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.
Similar Questions
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Important Questions from Coordinate Geometry
- If the line $\alpha x + 2y = 1$, where $\alpha \in \mathbb{R}$, does not meet the hyperbola $x^2 - 9y^2 = 9$, then a possible value of $\alpha$ is:
- Let $P(10, 2\sqrt{15})$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, whose foci are S and $S'$. If the length of its latus rectum is 8, then the square of the area of $\Delta PSS'$ is equal to :
- Let the locus of the mid-point of the chord through the origin O of the parabola $y^2 = 4x$ be the curve S. Let P be any point on S. Then the locus of the point, which internally divides OP in the ratio 3:1, is :
- Let a circle of radius 4 pass through the origin O, the points $A(-\sqrt{3}a, 0)$ and $B(0, -\sqrt{2}b)$, where $a$ and $b$ are real parameters and $ab \neq 0$. Then the locus of the centroid of $\Delta OAB$ is a circle of radius
- Let a line L passing through the point $P(1, 1, 1)$ be perpendicular to the lines $\frac{x-4}{4} = \frac{y-1}{1} = \frac{z-1}{1}$ and $\frac{x-17}{1} = \frac{y-71}{1} = \frac{z}{0}$. Let the line L intersect the yz-plane at the point Q. Another line parallel to L and passing through the point $S(1, 0, -1)$ intersects the yz-plane at the point R. Then the square of the area of the parallelogram PQRS is equal to ______.
