Difference between revisions of "2015 AMC 10A Problems/Problem 21"
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− | Problem | + | ==Problem 21== |
+ | Tetrahedron <math>ABCD</math> has <math>AB=5</math>, <math>AC=3</math>, <math>BC=4</math>, <math>BD=4</math>, <math>AD=3</math>, and <math>CD=\tfrac{12}5\sqrt2</math>. What is the volume of the tetrahedron? | ||
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+ | <math>\textbf{(A) }3\sqrt2\qquad\textbf{(B) }2\sqrt5\qquad\textbf{(C) }\dfrac{24}5\qquad\textbf{(D) }3\sqrt3\qquad\textbf{(E) }\dfrac{24}5\sqrt2</math> | ||
+ | |||
+ | ==Solution== | ||
+ | Let the midpoint of <math>CD</math> be <math>E</math>. We have <math>CE = \dfrac{6}{5} \sqrt{2}</math>, and so by the Pythagorean Theorem <math>AE = \dfrac{\sqrt{153}}{5}</math> and <math>BE = \dfrac{\sqrt{328}}{5}</math>. Because the altitude from <math>A</math> of tetrahedron <math>ABCD</math> passes touches plane <math>BCD</math> on <math>BE</math>, it is also an altitude of triangle <math>ABE</math>. The area of triangle <math>ABE</math> is, by Heron's Formula, given by | ||
+ | |||
+ | <cmath>16A^2 = 2a^2 b^2 + 2b^2 c^2 + 2c^2 a^2 - a^4 - b^4 - c^4 = -(a^2 + b^2 - c^2)^2 + 4a^2 b^2.</cmath> | ||
+ | Substituting <math>a = AE, b = BE, c = 5</math> and performing huge (but manageable) computations yield <math>A^2 = 18</math>, so <math>A = 3\sqrt{2}</math>. Thus, if <math>h</math> is the length of the altitude from <math>A</math> of the tetrahedron, <math>BE \cdot h = 2A = 6\sqrt{2}</math>. Our answer is thus | ||
+ | <cmath>V = \dfrac{1}{3} Bh = \dfrac{1}{3} h \cdot BE \cdot \dfrac{6\sqrt{2}}{5} = \dfrac{24}{5},</cmath> | ||
+ | and so our answer is <math>\textbf{(C)}</math>. |
Revision as of 19:49, 4 February 2015
Problem 21
Tetrahedron has , , , , , and . What is the volume of the tetrahedron?
Solution
Let the midpoint of be . We have , and so by the Pythagorean Theorem and . Because the altitude from of tetrahedron passes touches plane on , it is also an altitude of triangle . The area of triangle is, by Heron's Formula, given by
Substituting and performing huge (but manageable) computations yield , so . Thus, if is the length of the altitude from of the tetrahedron, . Our answer is thus and so our answer is .