A rectangular pyramid is a geometrical solid with a rectangular base bounded by four lateral faces meeting above the base at a common vertex, known as the apex. Each pair of opposite triangular lateral faces is congruent.
Rectangular Pyramid
How many faces,edges and vertices does a rectangular pyramid have?
A rectangular pyramid has 5 faces, 8 edges, and 5 vertices.
A net for a rectangular pyramid can illustrate its shape from a 2-D view. This net can be folded along the dotted lines to form a rectangular pyramid as shown in the diagram.
Rectangular Pyramid Net
Based on the position of its apex, a rectangular pyramid can be classified into 2 types – (1) Right rectangular pyramid, and (2) Oblique rectangular pyramid
Types
Right and Oblique
Right Rectangular Pyramid
A right rectangular pyramid is a pyramid whose apex is aligned right above its base center. An imaginary line drawn from the apex intersects the base at its center at a right angle. This line is its height.
In contrast, when the apex away from the base center, it is an oblique rectangular pyramid. Its height is a perpendicular line from the apex to the base.
Like all other polyhedrons, we can calculate the surface area and the volume of a rectangular pyramid.
Formulas
Volume
The formula is:
Volume (V) = ${\dfrac{1}{3}lwh}$, here l = base length, w = base width, h = height
Let us solve some examples to understand the concept better.
Find the volume of a rectangular-based pyramid with a base length of 19 cm, a base width of 14 cm, and a height of 20 cm.
Solution:
As we know, Volume (V) = ${\dfrac{1}{3}lwh}$, here l = 19 cm, w = 14 cm, h = 20 cm ∴ V = ${\dfrac{1}{3}\times 19\times 14\times 20}$ = 1773.33 cm3
Find the volume of a rectangular pyramid with bases of 15 cm and 11 cm and a height of 23 cm.
Solution:
As we know, Volume (V) = ${\dfrac{1}{3}lwh}$, here l = 15 cm, w = 11 cm, h = 23 cm ∴ V = ${\dfrac{1}{3}\times 15\times 15\times 23}$ = 1265 cm3
Finding the volume of a truncated rectangular pyramid when BOTTOM BASE LENGTH, BOTTOM BASE WIDTH, TOP BASE LENGTH, TOP BASE WIDTH, and HEIGHT are known
Find the volume of a truncated rectangular pyramid given in the figure.
Solution:
Here, we will use a formula for the truncated rectangular pyramid. Volume (V) = ${\dfrac{Ab+aB+2\left( ab+AB\right) }{6}\times h}$, here A = 9 cm, B = 8 cm, a = 4.5 cm, b = 4 cm, h = 5 cm ∴ V = ${\dfrac{9\times 4+4.5\times 8+2\left( 4.5\times 4+9\times 8\right) }{6}\times 5}$ [9 × 4 + 4.5 × 8 + 2(4.5 × 4 + 9 × 8)] × 5/6 = 210 cm3
Surface Area
The formula is:
Surface Area (SA) = ${lw+\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = base length, w = width, h = height
Also, ${\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$ = lateral surface area (LSA)
∴ SA = lw + LSA
Let us solve some examples to understand the concept better.
Find the lateral and total surface of a rectangular pyramid with a base length of 8 cm, a base width of 5 cm, and a height of 12 cm.
Solution:
As we know, Lateral Surface Area (LSA) = ${\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = 8 cm, w = 5 cm, h = 12 cm ∴ LSA = ${\dfrac{1}{2}\times 5\sqrt{4\times 12^{2}+8^{2}}+\dfrac{1}{2}\times 8\sqrt{4\times 12^{2}+5^{2}}}$ = 161.31 cm2 Total Surface Area (TSA) = lw + LSA, here l = 8 cm, w = 5 cm, LSA = 161.31 cm2 ∴ TSA = 8 × 5 + 161.31 = 201.31 cm2
Find the surface of a right rectangular pyramid with bases of 9 cm, 3 cm, and a height of 7 cm.
Solution:
As we know, Surface Area = Total Surface Area (TSA) ∴ Total Surface Area (TSA) = ${lw+\dfrac{1}{2}w\sqrt{4h^{2}+l^{2}}+\dfrac{1}{2}l\sqrt{4h^{2}+w^{2}}}$, here l = 9 cm, w = 3 cm, h = 7 cm ∴ TSA = ${9\times 3+\dfrac{1}{2}\times 3\sqrt{4\times 7^{2}+9^{2}}+\dfrac{1}{2}\times 9\sqrt{4\times 7^{2}+3^{2}}}$ = 116.4 cm2
Let us learn how to find the surface area of a rectangular pyramid with slant height. We will use the general formula, Total Surface Area (TSA) = ${B+\dfrac{1}{2}Ps}$, here B = base area, P = base perimeter, l = slant height
Finding the surface area of a rectangular pyramid when BASE LENGTH, BASE WIDTH, and SLANTHEIGHT are known
Find the surface of a rectangular pyramid with bases of 10 cm, 6 cm, and a slant height of 13 cm.
Solution:
Here, we will use the general formula. Total Surface Area (TSA) = ${B+\dfrac{1}{2}Ps}$,here B = base area, P = base perimeter, s = slant height B = l × w, here l = 10 cm, w = 6 cm, = 10 × 6 = 60 cm2 P = 2(l + w), here l = 10 cm, w = 6 cm = 2 × (10 + 6) = 32 cm Plugging the value of B and P in the general formula, Total Surface Area (TSA) = ${B+\dfrac{1}{2}Ps}$, here B = 60 cm2, P = 32 cm, s = 13 cm ∴ TSA = ${60+\dfrac{1}{2}\times 32\times 13}$ = 268 cm2
Find the volume of a truncated rectangular pyramid given in the figure.