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Posted on Jan 02, 2017

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SOURCE: find the length of the missing side of triangle

a. 49.93

b. 95.19

c. 50.93

d. 23.60

e. 122

Posted on Jul 06, 2009

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Hello Jacqueline;

My name Peter. I am a engineering college graduate. Later I changed careers and now, I am a retired field service appliance technician. So, I work on the side and it you have any appliances that need repair call me @ 414306-1362.

The length on one side of the end triangle is "a"

The area of the end triangle is: (a x a) x 0.433 = Surface Area

The surface area of the prism face = a x length.

The volume is: The surface area of the end x the surface area of the face.

My name Peter. I am a engineering college graduate. Later I changed careers and now, I am a retired field service appliance technician. So, I work on the side and it you have any appliances that need repair call me @ 414306-1362.

The length on one side of the end triangle is "a"

The area of the end triangle is: (a x a) x 0.433 = Surface Area

The surface area of the prism face = a x length.

The volume is: The surface area of the end x the surface area of the face.

Mar 06, 2015 | ixl.com

Presuming this is a triangular prism, just get the area of triangle and then multiply by the length of the prism. So, 1/2 * B * H of triangle times the L (length) of prism.

Oct 28, 2014 | Office Equipment & Supplies

I assume the length of a side is 16 in, and the height of the prism is 23 inches.

**Volume of a prism is V_prism=(Area of base)*(height of prism)**

Area of base is made up of the area of the 6 equilateral triangles with sides =16 inches. The height of such triangles is given by**8*SQRT(3)**.

Use the Pythagorean theorem or the trigonometric ratios to find the height of each of the 6 traingles.

Area of base =6*16*(8*SQRT(3))/2 sq in

**Volume =6*8*8*23*SQRT(3) cubic inches**

Calculate the value in cubic inches and convert to gallons: I leave that task to you.

Area of base is made up of the area of the 6 equilateral triangles with sides =16 inches. The height of such triangles is given by

Use the Pythagorean theorem or the trigonometric ratios to find the height of each of the 6 traingles.

Area of base =6*16*(8*SQRT(3))/2 sq in

Calculate the value in cubic inches and convert to gallons: I leave that task to you.

May 12, 2014 | Office Equipment & Supplies

Here is to get you started. To increase the size of the image do a CTRL Plus (+) in your browser.

You need to calculate the slant height of the pyramid for the formula of the lateral area. You should find a value of** (1/2)*SQRT(203) **or about 7.1239 cm

You need to calculate the altitude (height) of the pyramid from the apex (summit) to the center of the base triangle (center of inscribed circle, barycenter, orthocenter). The hypotenuse of such triangle is the slant height. One leg is the altitude (to be found),**the measure of the second leg is (1/3) the altitude** **of the equilateral triangle** that forms the base. You should find (1/3) m MH= (1/3)* **(11/2)*SQRT(3)**

1. Calculate the area of the base (use a formula for the equilateral triangle or the general formula for a triangle: you have its height MH ).

2. Lateral area = 3 times the area of triangle Triangle ECD (in yellow above).

3. Total area = area of base + lateral area.

4. Volume= (1/3)(Area of base)* (height of pyramid)

If you can see the details on the screen capture below, fine, Press CTRL + in your browser to increase the size.

You need to calculate the slant height of the pyramid for the formula of the lateral area. You should find a value of

You need to calculate the altitude (height) of the pyramid from the apex (summit) to the center of the base triangle (center of inscribed circle, barycenter, orthocenter). The hypotenuse of such triangle is the slant height. One leg is the altitude (to be found),

1. Calculate the area of the base (use a formula for the equilateral triangle or the general formula for a triangle: you have its height MH ).

2. Lateral area = 3 times the area of triangle Triangle ECD (in yellow above).

3. Total area = area of base + lateral area.

4. Volume= (1/3)(Area of base)* (height of pyramid)

If you can see the details on the screen capture below, fine, Press CTRL + in your browser to increase the size.

Mar 29, 2014 | Office Equipment & Supplies

2.0833 litres.

If this is homework, be sure to show your work.

If this is homework, be sure to show your work.

Feb 27, 2014 | Office Equipment & Supplies

Since it is rectangular prism, the base has the shape of a rectangle (Length L, Width W). Its area is

Conclusion: Volume of a right rectangular prism with dimensions L, W, H, is

Dec 18, 2013 | Computers & Internet

Right Rectangular prism, Length L, width W, Height H

Volume=2058 cm^3=L*W*H

However

L=3W, and H=2W

(3W)*W*(2W)=**6 W^3=2058**

W^3=(2058/6)=343

**W**=Cube root of 343=**(343)^(1/3)=7 cm**

**L=3*7=21 cm**

H=2*7=14 cm

Volume=2058 cm^3=L*W*H

However

L=3W, and H=2W

(3W)*W*(2W)=

W^3=(2058/6)=343

H=2*7=14 cm

Oct 29, 2013 | Computers & Internet

First a square has equal sides, it means s= square root of 225 = 15.

