Calculate A¹B in an Orthohedron: Given D¹C¹=10 and AA¹=12

Q: Why can I assume AB = 10 when only D¹C¹ is given?

In an orthohedron (rectangular prism), opposite faces are identical rectangles. Since D¹C¹ = 10 is on the top face, the corresponding edge AB = 10 on the bottom face has the same length.

Q: How do I know which edges form a right triangle with A¹B?

Look for perpendicular edges that connect to the endpoints of A¹B. Here, AA¹ (vertical) and AB (horizontal) meet at point A at a 90° angle, forming a right triangle with hypotenuse A¹B.

Q: Why don't I need the third dimension of the orthohedron?

A¹B lies in a 2D plane formed by edges AA¹ and AB. Since we're finding a face diagonal (not a space diagonal), only these two dimensions matter for this calculation.

Q: How do I simplify √244 to get 2√61?

Factor out perfect squares: $ \sqrt{244} = \sqrt{4 \times 61} = \sqrt{4} \times \sqrt{61} = 2\sqrt{61} $. Always look for the largest perfect square factor!

Q: What if I get confused about which vertices are which?

Use the prime notation as a guide: A and A¹ are connected vertically, while A and B are on the same horizontal level. The distance A¹B crosses between different levels.

3D Geometry with Pythagorean Theorem

Look at the orthohedron below.

$D^1C^1=10$

$AA^1=12$

Calculate $A^1B$ .

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find A1B

00:03 Face in box is rectangular therefore opposite sides are equal

00:10 Set the side value according to the given data

00:18 Draw the diagonal A1B

00:21 Use Pythagorean theorem in triangle A1AB to find A1B

00:27 Set appropriate values according to the given data and solve to find A1B

00:46 This is the length of diagonal A1B

00:53 Factorize 244 into factors 4 and 61

00:59 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Look at the orthohedron below.

$D^1C^1=10$

$AA^1=12$

Calculate $A^1B$ .

Step-by-step solution

From the given data, we know that:

$D_1C_1=A_1B_1=AB=10$

Let's draw a diagonal between A1 and B and focus on triangle AA1B.

We'll calculate A1B using the Pythagorean theorem:

$AA_1^2+AB^2=A_1B^2$

Then we will substitute in the known values:

$12^2+10^2=A_1B^2$

$A_1B^2=144+100=244$

Finally, we calculate square root:

$A_1B=\sqrt{244}$

$A_1B=\sqrt{4\times61}=\sqrt{4}\times\sqrt{61}$

$A_1B=2\sqrt{61}$

Final Answer

$2\sqrt{61}$

Key Points to Remember

Essential concepts to master this topic

Rule: In orthohedrons, use 3D Pythagorean theorem for space diagonals
Technique: Find face diagonal first: $12^2 + 10^2 = 244$
Check: Verify $\sqrt{244} = 2\sqrt{61}$ by squaring: $(2\sqrt{61})^2 = 244$ ✓

Common Mistakes

Avoid these frequent errors

Using only one dimension instead of two perpendicular edges
Don't just use AA¹ = 12 or AB = 10 alone to find A¹B = wrong answer! This ignores the 3D nature of the problem and treats it as a simple line segment. Always identify the right triangle formed by the two perpendicular edges and apply the Pythagorean theorem.

Practice Quiz

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What is the length of the hypotenuse?

FAQ

Everything you need to know about this question

Why can I assume AB = 10 when only D¹C¹ is given?

In an orthohedron (rectangular prism), opposite faces are identical rectangles. Since D¹C¹ = 10 is on the top face, the corresponding edge AB = 10 on the bottom face has the same length.

How do I know which edges form a right triangle with A¹B?

Look for perpendicular edges that connect to the endpoints of A¹B. Here, AA¹ (vertical) and AB (horizontal) meet at point A at a 90° angle, forming a right triangle with hypotenuse A¹B.

Why don't I need the third dimension of the orthohedron?

A¹B lies in a 2D plane formed by edges AA¹ and AB. Since we're finding a face diagonal (not a space diagonal), only these two dimensions matter for this calculation.

How do I simplify √244 to get 2√61?

Factor out perfect squares: $\sqrt{244} = \sqrt{4 \times 61} = \sqrt{4} \times \sqrt{61} = 2\sqrt{61}$ . Always look for the largest perfect square factor!

What if I get confused about which vertices are which?

Use the prime notation as a guide: A and A¹ are connected vertically, while A and B are on the same horizontal level. The distance A¹B crosses between different levels.

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