Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area

Q: Wait, why can we just assume $A$ is at the origin? Does that change the area?

That is a brilliant question! In geometry, the area of a shape is an intrinsic property—it depends on the relative distances between points, not their absolute position in space. By placing $A$ at $(0,0,0)$, we are simply choosing a convenient frame of reference to make the math easier. It does not change the shape or the area of the tetrahedron at all!

Q: Is De Gua's Theorem part of the standard JEE syllabus, or is it an extra trick?

While it is not always explicitly taught in textbooks, it is a direct consequence of vector cross products. You can derive it by defining the vectors $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$ and calculating the area of $\Delta BCD$ using the cross product formula $\frac{1}{2} |\vec{BC} \times \vec{BD}|$. Knowing it makes you faster, but understanding the vector derivation makes you a master.

Q: Why are the areas of the triangles $\frac{1}{2}ab$ and not something more complex?

Because the edges are mutually perpendicular! In a right-angled triangle, the area is simply $\frac{1}{2} \times \text{base} \times \text{height}$. Since $AB \perp AC$, $AC$ acts as the height for base $AB$. It is the simplest, most elegant way to define area in 3D space.

Q: Can I solve this without the theorem by finding the coordinates of $B$, $C$, and $D$?

Absolutely! If you find $a$, $b$, and $c$ by solving the system of equations $\frac{1}{2}ab=5, \frac{1}{2}bc=6, \frac{1}{2}ac=7$, you can find the coordinates of $B(a,0,0)$, $C(0,b,0)$, and $D(0,0,c)$. Then, use the vector area formula. It is a longer path, but it is a fantastic way to verify your answer.

Q: Why is the answer $\sqrt{110}$ and not something simpler like $12$?

In JEE problems, we often expect 'clean' integers, but nature—and geometry—is rarely so tidy! The square root is perfectly valid. Trust your derivation. If the math leads you to $\sqrt{110}$, then $\sqrt{110}$ is the absolute truth of the geometry.

Question

Let $ABCD$ be a tetrahedron such that the edges $AB, AC$ and $AD$ are mutually perpendicular. Let the areas of the triangles $ABC, ACD$ and $ADB$ be $5, 6$ and $7$ square units respectively. Then the area (in square units) of the $\Delta BCD$ is equal to :

Options:

(A) $\sqrt{340}$

(B) $12$

(C) $\sqrt{110}$

(D) $7\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the corner of a perfectly square room. You look at the point where the two walls meet the floor. That corner is a point of perfect symmetry, a place where three lines—the two wall-floor intersections and the vertical corner line—meet at right angles.

This is the essence of our tetrahedron $ABCD$. In this problem, we are not just dealing with a flat shape; we are looking at a 3D object that captures the very essence of orthogonality.

### Setting the Stage: The Coordinate System

When we face a problem like this, the first step is to simplify our perspective. We have a tetrahedron where edges $AB$, $AC$, and $AD$ are mutually perpendicular. This is our 'anchor.'

Let us place vertex $A$ at the origin $(0,0,0)$ of a 3D Cartesian coordinate system. By aligning $AB$ along the $x$-axis, $AC$ along the $y$-axis, and $AD$ along the $z$-axis, we transform a complex spatial problem into a clean algebraic one.

Let the lengths of these edges be $a$, $b$, and $c$ respectively. Now, the vertices are simply $B(a,0,0)$, $C(0,b,0)$, and $D(0,0,c)$.

### The Beauty of Orthogonal Faces

Because our edges are perpendicular, the faces $\Delta ABC$, $\Delta ACD$, and $\Delta ADB$ are all right-angled triangles. The area of a right-angled triangle is the simplest, most elegant calculation in geometry: $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.

For $\Delta ABC$ in the $xy$-plane, the area is:
$$\frac{1}{2}ab = 5$$

For $\Delta ACD$ in the $yz$-plane, the area is:
$$\frac{1}{2}bc = 6$$

For $\Delta ADB$ in the $zx$-plane, the area is:
$$\frac{1}{2}ac = 7$$

These three values are the 'projections' of our tetrahedron onto the coordinate planes. They hold the secret to the area of the slanted face, $\Delta BCD$.

### De Gua's Theorem: The 3D Pythagorean Masterpiece

Now, we arrive at the heart of the problem. You might be familiar with the Pythagorean theorem for 2D triangles, $a^2 + b^2 = c^2$. But did you know there is a 3D version?

It is called **De Gua's Theorem**. It states that for a right-corner tetrahedron, the square of the area of the face opposite the right-corner vertex is equal to the sum of the squares of the areas of the three right-angled faces. Mathematically, this is expressed as:

$$[Area(\Delta BCD)]^2 = [Area(\Delta ABC)]^2 + [Area(\Delta ACD)]^2 + [Area(\Delta ADB)]^2$$

This is not just a formula; it is a profound statement about how area behaves in three dimensions. It tells us that the 'tilted' area is composed of the shadows cast on the three primary planes.

### The Final Calculation

With the theorem in our toolkit, the rest is a beautiful exercise in arithmetic. We substitute our known values into the equation:

$$[Area(\Delta BCD)]^2 = 5^2 + 6^2 + 7^2$$

Calculating the squares, we get:

$$[Area(\Delta BCD)]^2 = 25 + 36 + 49$$

Summing these up, we find:

$$[Area(\Delta BCD)]^2 = 110$$

Taking the square root of both sides, we arrive at our final result:

**$$Area(\Delta BCD) = \sqrt{110}$$**

### Reflecting on the Journey

Look at what we have achieved. We took a 3D object, anchored it in a coordinate system, identified the right-angled nature of its faces, and applied a powerful geometric theorem to find the area of a face that seemed impossible to measure at first glance.

This is the power of physics and mathematics—taking the complex, the abstract, and the intimidating, and breaking it down into simple, logical steps. Never fear the geometry; embrace it, visualize it, and let the math guide you to the truth.

Select Answer:

Visualized Solution

Visualizing the Tetrahedron $A B C D$

Setting up the Coordinate System

Defining Edge Lengths

Area of $Δ A B C$

Area of $Δ A C D$

Area of $Δ A D B$

De Gua's Theorem

Substituting the Values

Calculating Squares

Summing the Squares

Final Area Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Setting the Stage

The Beauty of Orthogonal Faces

De Gua's Theorem

The Final Calculation

Reflecting on the Journey

Similar Questions

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The length of the perpendicular drawn from the point $(2, 1, 4)$ to the plane containing the lines $r = (\hat{i} + \hat{j}) + λ (\hat{i} + 2 \hat{j} - \hat{k})$ and $r = (\hat{i} + \hat{j}) + μ (- \hat{i} + \hat{j} - 2 \hat{k})$ is :

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Select Answer:

Visualized Solution

Visualizing the Tetrahedron .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

Setting up the Coordinate System

Defining Edge Lengths

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔABC

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔACD

Area of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔADB

De Gua's Theorem

Substituting the Values

Calculating Squares

Summing the Squares

Final Area Calculation

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Setting the Stage

The Beauty of Orthogonal Faces

De Gua's Theorem

The Final Calculation

Reflecting on the Journey

Similar Questions

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Visualizing the Tetrahedron $A B C D$

Area of $Δ A B C$

Area of $Δ A C D$

Area of $Δ A D B$

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