Height of the base is equal to 15. Volume of the prism = Base area x Height = 225(15) = 3375 cu. cm x 1 liter / 1000 cu. cm = 3.375 liters

Ans = 3.375 liters

Height of the base is equal to 15. Volume of the prism = Base area x Height = 225(15) = 3375 cu. cm x 1 liter / 1000 cu. cm = 3.375 liters

Ans = 3.375 liters

May 11, 2013 | Cars & Trucks

Find the vol of the water in cu/in

15 * 231 = 3465 cu/in

:

Change 11 ft: 11*12 = 132 inches

:

Find the triangular area of the end of the trough; 3465/132 = 26.25 sq/inches

:

Let x = width of the trough; A right triangle area = [img width="16" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image001.jpg" alt="1%2F2" v:shapes="Picture_x0020_1"> x^2 = 2(26.25)

x =[img width="53" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image002.jpg" alt="sqrt%2852.5%29" v:shapes="Picture_x0020_2"> x = 7.2457 inches the width of each board

Find the width of the trough which is the the hypotenuse

h =[img width="69" height="40" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image003.jpg" alt="sqrt%28x%5E2+%2B+x%5E2%29" v:shapes="Picture_x0020_5"> h =[img width="110" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image004.jpg" alt="sqrt%2852.5+%2B+52.5%29" v:shapes="Picture_x0020_6"> h =[img width="43" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image005.jpg" alt="sqrt%28105%29" v:shapes="Picture_x0020_7"> h = 10.247 inches wide, also the base of the triangle

:

Find the altitude of the triangle, which is the depth of the water; we know it's area[img width="16" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image001.jpg" alt="1%2F2" v:shapes="Picture_x0020_8"> 10.247d = 2(26.25)

10.247d = 52.5

d =[img width="56" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image006.jpg" alt="52.5%2F10.247" v:shapes="Picture_x0020_9"> d = 5.12 inches is the depth of the water

15 * 231 = 3465 cu/in

:

Change 11 ft: 11*12 = 132 inches

:

Find the triangular area of the end of the trough; 3465/132 = 26.25 sq/inches

:

Let x = width of the trough; A right triangle area = [img width="16" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image001.jpg" alt="1%2F2" v:shapes="Picture_x0020_1"> x^2 = 2(26.25)

x =[img width="53" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image002.jpg" alt="sqrt%2852.5%29" v:shapes="Picture_x0020_2"> x = 7.2457 inches the width of each board

Find the width of the trough which is the the hypotenuse

h =[img width="69" height="40" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image003.jpg" alt="sqrt%28x%5E2+%2B+x%5E2%29" v:shapes="Picture_x0020_5"> h =[img width="110" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image004.jpg" alt="sqrt%2852.5+%2B+52.5%29" v:shapes="Picture_x0020_6"> h =[img width="43" height="22" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image005.jpg" alt="sqrt%28105%29" v:shapes="Picture_x0020_7"> h = 10.247 inches wide, also the base of the triangle

:

Find the altitude of the triangle, which is the depth of the water; we know it's area[img width="16" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image001.jpg" alt="1%2F2" v:shapes="Picture_x0020_8"> 10.247d = 2(26.25)

10.247d = 52.5

d =[img width="56" height="39" src="file:///C:\Users\user\AppData\Local\Temp\msohtmlclip1\01\clip_image006.jpg" alt="52.5%2F10.247" v:shapes="Picture_x0020_9"> d = 5.12 inches is the depth of the water

Jul 24, 2011 | Computers & Internet

Hello

**Rectangular Prism/Cuboid Definition:**

A Rectangular Prism/Cuboid is a solid figure bounded by six rectangular faces, a rectangular box. All angles are right angles, and opposite faces of a cuboid are equal. It is also a right rectangular prism.

**Rectangular Prism/Cuboid Formula**:

Area of Base(A) = l * w

Perimeter of Base(P) = 2l + 2w

Surface Area of Prism = 2(lw) + (2l + 2w)h = 2A + Ph

Volume of Prism = lwh = Ah

Diagonal of Prism = Sqrt(l² + w² + h²)

where

l = length, w = width, h = height

Hope this helps, if so do rate the solution

A Rectangular Prism/Cuboid is a solid figure bounded by six rectangular faces, a rectangular box. All angles are right angles, and opposite faces of a cuboid are equal. It is also a right rectangular prism.

Area of Base(A) = l * w

Perimeter of Base(P) = 2l + 2w

Surface Area of Prism = 2(lw) + (2l + 2w)h = 2A + Ph

Volume of Prism = lwh = Ah

Diagonal of Prism = Sqrt(l² + w² + h²)

where

l = length, w = width, h = height

Hope this helps, if so do rate the solution

Jan 15, 2011 | MathRescue Word Problems Of Algebra Lite

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