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 Tetrahedron Volume Travail (Posted on 2016-03-12)
Faces PQR and QRS of tetrahedron PQRS intersect at an angle of 30o.

The area of face PQR is 120.
The area of face QRS is 80, and QR = 10.
Determine the volume of the tetrahedron PQRS.

 No Solution Yet Submitted by K Sengupta No Rating

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Let T be the orthogonal projection of S onto the plane containing PQR.  Let U be the point on the line containing QR such that TU is perpendicular to QR.

Then STU is a right triangle.  It lies on a plane perpendicular to line QR.  Then ST is the height of the tetrahedron relative to face PQR, angle STU is a right angle, and angle SUT is the dihedral angle between faces PQR and QRS.

SU equals the altitude of QRS perpendicular to QR, which equals 2*area(QRS)/QR.  ST equals SU*sin(angle SUT).  Then the volume of the tetrahedron PQRS is given by Volume = (1/3)*area(PRQ)*(2*area(QRS)/QR)*sin(angle SUT)

This simplifies to Volume = (2/3) * area(PRQ) * area(QRS) * sin(dihedral angle) / QR

Then plug in the given values: Volume = (2/3) * 80 * 120 * sin(30 deg) / 10 = 320

 Posted by Brian Smith on 2016-03-12 12:44:21

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