Difference between revisions of "1992 AIME Problems/Problem 7"
I_like_pie (talk | contribs) |
m (→Solution) |
||
(6 intermediate revisions by 6 users not shown) | |||
Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
+ | Faces <math>ABC^{}_{}</math> and <math>BCD^{}_{}</math> of tetrahedron <math>ABCD^{}_{}</math> meet at an angle of <math>30^\circ</math>. The area of face <math>ABC^{}_{}</math> is <math>120^{}_{}</math>, the area of face <math>BCD^{}_{}</math> is <math>80^{}_{}</math>, and <math>BC=10^{}_{}</math>. Find the volume of the tetrahedron. | ||
== Solution == | == Solution == | ||
− | {{ | + | Since the area <math>BCD=80=\frac{1}{2}\cdot10\cdot16</math>, the perpendicular from <math>D</math> to <math>BC</math> has length <math>16</math>. |
+ | |||
+ | The perpendicular from <math>D</math> to <math>ABC</math> is <math>16 \cdot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>. | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1992|num-b=6|num-a=8}} | |
− | |||
− | |||
− | + | [[Category:Intermediate Geometry Problems]] | |
+ | {{MAA Notice}} |
Latest revision as of 17:40, 14 March 2017
Problem
Faces and
of tetrahedron
meet at an angle of
. The area of face
is
, the area of face
is
, and
. Find the volume of the tetrahedron.
Solution
Since the area , the perpendicular from
to
has length
.
The perpendicular from to
is
. Therefore, the volume is
.
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.