Other Polyhedrons - Math

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Find the surface area of the following polyhedron.

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The formula for the surface area of a polyhedron is:

$SA = (cone)+ $\frac{1}{2}$(sphere)$

$SA = $\frac{1}{2}$(2\pi r)(h_s)+ $\frac{1}{2}$(4\pi $r^2$)$

$SA = \pi rh_s+ 2\pi $r^2$$

where is the radius of the polyhedron and is the slant height of the cone.

Use the formula for a triangle to find the length of the radius:

$a-a-a$\sqrt{2}$$

$4$\sqrt{2}$m-4$\sqrt{2}$m-8m$

Plugging in our values, we get:

$SA = \pi (4$\sqrt{2}$m)(8m)+ 2\pi $(4$\sqrt{2}$m)^2$$

$SA = 32$\sqrt{2}$\pi + 64 \pi $m^2$$

Find the surface area of the following half-cylinder.

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The formula for the surface area of a half-cylinder must include one-half of the surface area of a cylinder, which would be:

$SA = $\frac{1}{2}$ (lateral\ area + 2 \cdot (base))$

We also need to add the area of the new rectangular face that is created by cutting the cylinder in half. The area of this rectangle would be:

$A=l\times w$

where the length of the rectangle is the same as the height of the half-cylinder, and the width of the rectangle is the same as the diameter of the base of the half-cylinder. So we can rewrite the area of the rectangle as:

$A=h\p\times d$

Now we can combine the two area formulas to find the total surface area of the half-cylinder:

$SA = $\frac{1}{2}$ (2 \pi r (h) + 2 (\pi $r^2$))+h\cdot d$

$SA = \pi r (h) + \pi $r^2$+hd$

where is the radius of the base and is the length of the height, and is the diameter of the base.

Plugging in our values, we get:

$SA = \pi (5m) (15m) + \pi $(5m)^2$+15\cdot 10$

$SA = 75 \pi $m^2$ + 25 \pi $m^2$+150 = 150+100 \pi $m^2$$

Find the surface area of the following polyhedron.

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The formula for the surface area of the polyhedron is:

$SA = $SA_{cone}$$+\frac{1}{2}$SA_{sphere}$$

$SA = \pi r l+\frac{1}{2}$(4 \pi $r^2$)$

$SA = \pi r l+2 \pi $r^2$$

Where is the radius of the cone, is the slant height of the cone, and is the radius of the sphere

Use the formula for a triangle to find the radius and slant height:

$a-a$\sqrt{2}$-2a$

$3$\sqrt{3}$m-9m-6$\sqrt{3}$m$

Plugging in our values, we get:

$SA = \pi r l+2 \pi $r^2$$

$SA = $(\pi)(3$\sqrt{3}$m)(6$\sqrt{3}$m)+2(\pi)(3$\sqrt{3}$m)^2$$

$SA = 54 \pi $m^2$ + 54 \pi $m^2$ = 108 \pi $m^2$$

Find the surface area of the following polyhedron.

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The formula for the surface area of the polyhedron is:

$SA = (cone:no\ base)+(cylinder:one\ base)$

$SA = $\frac{1}{2}$(circumference)(slant\ height)+(base)+(circumference)(height)$

$SA = $\frac{1}{2}$(2 \pi r)(h_s)+(\pi $r^2$)+(2 \pi r)(h)$

where is the radius of the cone, is the slant height of the cone, is the radius of the cylinder, and is the height of the cylinder.

Use the formula for a triangle to find the length of the radius:

$a-a$\sqrt{3}$-2a$

$3m-3$\sqrt{3}$m-6m$

Plugging in our values, we get:

$SA = $\frac{1}{2}$(2\pi $(3m))(6m)+(\pi)(3m)^2$+(2\pi (3m)(9m))$

$SA = 81 \pi $m^2$$

Find the volume of the following half cylinder.

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The formula for the volume of a half-cylinder is:

$V = $\frac{1}{2}$ (\pi $r^2$ h)$

where is the radius of the base and is the length of the height.

