Question

8. Find the surface area of this rectangular pyramid

15.6 mm height

18 mm width

18 mm length

Answer 1

Answer :

• 972.4 mm^2

Explanation :

The total surface area of a rectangular pyramid is the sum of the area of the base,the area of the front,back and the side triangles.

The total surface area of a rectangle pyramid is given by,

• wl + l√((w/2)^2 + h^2) + w√((l/2)^2 + h^2)

where,

• w = width
• l = length
• h = height

plugging in,

• (18*18 + 18√((18/2)^2 + 15.6^2) + 18√((18/2)^2 + 15.6^2)) mm^2
• 972.4 mm^2

Answer 2

Answer:

972.36 mm² (2 d.p.)

Step-by-step explanation:

The formula to calculate the surface area of a rectangular pyramid is:

$\boxed{\begin{array}{l}\underline{\textsf{Surface Area of a Rectangular Pyramid}}\\\\SA =wl+\dfrac{1}{2}w\sqrt{4h^2+l^2}+\dfrac{1}{2}l\sqrt{4h^2+w^2}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{ SA is the surface area.}\\\phantom{ww}\bullet\;\textsf{ w is the width of the rectangular base.}\\\phantom{ww}\bullet\;\textsf{ l is the length of the rectangular base.}\\\phantom{ww}\bullet\;\textsf{ h is the height perpendicular to the base.}\end{array}}$

In this case:

• w = 18 mm
• l = 18 mm
• h = 15.6 mm

Substitute the given values into the formula and solve for SA:

$SA =(18)(18)+\dfrac{1}{2}(18)\sqrt{4(15.6)^2+(18)^2}+\dfrac{1}{2}(18)\sqrt{4(15.6)^2+(18)^2}\\\\\\SA =324+9\sqrt{4(243.36)+324}+9\sqrt{4(243.36)+324}\\\\\\SA =324+9\sqrt{973.44+324}+9\sqrt{973.44+324}\\\\\\SA =324+9\sqrt{1297.44}+9\sqrt{1297.44}\\\\\\SA =972.3599000555...\\\\\\SA=972.36\; \sf mm^2\;(2\; d.p.)$

Therefore, the surface area of the given rectangular pyramid rounded to the nearest thousandth is:

$\Large\boxed{\boxed{972.36\; \sf mm^2}}$