Pythagorean Theorem in Cuboids Practice Problems

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:

$A_1B=\sqrt{244}$

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

$A_1B=2\sqrt{61}$

We will use the Pythagorean theorem to find diagonal AD1:

$AA_1+A_1D_1=AD_1$

Let's input the known data:

$5^2+7^2=D_1A^2$

$D_1A^2=25+49=74$

Let's find the square root:

$AD_1=\sqrt{74}$

From the data we can see that:

$AA_1=DD_1=5$

Now let's look at triangle DD1C1 and calculate DC1 using the Pythagorean theorem:

$D_1D^2+D_1C_1^2=C_1D^2$

Let's input the existing data:

$5^2+4^2=C_1D^2$

$C_1D^2=25+16=41$

Let's find the square root:

$DC_1=\sqrt{41}$

Now let's focus on triangle A1D1C1 and find diagonal A1C1:

$A_1D_1^2+D_1C_1^2=A_1C_1^2$

Let's input the known data:

$7^2+4^2=A_1C_1^2$

$A_1C_1^2=49+16=65$

Let's find the square root:

$A_1C_1=\sqrt{65}$

Now we have all 3 lengths of all possible diagonal corners in the box:

$\sqrt{74},\sqrt{41},\sqrt{65}$

Let's look at face CC1D1D and use the Pythagorean theorem to find the diagonal of the face:

$D_1C_1^2+CC_1^2=D_1C^2$

Let's insert the known data:

$10^2+4^2=D_1C^2$

$116=D_1C^2$

Let's focus a bit on triangle BCD1 and use the Pythagorean theorem to find diagonal BD1:

$D_1C^2+CB^2=BD_1^2$

Let's insert the known data:

$116+7^2=BD_1^2$

$116+49=BD_1^2$

$165=BD_1^2$

Let's find the root:

$\sqrt{165}=BD_1$

To find the length of the diagonal $AC$ on a cuboid, we will use the Pythagorean theorem twice:

• Step 1: Calculate $AB$, the diagonal of the face of the cuboid:

We assume the cuboid's known dimensions, with width $8 \, \text{cm}$ and height $4 \, \text{cm}$. Assume $l$ for one dimension along the base.

Since $AB = \sqrt{w^2 + l^2}$, solving for AB means understanding both directions. As the length isn't given, we solve specifically for vertically so $AC$ depends on full volume space:

• Step 2: Calculate diagonal $AC$:

$AC = \sqrt{AB^2 + \text{height}^2} = \sqrt{8^2 + 4^2} = \sqrt{64 + 16} = \sqrt{80} \, \text{cm}$

In this way, we determine the length of diagonal $AC$ is $\sqrt{80}$ cm.

The correct choice corresponding to this calculation is Choice 3: $\sqrt{80}$ cm.

The problem asks us to find the height of a rectangular prism. Given the space diagonal $d = 2\sqrt{14}$ cm and base dimensions (length = 6 cm, width = 4 cm), we can use the formula for the space diagonal:

$d = \sqrt{l^2 + w^2 + h^2}$

Substitute the known values:

$2\sqrt{14} = \sqrt{6^2 + 4^2 + h^2}$

First, calculate $6^2 + 4^2$:

$6^2 = 36, \quad 4^2 = 16$

Thus, $6^2 + 4^2 = 52$.

Substitute back into the equation:

$(2\sqrt{14})^2 = 52 + h^2$

Calculate $(2\sqrt{14})^2$:

$2^2 \times (\sqrt{14})^2 = 4 \times 14 = 56$

Substitute and solve for $h^2$:

$56 = 52 + h^2$

Subtract 52 from both sides:

$h^2 = 4$

Take the square root of both sides:

$h = \sqrt{4} = 2$

Therefore, the height of the rectangular prism $BB^1$ is $2$ cm.