AMC 10 · 2008 · #18
Grade 8 geometry-2dA right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?
Pick an answer.
AMC 10 2008 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.
Toolkit + CCSS Solution
Understand
Restated: A right triangle has perimeter $32$ and area $20$. Find the length of its hypotenuse.
Givens: The triangle has a right angle; Its perimeter is $32$; Its area is $20$; Answer choices: (A) $\tfrac{57}{4}$, (B) $\tfrac{59}{4}$, (C) $\tfrac{61}{4}$, (D) $\tfrac{63}{4}$, (E) $\tfrac{65}{4}$
Unknowns: The length of the hypotenuse
Understand
Restated: A right triangle has perimeter $32$ and area $20$. Find the length of its hypotenuse.
Givens: The triangle has a right angle; Its perimeter is $32$; Its area is $20$; Answer choices: (A) $\tfrac{57}{4}$, (B) $\tfrac{59}{4}$, (C) $\tfrac{61}{4}$, (D) $\tfrac{63}{4}$, (E) $\tfrac{65}{4}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #16 Change Focus / Count the Complement, #15 Organize Information in More Ways
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.
Execute — Answer: B
6.EE.B.6 Step 1 Name the sides and write the three facts
- Let the two legs be $a$ and $b$, and let the hypotenuse be $c$.
- The perimeter gives $a+b+c=32$.
- The area of a right triangle is half the product of its legs, so $\tfrac{1}{2}ab=20$, which means $ab=40$.
- The right angle means the Pythagorean theorem holds: $a^2+b^2=c^2$.
💡 Giving each side a letter turns three geometry facts into three equations you can push around.
8.G.B.7 Step 2 Aim at the hypotenuse, not the legs
- The question wants only $c$.
- Two of the equations already point that way: by the Pythagorean theorem $a^2+b^2$ is exactly $c^2$, and the perimeter rearranges to $a+b=32-c$.
- So if the sum $a+b$ and the sum $a^2+b^2$ can be linked, $c$ will be the only unknown left standing.
💡 You rarely need every piece; here the legs can stay unknown as long as the hypotenuse comes out alone.
6.EE.A.3 Step 3 Square the leg-sum to combine what you know
- Squaring $a+b$ gives $(a+b)^2 = a^2+2ab+b^2$.
- The right side is built entirely from known quantities: $a^2+b^2=c^2$ and $ab=40$, so $(a+b)^2 = c^2 + 2(40) = c^2+80$.
- But $a+b$ also equals $32-c$, so $(a+b)^2 = (32-c)^2 = 1024-64c+c^2$.
- Setting the two expressions for $(a+b)^2$ equal links everything in one equation.
💡 Squaring the sum is the bridge — it manufactures the $a^2+b^2$ and $ab$ you already know.
8.EE.C.7 Step 4 Cancel the squares and solve for c
- The $c^2$ appears on both sides and cancels, leaving a linear equation: $80 = 1024 - 64c$.
- Then $64c = 1024-80 = 944$, so $c = \tfrac{944}{64} = \tfrac{59}{4}$.
- The hypotenuse has length $\tfrac{59}{4}$, which is (B).
💡 Once the squares cancel, what looked quadratic is really a one-step linear solve.
6.EE.B.6 Let the two legs be $a$ and $b$, and let the hypotenuse be $c$. The perimeter gi 8.G.B.7 The question wants only $c$. Two of the equations already point that way: by the 6.EE.A.3 Squaring $a+b$ gives $(a+b)^2 = a^2+2ab+b^2$. The right side is built entirely f 8.EE.C.7 The $c^2$ appears on both sides and cancels, leaving a linear equation: $80 = 10 Review
Reasonableness: The hypotenuse must be shorter than the two legs combined, so it is less than half the perimeter, i.e. less than $16$; $\tfrac{59}{4}=14.75<16$ passes, while the largest choice $\tfrac{65}{4}=16.25$ would break that bound, ruling out (E). Real legs also exist: $a+b=32-\tfrac{59}{4}=\tfrac{69}{4}$ and $ab=40$ give discriminant $\left(\tfrac{69}{4}\right)^2-4(40)=\tfrac{2201}{16}>0$, so two positive legs are possible. Everything is consistent with (B).
Alternative: Use the right-triangle inradius shortcut. For any triangle, area $= r\cdot s$ where $s$ is the semiperimeter; here $s=16$ and area $20$ give $r=\tfrac{20}{16}=\tfrac{5}{4}$. For a right triangle the inradius satisfies $r = s - c$, so $c = s - r = 16 - \tfrac{5}{4} = \tfrac{59}{4}$ — the same answer (B) without ever touching the legs.
CCSS standards used (min grade 8)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Naming the legs $a,b$ and hypotenuse $c$ and writing the perimeter, area, and Pythagorean facts as three equations.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using $a^2+b^2=c^2$ so the sum of the squared legs can be replaced by $c^2$, keeping the hypotenuse as the target.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Expanding $(a+b)^2 = a^2+2ab+b^2$ and $(32-c)^2 = 1024-64c+c^2$ to combine the known quantities into one equation.)8.EE.C.7Solve linear equations in one variable (Cancelling $c^2$ and solving $80 = 1024-64c$ to get $c=\tfrac{59}{4}$.)
⭐ 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.
⭐ 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.
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