AMC 10 · 2008 · #18

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

pythagorean-theoremarea-trianglessystems-of-equations convert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights

Problem

A right triangle has perimeter 32 and area 20. Find the length of its hypotenuse.

Pick an answer.

(A)

$\ \frac{57}{4}$

(B)

$\ \frac{59}{4}$

(C)

$\ \frac{61}{4}$

(D)

$\ \frac{63}{4}$

(E)

$\ \frac{65}{4}$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Three sentences of geometry (right angle, perimeter, area) turn into three equations the moment you name the sides, so Tool #4 (Introduce a Variable) is the starting move. The trap is trying to solve for the two legs. Tool #16 (Change Focus) says: aim only at the hypotenuse and let the legs stay hidden. Tool #15 (Organize Information in More Ways) supplies the bridge — squaring the sum of the legs re-packages the perimeter and area into a single equation whose only unknown is the hypotenuse.

1STEP 1

Name the sides and write the three facts

Name the legs a, b and the hypotenuse c. Perimeter gives a+b+c=32, area gives ab=40, and the right angle gives a²+b²=c².

a+b+c=32, ab=40, a²+b²=c²

2STEP 2

Aim at the hypotenuse, not the legs

Only c is wanted. The perimeter rearranges to a+b=32-c, and Pythagoras already says a²+b²=c², so linking the two leaves c alone.

a+b = 32-c, a²+b² = c²

3STEP 3

Square the leg-sum to combine what you know

Square the leg sum: as a²+2ab+b² it is c²+80, while a+b=32-c makes it 1024-64c+c².

c² + 80 = (32-c)² = 1024 - 64c + c²

4STEP 4

Cancel the squares and solve for c

The c² cancels: 80 = 1024-64c, so 64c = 944 and c = 944/64 = 59/4, choice (B).

80 = 1024 - 64c → 64c = 944 → c = 944/64 = 59/4 → (B)

Answer

59/4

The hypotenuse must be shorter than the two legs combined, so it is less than half the perimeter, i.e. less than 16; 59/4=14.75 ＜ 16 passes, while the largest choice 65/4=16.25 would break that bound, ruling out (E). Real legs also exist: a+b=32-59/4=69/4 and ab=40 give discriminant (69/4)²-4(40)=2201/16 ＞ 0, so two positive legs are possible. Everything is consistent with (B).

💡Key takeaway

To find just the hypotenuse, square the sum of the legs: it turns the perimeter and area you already know into one equation with only the hypotenuse left.

Name the sides and write the three facts
Aim at the hypotenuse, not the legs
Square the leg-sum to combine what you know
Cancel the squares and solve for c

Name the sides and write the three facts

Aim at the hypotenuse, not the legs

Square the leg-sum to combine what you know

Cancel the squares and solve for c

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