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:
-
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)
-
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 $
-
Substitute Given Values: Plug the known areas into the formula:
$ A_4^2 = 5^2 + 6^2 + 7^2 $
-
Calculate the Sum of Squares:
$ A_4^2 = 25 + 36 + 49 $
$ A_4^2 = 110 $
-
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
- 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 ______.
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
- The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
- Let $\vec{a} = 2\hat{i} - 5\hat{j} + 5\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 3\hat{k}$. If $\vec{c}$ is a vector such that $2(\vec{a} \times \vec{c}) + 3(\vec{b} \times \vec{c}) = \vec{0}$ and $(\vec{a} - \vec{b}) \cdot \vec{c} = -97$, then $|\vec{c} \times \hat{k}|^2$ is equal to
- Let the image of parabola $x^2 = 4y$, in the line $x - y = 1$ be $(y + a)^2 = b(x - c)$, $a, b, c \in \mathbb{N}$. Then $a + b + c$ is equal to
- Let $(h, k)$ lie on the circle $C : x^2 + y^2 = 4$ and the point $(2h + 1, 3k + 2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac{5}{e^2}$ is equal to _______.
- If $P$ is a point on the circle $x^2 + y^2 = 4$, $Q$ is a point on the straight line $5x + y + 2 = 0$ and $x - y + 1 = 0$ is the perpendicular bisector of $PQ$, then $13$ times the sum of abscissa of all such points $P$ is __________.
- If the distance of the point $\text{P}(43, \alpha, \beta)$, $\beta < 0$, from the line $\overrightarrow{\text{r}} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k})$, $\mu \in \mathbf{R}$ along a line with direction ratios $3, -1, 0$ is $13\sqrt{10}$, then $\alpha^{2} + \beta^{2}$ is equal to _________.
Important Questions from Coordinate Geometry
- 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 ______.
- Let the angles made with the positive $x$-axis by two straight lines drawn from the point $P(2, 3)$ and meeting the line $x + y = 6$ at a distance $\sqrt{\frac{2}{3}}$ from the point $P$ be $\theta_1$ and $\theta_2$. Then the value of $(\theta_1 + \theta_2)$ is:
- The sum of all values of $\alpha$, for which the shortest distance between the lines $\frac{x + 1}{\alpha} = \frac{y - 2}{-1} = \frac{z - 4}{-\alpha}$ and $\frac{x}{\alpha} = \frac{y - 1}{2} = \frac{z - 1}{2\alpha}$ is $\sqrt{2}$, is
- Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, (a > b)$, be 30. If its eccentricity is the maximum value of the function $f(t) = -\frac{3}{4} + 2t - t^2$, then $(a^2 + b^2)$ is equal to
- Let $\vec{a} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b} = \hat{i} + 3\hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} + \hat{j} + 3\hat{k}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