Plugging in our values, we get:

$V = $\frac{1}{2}$ \pi $(5m)^2$ (15m)$

$V = 187.5 \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = $V_{cone}$$+\frac{1}{2}$V_{sphere}$$

$V = $\frac{\pi $r^2$ h}{3}$ +\frac{1}{2}$ $\frac{4\pi $r^3$}{3}$$

$V = $\frac{\pi $r^2$ h}{3}$ +\frac{2\pi $r^3$}{3}$$

Where is the radius of the cone, is the height of the cone, and is the radius of the sphere.

Use the formula for a triangle to find the length of the radius:

$a-a$\sqrt{2}$-2a$

$3$\sqrt{3}$m-9m-6$\sqrt{3}$m$

Plugging in our values, we get:

$V = $\frac{\pi $(3\sqrt{3}$m)^29m$}{3}+\frac{2 \pi $(3\sqrt{3}$m)^3$}{3}$

$V = $\frac{\pi $(27m^2$)9m}{3}$+\frac{2 \pi $(81\sqrt{3}$m^3$)}{3}$

$V = 81 \pi $m^3$+54$\sqrt{3}$ \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = $\frac{\pi $r^2$ h}{3}$+\pi $r^2$ h$

where is the radius of the cone, is the height of the cone, is the radius of the cylinder, and is the height of the cylinder.

Use the formula for a triangle to find the length of the radius and height of the cone:

$a-a$\sqrt{3}$-2a$

$3m-3$\sqrt{3}$m-6m$

Plugging in our values, we get:

$V = $\frac{\pi $(3m)^2$ (3\sqrt{3}$m)}{3}+\pi $(3m)^2$ (9m)$

$V = 9 \pi $$\sqrt{3}$m^3$ + 81 \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = (cone) + $\frac{1}{2}$(sphere)$

$V = $\frac{1}{3}$(\pi $r^2$)(h)+\frac{1}{2}$\left($\frac{4\pi $r^3$}{3}\right)$

where is the radius of the polyhedron and is the height of the cone.

Use the formula for a triangle to find the length of the radius and height:

$a-a-a$\sqrt{2}$$

$4$\sqrt{2}$m-4$\sqrt{2}$m-8m$

Plugging in our values, we get:

$V = $\frac{1}{3}$(\pi $(4$\sqrt{2}$m)^2$)(4$\sqrt{2}$m)+\frac{1}{2}$\left($\frac{4\pi $(4\sqrt{2}$m)^3$}{3}\right)$

$V = 128 $\sqrt{2}$ \pi $m^3$$

Find the surface area of the following polyhedron.

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The formula for the surface area of a polyhedron is:

$SA = (cone)+ $\frac{1}{2}$(sphere)$

$SA = $\frac{1}{2}$(2\pi r)(h_s)+ $\frac{1}{2}$(4\pi $r^2$)$

$SA = \pi rh_s+ 2\pi $r^2$$

where is the radius of the polyhedron and is the slant height of the cone.

Use the formula for a triangle to find the length of the radius:

$a-a-a$\sqrt{2}$$

$4$\sqrt{2}$m-4$\sqrt{2}$m-8m$

Plugging in our values, we get:

$SA = \pi (4$\sqrt{2}$m)(8m)+ 2\pi $(4$\sqrt{2}$m)^2$$

$SA = 32$\sqrt{2}$\pi + 64 \pi $m^2$$

Find the surface area of the following half-cylinder.

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The formula for the surface area of a half-cylinder must include one-half of the surface area of a cylinder, which would be:

$SA = $\frac{1}{2}$ (lateral\ area + 2 \cdot (base))$

We also need to add the area of the new rectangular face that is created by cutting the cylinder in half. The area of this rectangle would be:

$A=l\times w$

where the length of the rectangle is the same as the height of the half-cylinder, and the width of the rectangle is the same as the diameter of the base of the half-cylinder. So we can rewrite the area of the rectangle as:

$A=h\p\times d$

Now we can combine the two area formulas to find the total surface area of the half-cylinder:

$SA = $\frac{1}{2}$ (2 \pi r (h) + 2 (\pi $r^2$))+h\cdot d$

$SA = \pi r (h) + \pi $r^2$+hd$

where is the radius of the base and is the length of the height, and is the diameter of the base.

Plugging in our values, we get:

$SA = \pi (5m) (15m) + \pi $(5m)^2$+15\cdot 10$

$SA = 75 \pi $m^2$ + 25 \pi $m^2$+150 = 150+100 \pi $m^2$$

Find the surface area of the following polyhedron.

