2019 AMC 12B Problems/Problem 18
Problem
Square pyramid has base , which measures cm on a side, and altitude perpendicular to the base, which measures cm. Point lies on , one third of the way from to ; point lies on , one third of the way from to ; and point lies on , two thirds of the way from to . What is the area, in square centimeters, of ?
Solution 1 (coordinate bash)
Using the given data, we can label the points and . We can also find the points . Similarly, and .
Using the distance formula, , , and . Using Heron's formula, or by dropping an altitude from to find the height, we can then find that the area of is .
Note: After finding the coordinates of and , we can alternatively find the vectors and , then apply the formula . In this case, the cross product equals , which has magnitude , giving the area as like before.
Solution 2
As in Solution 1, let and , and calculate the coordinates of , , and as . Now notice that the plane determined by is perpendicular to the plane determined by . To see this, consider the bird's-eye view, looking down upon , , and projected onto : Additionally, we know that is parallel to the plane determined by , since and have the same -coordinate. Hence the height of is equal to the -coordinate of minus the -coordinate of , giving . By the distance formula, , so the area of is .
Solution 3 (geometry)
By the Pythagorean Theorem, we can calculate and . Now by the Law of Cosines in , we have .
Similarly, by the Law of Cosines in , we have , so . Observe that (by side-angle-side), so .
Next, notice that is parallel to , and therefore is similiar to . Thus we have . Since , this gives .
Now we have the three side lengths of isosceles : , . Letting the midpoint of be , is the perpendicular bisector of , and so can be used as a height of (taking as the base). Using the Pythagorean Theorem again, we have , so the area of is .
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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