2012 AMC 12B Problem 10 — Solution | LIVE by Po-Shen Loh

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2012 AMC 12B Problem 10

Below is the professionally curated solution for Problem 10 of the 2012 AMC 12B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 12B solutions, or check the answer key.

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Concepts:system of equations substitution triangle area

Difficulty rating: 1500

10.

What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 = 25$ and $(x - 4)^2 + 9y^2 = 81?$

$24$

$27$

$36$

$37.5$

$42$

Solution:

From $x^2+y^2=25$ we get $y^2=25-x^2.$ Substituting into $(x-4)^2+9y^2=81$ gives $x^2+x-20=0,$ so $x=4$ or $x=-5.$

The intersection points are $(-5,0),$ $(4,3),$ and $(4,-3).$

The vertical side from $(4,3)$ to $(4,-3)$ has length $6,$ and the horizontal distance to $(-5,0)$ is $9,$ so the area is $\tfrac12\cdot6\cdot9=27.$

Thus, the correct answer is B.

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