What will be the length of longest diagonal of the cuboid having length 13 cm width 11 cm and height 20 cm?
26.27 cm
A cuboid is a three-dimensional solid shape bounded by six rectangular faces. It is also known as a rectangular prism. The longest diagonal of a cuboid is a line segment connecting two opposite vertices that do not lie on the same face.
The length of the longest diagonal (\(d\)) of a cuboid with length (\(l\)), width (\(w\)), and height (\(h\)) is given by the formula derived from extending the Pythagorean theorem to three dimensions:
\[ d = \sqrt{l^2 + w^2 + h^2} \]
In this problem, we are given the dimensions of the cuboid:
Now, we substitute these values into the formula:
\[ d = \sqrt{(13 \, \text{cm})^2 + (11 \, \text{cm})^2 + (20 \, \text{cm})^2} \]
Calculate the squares of each dimension:
Sum the squared values:
\[ 13^2 + 11^2 + 20^2 = 169 + 121 + 400 = 690 \]
Now, take the square root of the sum:
\[ d = \sqrt{690} \, \text{cm} \]
Calculating the square root of 690 gives an approximate value:
\[ \sqrt{690} \approx 26.2679 \, \text{cm} \]
Rounding this value to two decimal places, we get:
\[ d \approx 26.27 \, \text{cm} \]
The length of the longest diagonal of the cuboid is approximately 26.27 cm.
Comparing this result with the given options:
The calculated length matches Option 2.
| Concept | Formula | Description |
|---|---|---|
| Volume (V) | \(V = l \times w \times h\) | Space occupied by the cuboid |
| Surface Area (SA) | \(SA = 2(lw + lh + wh)\) | Total area of all faces |
| Longest Diagonal (d) | \(d = \sqrt{l^2 + w^2 + h^2}\) | Length of the longest distance between two vertices |
The formula for the longest diagonal of a cuboid is a direct application of the Pythagorean theorem in three dimensions. Consider a cuboid with vertices at (0,0,0) and (l,w,h). The distance between these two points is the length of the longest diagonal. Using the distance formula in 3D (which is derived from the Pythagorean theorem), the distance is \(\sqrt{(l-0)^2 + (w-0)^2 + (h-0)^2}\), which simplifies to \(\sqrt{l^2 + w^2 + h^2}\). This shows how the familiar theorem extends from 2D right triangles to 3D rectangular solids.
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