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The Pythagorean Theorem in Three Dimensions Practice Complete this assessment to review what you've learned. It will not count...Question
The Pythagorean Theorem in Three Dimensions Practice
Complete this assessment to review what you've learned. It will not count toward your grade.
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An illustration shows a three dimensional rectangular prism with labeled parts. The top, front and right sides are visible. The bottom, left, and rear sides are not visible, but their positions are indicated by dashed lines. The vertical edge joining the front and left sides is labeled h. A right angle is indicated where that edge meets the front base edge. The front base edge is labeled l. A right angle is indicated where that edge meets the right base edge. The right base edge is labeled w. Two dotted lines extend from the vertex at the back right side of the prism through the interior. The first dotted line extends up to the vertex at the front left side and is labeled d. The second dotted line extends diagonally across the base to the vertex at the front left side where the perpendicular height line meets the base edge and is labeled x.
Given l=30 in.
, w=20 in.
, h=18 in.
, and x=36.1 in.
, find the length of d
. Round the answer to the nearest tenth.
(1 point)
in.
Remaining Attempts : 3
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                    Answered by
            GPT-4o mini
            
    To find the length of \( d \) in the three-dimensional rectangular prism using the Pythagorean theorem, we'll apply the theorem in two stages. 
1. First, we find the diagonal length across the base, which we'll call \( x \). This can be calculated using the lengths of \( l \) and \( w \):
\[
x = \sqrt{l^2 + w^2}
\]
Given \( l = 30 \, \text{in} \) and \( w = 20 \, \text{in} \):
\[
x = \sqrt{30^2 + 20^2} = \sqrt{900 + 400} = \sqrt{1300} \approx 36.1 \, \text{in}
\]
This aligns with the given value for \( x \).
2. Next, we find the diagonal length \( d \) using the length \( x \) and the height \( h \):
\[
d = \sqrt{x^2 + h^2}
\]
With \( h = 18 \, \text{in} \):
\[
d = \sqrt{(36.1)^2 + (18)^2} = \sqrt{1300 + 324} = \sqrt{1624}
\]
Now, we calculate that:
\[
\sqrt{1624} \approx 40.3 \, \text{in}
\]
Thus, rounding to the nearest tenth, the length of \( d \) is:
\[
\boxed{40.3} \, \text{in}
\]
    
1. First, we find the diagonal length across the base, which we'll call \( x \). This can be calculated using the lengths of \( l \) and \( w \):
\[
x = \sqrt{l^2 + w^2}
\]
Given \( l = 30 \, \text{in} \) and \( w = 20 \, \text{in} \):
\[
x = \sqrt{30^2 + 20^2} = \sqrt{900 + 400} = \sqrt{1300} \approx 36.1 \, \text{in}
\]
This aligns with the given value for \( x \).
2. Next, we find the diagonal length \( d \) using the length \( x \) and the height \( h \):
\[
d = \sqrt{x^2 + h^2}
\]
With \( h = 18 \, \text{in} \):
\[
d = \sqrt{(36.1)^2 + (18)^2} = \sqrt{1300 + 324} = \sqrt{1624}
\]
Now, we calculate that:
\[
\sqrt{1624} \approx 40.3 \, \text{in}
\]
Thus, rounding to the nearest tenth, the length of \( d \) is:
\[
\boxed{40.3} \, \text{in}
\]
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