Calculate Orthohedron Dimensions: Solving √(5a²+6a+b⁴+9) Diagonal Problem

Q: Why do we use the 3D Pythagorean theorem here?

An orthohedron is a 3D rectangular box, so its main diagonal passes through all three dimensions. The regular Pythagorean theorem only works for 2D triangles, but for 3D boxes we need $ d^2 = l^2 + w^2 + h^2 $.

Q: How do I expand (a+3)² correctly?

Use the formula $ (x+y)^2 = x^2 + 2xy + y^2 $. So $ (a+3)^2 = a^2 + 2(a)(3) + 3^2 = a^2 + 6a + 9 $. Don't forget the middle term!

Q: Why does the height equal b² and not just b?

When we solve $ b^4 = h^2 $, we take the square root of both sides. Since $ \sqrt{b^4} = b^2 $ (assuming positive values), the height is b², not b.

Q: What if I get negative values for the dimensions?

Dimensions of geometric shapes must be positive. If you get negative values, check your algebra or consider that the problem might have no real solution for those parameter values.

Q: How can I verify my answer is correct?

Substitute your dimensions back into the diagonal formula: $ \sqrt{(2a)^2 + (a+3)^2 + (b^2)^2} $ should equal $ \sqrt{5a^2+6a+b^4+9} $. Expand and simplify to check!

3D Pythagorean Theorem with Algebraic Expressions

An orthohedron has a diagonal that is $\sqrt{5a^2+6a+b^4+9}$ long.

Its length is $2a$ and its width is $a+3$ .

Calculate the dimensions of the orthohedron.

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Calculate the dimensions of the box

00:03 Use the Pythagorean theorem in triangle A1D1C1 to find A1C1

00:16 Substitute appropriate values according to the given data and solve for A1C1

00:26 Open brackets properly

00:37 Collect terms and arrange

00:41 This is the diagonal of the face squared

00:48 Now use the Pythagorean theorem in triangle CC1A1 to find CC1

01:05 Substitute appropriate values according to the given data and solve for CC1

01:12 Substitute the value of A1C according to the given data

01:21 Simplify what's possible

01:34 This is the size of box CC1

01:44 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

An orthohedron has a diagonal that is $\sqrt{5a^2+6a+b^4+9}$ long.

Its length is $2a$ and its width is $a+3$ .

Calculate the dimensions of the orthohedron.

Step-by-step solution

The problem involves an orthohedron with a given diagonal length expressed as $\sqrt{5a^2+6a+b^4+9}$ . The known dimensions are length $2a$ and width $a+3$ , and we need to calculate the unknown height.

Using the Pythagorean theorem in three dimensions: $d^2 = (\text{length})^2 + (\text{width})^2 + (\text{height})^2$ .

Substitute the given values: $\left( \sqrt{5a^2 + 6a + b^4 + 9} \right)^2 = (2a)^2 + (a+3)^2 + h^2$ .

Simplifying gives: $5a^2 + 6a + b^4 + 9 = 4a^2 + (a^2 + 6a + 9) + h^2$ .

Combine like terms on the right: $5a^2 + 6a + b^4 + 9 = 5a^2 + 6a + 9 + h^2$ .

Subtract $5a^2 + 6a + 9$ from both sides: $b^4 = h^2$ .

This gives the height as $h = b^2$ .

Thus, the dimensions of the orthohedron are $(2a, a+3, b^2)$ .

Therefore, the solution to the problem is $\boxed{(2a, a+3, b^2)}$ .

Final Answer

$2a,a+3,b^2$

Key Points to Remember

Essential concepts to master this topic

Rule: For orthohedron diagonal: $d^2 = length^2 + width^2 + height^2$
Technique: Expand $(a+3)^2 = a^2 + 6a + 9$ then combine like terms
Check: Verify $5a^2 + 6a + b^4 + 9 = 4a^2 + a^2 + 6a + 9 + (b^2)^2$ ✓

Common Mistakes

Avoid these frequent errors

Forgetting to expand (a+3)² completely
Don't write (a+3)² = a² + 9 and skip the middle term = missing 6a in your calculation! This gives you the wrong equation and wrong height. Always expand binomial squares using (x+y)² = x² + 2xy + y².

Practice Quiz

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Look at the triangle in the diagram. How long is side AB?

FAQ

Everything you need to know about this question

Why do we use the 3D Pythagorean theorem here?

An orthohedron is a 3D rectangular box, so its main diagonal passes through all three dimensions. The regular Pythagorean theorem only works for 2D triangles, but for 3D boxes we need $d^2 = l^2 + w^2 + h^2$ .

How do I expand (a+3)² correctly?

Use the formula $(x+y)^2 = x^2 + 2xy + y^2$ . So $(a+3)^2 = a^2 + 2(a)(3) + 3^2 = a^2 + 6a + 9$ . Don't forget the middle term!

Why does the height equal b² and not just b?

When we solve $b^4 = h^2$ , we take the square root of both sides. Since $\sqrt{b^4} = b^2$ (assuming positive values), the height is b², not b.

What if I get negative values for the dimensions?

Dimensions of geometric shapes must be positive. If you get negative values, check your algebra or consider that the problem might have no real solution for those parameter values.

How can I verify my answer is correct?

Substitute your dimensions back into the diagonal formula: $\sqrt{(2a)^2 + (a+3)^2 + (b^2)^2}$ should equal $\sqrt{5a^2+6a+b^4+9}$ . Expand and simplify to check!

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