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The formula for the surface area of the polyhedron is:

$SA = $SA_{cone}$$+\frac{1}{2}$SA_{sphere}$$

$SA = \pi r l+\frac{1}{2}$(4 \pi $r^2$)$

$SA = \pi r l+2 \pi $r^2$$

Where is the radius of the cone, is the slant height of the cone, and is the radius of the sphere

Use the formula for a triangle to find the radius and slant height:

$a-a$\sqrt{2}$-2a$

$3$\sqrt{3}$m-9m-6$\sqrt{3}$m$

Plugging in our values, we get:

$SA = \pi r l+2 \pi $r^2$$

$SA = $(\pi)(3$\sqrt{3}$m)(6$\sqrt{3}$m)+2(\pi)(3$\sqrt{3}$m)^2$$

$SA = 54 \pi $m^2$ + 54 \pi $m^2$ = 108 \pi $m^2$$

Find the surface area of the following polyhedron.

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The formula for the surface area of the polyhedron is:

$SA = (cone:no\ base)+(cylinder:one\ base)$

$SA = $\frac{1}{2}$(circumference)(slant\ height)+(base)+(circumference)(height)$

$SA = $\frac{1}{2}$(2 \pi r)(h_s)+(\pi $r^2$)+(2 \pi r)(h)$

where is the radius of the cone, is the slant height of the cone, is the radius of the cylinder, and is the height of the cylinder.

Use the formula for a triangle to find the length of the radius:

$a-a$\sqrt{3}$-2a$

$3m-3$\sqrt{3}$m-6m$

Plugging in our values, we get:

$SA = $\frac{1}{2}$(2\pi $(3m))(6m)+(\pi)(3m)^2$+(2\pi (3m)(9m))$

$SA = 81 \pi $m^2$$

Find the volume of the following half cylinder.

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The formula for the volume of a half-cylinder is:

$V = $\frac{1}{2}$ (\pi $r^2$ h)$

where is the radius of the base and is the length of the height.

Plugging in our values, we get:

$V = $\frac{1}{2}$ \pi $(5m)^2$ (15m)$

$V = 187.5 \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = $V_{cone}$$+\frac{1}{2}$V_{sphere}$$

$V = $\frac{\pi $r^2$ h}{3}$ +\frac{1}{2}$ $\frac{4\pi $r^3$}{3}$$

$V = $\frac{\pi $r^2$ h}{3}$ +\frac{2\pi $r^3$}{3}$$

Where is the radius of the cone, is the height of the cone, and is the radius of the sphere.

Use the formula for a triangle to find the length of the radius:

$a-a$\sqrt{2}$-2a$

$3$\sqrt{3}$m-9m-6$\sqrt{3}$m$

Plugging in our values, we get:

$V = $\frac{\pi $(3\sqrt{3}$m)^29m$}{3}+\frac{2 \pi $(3\sqrt{3}$m)^3$}{3}$

$V = $\frac{\pi $(27m^2$)9m}{3}$+\frac{2 \pi $(81\sqrt{3}$m^3$)}{3}$

$V = 81 \pi $m^3$+54$\sqrt{3}$ \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = $\frac{\pi $r^2$ h}{3}$+\pi $r^2$ h$

where is the radius of the cone, is the height of the cone, is the radius of the cylinder, and is the height of the cylinder.

Use the formula for a triangle to find the length of the radius and height of the cone:

$a-a$\sqrt{3}$-2a$

$3m-3$\sqrt{3}$m-6m$

Plugging in our values, we get:

$V = $\frac{\pi $(3m)^2$ (3\sqrt{3}$m)}{3}+\pi $(3m)^2$ (9m)$

$V = 9 \pi $$\sqrt{3}$m^3$ + 81 \pi $m^3$$

Find the volume of the following polyhedron.

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The formula for the volume of the polyhedron is:

$V = (cone) + $\frac{1}{2}$(sphere)$

$V = $\frac{1}{3}$(\pi $r^2$)(h)+\frac{1}{2}$\left($\frac{4\pi $r^3$}{3}\right)$

where is the radius of the polyhedron and is the height of the cone.

