AMC 10 · 2015 · #16
Grade 8 geometry-3dPick an answer.
The six edges hide two identical 3-4-5 right triangles glued along the edge AB. The whole problem turns into picturing how the two triangles fold across AB. If we can find the height of D above the plane of triangle ABC, the volume is just one-third base times height, so the spatial picture is the key, supported by breaking the solid into familiar right triangles.
Spot the two 3-4-5 triangles
Two faces are familiar right triangles.
Any triangle whose sides are 3, 4, 5 is automatically a right triangle.
Any triangle whose sides are three, four and five is automatically a right triangle.
▸ Why?
The relation between the squares holds exactly when the angle is right, so it reads backwards as well.
▸ Why?
Any triangle with those side ratios is the same shape, so the right angle comes along whatever the scale.
Drop both altitudes to AB
Each drops the same altitude length.
From the right angle, the shortest path to the hypotenuse is the product of the legs divided by the hypotenuse.
8.G.B.7Draw A DiagramShow CH and DH are perpendicular
The last edge makes those altitudes perpendicular.
If two segments satisfy a² + b² = c², the corner between them is a perfect right angle.
8.G.B.6Visualize Spatial RelationshipsRead off the height of the solid
So one altitude is the solid's height.
A segment perpendicular to two crossing lines of a flat surface stands straight up out of that surface.
6.G.A.1Visualize Spatial RelationshipsCompute the volume
The volume comes to 24/5, choice (C).
A pyramid holds exactly one-third of the box with the same base and height.
8.G.C.9Visualize Spatial RelationshipsWhen edges hide 3-4-5 right triangles, fold them flat in your mind, find where the heights meet, and the volume falls out of one-third base times height.
- Spot the two 3-4-5 triangles
- Drop both altitudes to AB
- Show CH and DH are perpendicular
- Read off the height of the solid
- Compute the volume
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