In triangle ABC, point D lies on side AB and point E lies on side AC such that segment DE … — SAT

In triangle $ABC$ , point $D$ lies on side $AB$ and point $E$ lies on side $AC$ such that segment $DE$ is parallel to segment $BC$ . The length of segment $AD$ is $2x + 1$ , the length of segment $DB$ is $x + 1$ , the length of segment $AE$ is $10$ , and the length of segment $EC$ is $6$ . What is the length of segment $AB$ ?

Cevap: 8

Cevap

The length of segment

AB

is 8.

By the Triangle Proportionality Theorem, since segment

DE

is parallel to segment

BC

, the segments on the transversal sides are proportional:

\frac{AD}{DB} = \frac{AE}{EC}

. Substituting the given expressions and values yields

\frac{2x + 1}{x + 1} = \frac{10}{6}

. Simplifying the fraction on the right side to

\frac{5}{3}

and cross-multiplying gives

3(2x + 1) = 5(x + 1)

. Solving this equation yields

x = 2

. The length of segment

AB

is the sum of

AD

and

DB

, which is

(2x + 1) + (x + 1) = 3x + 2

. Substituting

x = 2

gives

AB = 8

Adım Adım Çözüm

Set up the proportion using the Triangle Proportionality Theorem.

\frac{AD}{DB} = \frac{AE}{EC}

Since segment

DE

is parallel to segment

BC

, it divides the sides of triangle

ABC

proportionally.

Substitute the given values and simplify the constant ratio.

\frac{2x + 1}{x + 1} = \frac{5}{3}

The length of segment

AE

is 10 and

EC

is 6, so

\frac{AE}{EC} = \frac{10}{6} = \frac{5}{3}

Cross-multiply and solve the linear equation for

x

x = 2

Cross-multiplying gives

3(2x + 1) = 5(x + 1)

, which simplifies to

6x + 3 = 5x + 5

. Subtracting

5x

and 3 from both sides yields

x = 2

Calculate the total length of segment

AB

AB = 8

The total length

AB

is the sum of

AD

and

DB

. Thus,

AB = (2x + 1) + (x + 1) = 3x + 2

. Substituting

x = 2

gives

3(2) + 2 = 8

Anahtar Kavram

Triangle Proportionality Theorem and Similar Triangles

Alternatif Yöntem

Instead of using the Triangle Proportionality Theorem directly, we can use the similarity of triangles

ADE

and

ABC

. Since

DE \parallel BC

, we have

\triangle ADE \sim \triangle ABC

by AA similarity. This gives the ratio of corresponding side lengths:

\frac{AD}{AB} = \frac{AE}{AC}

. Substituting the expressions yields

\frac{2x + 1}{3x + 2} = \frac{10}{16} = \frac{5}{8}

. Cross-multiplying gives

8(2x + 1) = 5(3x + 2) \implies 16x + 8 = 15x + 10 \implies x = 2

. Then,

AB = 3x + 2 = 3(2) + 2 = 8

Tahmini Süre:1m 30s

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