Use the formula for a triangle to find the length of the radius and height:

$a-a-a$\sqrt{2}$$

$4$\sqrt{2}$m-4$\sqrt{2}$m-8m$

Plugging in our values, we get:

$V = $\frac{1}{3}$(\pi $(4$\sqrt{2}$m)^2$)(4$\sqrt{2}$m)+\frac{1}{2}$\left($\frac{4\pi $(4\sqrt{2}$m)^3$}{3}\right)$

$V = 128 $\sqrt{2}$ \pi $m^3$$

Find the surface area of the following half-cylinder.

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The formula for the surface area of a half-cylinder must include one-half of the surface area of a cylinder, which would be:

$SA = $\frac{1}{2}$ (lateral\ area + 2 \cdot (base))$

We also need to add the area of the new rectangular face that is created by cutting the cylinder in half. The area of this rectangle would be:

$A=l\times w$

where the length of the rectangle is the same as the height of the half-cylinder, and the width of the rectangle is the same as the diameter of the base of the half-cylinder. So we can rewrite the area of the rectangle as:

$A=h\p\times d$

Now we can combine the two area formulas to find the total surface area of the half-cylinder:

$SA = $\frac{1}{2}$ (2 \pi r (h) + 2 (\pi $r^2$))+h\cdot d$

$SA = \pi r (h) + \pi $r^2$+hd$

where is the radius of the base and is the length of the height, and is the diameter of the base.

Plugging in our values, we get:

$SA = \pi (5m) (15m) + \pi $(5m)^2$+15\cdot 10$

$SA = 75 \pi $m^2$ + 25 \pi $m^2$+150 = 150+100 \pi $m^2$$

Find the surface area of the following polyhedron.

Tap to see back →

The formula for the surface area of the polyhedron is:

$SA = $SA_{cone}$$+\frac{1}{2}$SA_{sphere}$$

$SA = \pi r l+\frac{1}{2}$(4 \pi $r^2$)$

$SA = \pi r l+2 \pi $r^2$$

Where is the radius of the cone, is the slant height of the cone, and is the radius of the sphere

Use the formula for a triangle to find the radius and slant height:

$a-a$\sqrt{2}$-2a$

$3$\sqrt{3}$m-9m-6$\sqrt{3}$m$

Plugging in our values, we get:

$SA = \pi r l+2 \pi $r^2$$

$SA = $(\pi)(3$\sqrt{3}$m)(6$\sqrt{3}$m)+2(\pi)(3$\sqrt{3}$m)^2$$

$SA = 54 \pi $m^2$ + 54 \pi $m^2$ = 108 \pi $m^2$$

Find the surface area of the following polyhedron.

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The formula for the surface area of the polyhedron is:

$SA = (cone:no\ base)+(cylinder:one\ base)$

$SA = $\frac{1}{2}$(circumference)(slant\ height)+(base)+(circumference)(height)$

$SA = $\frac{1}{2}$(2 \pi r)(h_s)+(\pi $r^2$)+(2 \pi r)(h)$

where is the radius of the cone, is the slant height of the cone, is the radius of the cylinder, and is the height of the cylinder.

Use the formula for a triangle to find the length of the radius:

$a-a$\sqrt{3}$-2a$

$3m-3$\sqrt{3}$m-6m$

Plugging in our values, we get:

$SA = $\frac{1}{2}$(2\pi $(3m))(6m)+(\pi)(3m)^2$+(2\pi (3m)(9m))$

$SA = 81 \pi $m^2$$

Find the surface area of the following polyhedron.

Tap to see back →

The formula for the surface area of a polyhedron is:

$SA = (cone)+ $\frac{1}{2}$(sphere)$

$SA = $\frac{1}{2}$(2\pi r)(h_s)+ $\frac{1}{2}$(4\pi $r^2$)$

$SA = \pi rh_s+ 2\pi $r^2$$

where is the radius of the polyhedron and is the slant height of the cone.

Use the formula for a triangle to find the length of the radius:

$a-a-a$\sqrt{2}$$

$4$\sqrt{2}$m-4$\sqrt{2}$m-8m$

Plugging in our values, we get:

$SA = \pi (4$\sqrt{2}$m)(8m)+ 2\pi $(4$\sqrt{2}$m)^2$$

$SA = 32$\sqrt{2}$\pi + 64 \pi $m^2$$