AMC 10 · 2009 · #17

Grade 8 geometry-2d

similar-trianglespythagorean-theorem identify-subproblems ↑ Prerequisites: similar-triangles

📏 Medium solution 💡 3 insights

Problem

Rectangle $ABCD$ has $AB=4$ and $BC=3$ . Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$ , and $A$ and $C$ lie on $DE$ and $DF$ , respectively. What is $EF$ ?

Pick an answer.

(A)

$\ 9$

(B)

$\ 10$

(C)

$\ \frac {125}{12}$

(D)

$\ \frac {103}{9}$

(E)

$\ 12$

AMC 10 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: In rectangle ABCD with AB = 4 and BC = 3, a line EF passes through corner B and is perpendicular to the diagonal DB. The two sides DA and DC, when extended, meet this line: A sits on segment DE and C sits on segment DF. We must find the length of EF.

Givens: ABCD is a rectangle with AB = 4 and BC = 3; EF passes through B and is perpendicular to diagonal DB; A lies on segment DE, and C lies on segment DF

Unknowns: The length of segment EF

Understand

Givens: ABCD is a rectangle with AB = 4 and BC = 3; EF passes through B and is perpendicular to diagonal DB; A lies on segment DE, and C lies on segment DF

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram, #4 Introduce a Variable

EF is hard to measure directly, but B splits it into two pieces EB and BF. Each piece lives in its own right triangle that is similar to a triangle made from the rectangle. So the plan is: draw the figure, find the diagonal, then solve two small similar-triangle subproblems and add the pieces.

Execute — Answer: C

#1 Draw a Diagram 8.G.B.7 Step 1

Find the diagonal DB

Diagonal DB cuts the rectangle into right triangle DAB with legs DA = BC = 3 and AB = 4.
Apply the Pythagorean theorem to get DB.

$$DB=\sqrt{AB^2+DA^2}=\sqrt{4^2+3^2}=\sqrt{25}=5$$

💡 A 3-4-5 right triangle always has hypotenuse 5, so the diagonal is exactly 5.

#7 Identify Subproblems 8.G.A.5 Step 2

Split EF at B

Because DA and DC are perpendicular sides of the rectangle and E is on line DA while F is on line DC, angle EDF equals angle ADC = 90 degrees.
So triangle DEF has a right angle at D, and DB (perpendicular to EF) is its altitude from that right angle.
The point B breaks EF into two parts, so EF = EB + BF; find each part separately.

$$\angle EDF=\angle ADC=90^\circ,\qquad EF=EB+BF$$

💡 The right angle of the rectangle becomes the right angle of the big triangle, so B just divides the far side into two chunks.

#4 Introduce a Variable 8.G.A.4 Step 3

Similar triangles give EB

Triangles DAB and DBE share the angle at D, and each has a right angle (at A in DAB, at B in DBE), so they are similar.
Matching sides gives a proportion that solves for EB.

$$\triangle DAB\sim\triangle DBE\;\Rightarrow\;\frac{EB}{AB}=\frac{DB}{DA}\;\Rightarrow\;EB=\frac{4\cdot 5}{3}=\frac{20}{3}$$

💡 Same-shaped triangles keep side ratios equal, so EB scales up from AB by the same factor DB grows from DA.

#4 Introduce a Variable 8.G.A.4 Step 4

Similar triangles give BF

In the same way, triangles DCB and DBF share the angle at D and each has a right angle (at C in DCB, at B in DBF), so they are similar.
The matching proportion solves for BF, using DC = AB = 4 and CB = 3.

$$\triangle DCB\sim\triangle DBF\;\Rightarrow\;\frac{BF}{CB}=\frac{DB}{DC}\;\Rightarrow\;BF=\frac{3\cdot 5}{4}=\frac{15}{4}$$

💡 The other half of the figure is the same trick with the sides swapped, so BF scales from CB by the factor DB grows from DC.

#7 Identify Subproblems 5.NF.A.1 Step 5

Add the two pieces

Add EB and BF over a common denominator to get the full length EF.
This matches answer choice (C).

$$EF=\frac{20}{3}+\frac{15}{4}=\frac{80}{12}+\frac{45}{12}=\frac{125}{12}$$

💡 The whole segment is just the two chunks put back together, so add the fractions.

[1] #1 8.G.B.7 Diagonal DB cuts the rectangle into right triangle DAB with legs DA = BC = 3 and

[2] #7 8.G.A.5 Because DA and DC are perpendicular sides of the rectangle and E is on line DA w

[3] #4 8.G.A.4 Triangles DAB and DBE share the angle at D, and each has a right angle (at A in

[4] #4 8.G.A.4 In the same way, triangles DCB and DBF share the angle at D and each has a right

[5] #7 5.NF.A.1 Add EB and BF over a common denominator to get the full length EF. This matches

Review

Reasonableness: EF = 125/12 is about 10.4, which is a bit longer than the diagonal DB = 5 and sensibly larger than either side of the rectangle, since EF stretches beyond both A and C. Choice (C) = 125/12 fits, and the nearby traps 10 and 103/9 are close decoys that the exact fractions rule out.

Alternative: Use the altitude-on-hypotenuse relation directly: in right triangle DEF, DB is the altitude to hypotenuse EF, so area gives DE * DF = DB * EF. Finding DE = 25/3 and DF = 25/4 from the two similar triangles and multiplying yields the same EF = 125/12.

CCSS standards used (min grade 8)

8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the diagonal DB = 5 from the legs 3 and 4)
8.G.A.5 Use informal arguments to establish facts about angle sum and exterior angles (Arguing that angle EDF = angle ADC = 90 degrees so DEF is a right triangle)
8.G.A.4 Understand that a two-dimensional figure is similar to another using transformations (Setting up the similar-triangle proportions that give EB and BF)
5.NF.A.1 Add and subtract fractions with unlike denominators (Adding 20/3 and 15/4 to get the final length EF)

⭐ Cut the mystery segment where the diagonal hits it, match each half to a look-alike right triangle, and add the two halves.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Find the diagonal DB

Split EF at B

Similar triangles give EB

Similar triangles give BF

Add the two pieces

Review

CCSS standards used (min grade 8)

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