Tetrahedron Area Calculation
Problem Setup: We are given a tetrahedron ABCD where the edges originating from vertex A (AB, AC, AD) are mutually perpendicular. The areas of the three faces meeting at A are provided: Area($\triangle ABC$) = 5, Area($\triangle ACD$) = 6, and Area($\triangle ADB$) = 7. We need to find the area of the fourth face, $\triangle BCD$.
Applying De Gua's Theorem
This problem can be solved efficiently using a generalization of De Gua's theorem, which relates the areas of the faces of a right-angled tetrahedron (one where three faces meeting at a vertex are mutually perpendicular). The theorem states that the square of the area of the face opposite the right-angled vertex is equal to the sum of the squares of the areas of the other three faces.
Solution Steps:
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Identify Given Areas: Let the areas of the faces be:
- $A_1 = \text{Area}(\triangle ABC) = 5$
- $A_2 = \text{Area}(\triangle ACD) = 6$
- $A_3 = \text{Area}(\triangle ADB) = 7$
- $A_4 = \text{Area}(\triangle BCD)$ (The area we need to find)
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State De Gua's Theorem: For a tetrahedron with mutually perpendicular edges at vertex A, the relationship is:
$ A_4^2 = A_1^2 + A_2^2 + A_3^2 $
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Substitute Given Values: Plug the known areas into the formula:
$ A_4^2 = 5^2 + 6^2 + 7^2 $
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Calculate the Sum of Squares:
$ A_4^2 = 25 + 36 + 49 $
$ A_4^2 = 110 $
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Find the Area $A_4$: Take the square root of the result:
$ A_4 = \sqrt{110} $
The area of the triangle $\triangle BCD$ is $\sqrt{110}$ square units.
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