k)^2} = \sqrt{169 - 25k^2}

. In right

\triangle ADC

AC = \sqrt{AD^2 + DC^2} = \sqrt{(169 - 25k^2) + (16k)^2} = \sqrt{169 + 231k^2}

Since

AD \perp BC

, both

\triangle ABD

and

\triangle ADC

are right triangles.

Set up and solve the perimeter equation for

k

Perimeter

= AB + BC + AC = 13 + 21k + \sqrt{169 + 231k^2} = 54 \implies \sqrt{169 + 231k^2} = 41 - 21k

. Squaring both sides yields

169 + 231k^2 = 1681 - 1722k + 441k^2 \implies 210k^2 - 1722k + 1512 = 0 \implies 5k^2 - 41k + 36 = 0

. Factoring gives

(5k - 36)(k - 1) = 0

, so

k = 1

k = 7.2

The given perimeter allows over-constraining the side length expressions to a quadratic in

k

Test roots for validity and calculate final triangle area.

For

k = 7.2

41 - 21(7.2) = -110.2 < 0

, which is extraneous. For

k = 1

BD = 5

DC = 16

BC = 21

AD = 12

, and

AC = 20

. Area

(\triangle ABC) = \frac{1}{2} \cdot BC \cdot AD = \frac{1}{2} \cdot 21 \cdot 12 = 126

Extraneous roots introduced by squaring must be discarded, and the valid root gives the exact area.

Key Concept

Decomposing triangles into adjacent right triangles, leveraging shared altitudes for area ratios, and applying algebraic perimeter constraints with Pythagorean equations.

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