Anew building is being constructed at ut austin. it is 200 feet tall, and has a
rectangular 100 ft by 150 ft footprint. the building has a walkway that is x feet wide
surrounding all four sides of the building. if the total area (around and including the
building) is 16,536 square feet, how wide is the walkway?

Walkway is 3 feet wide.

Step-by-step explanation:

Given:

Length of the building is 150 ft

width of the building is 100 ft

walkway is x ft wide.

Total area(around and including the building) = 16536 square feet

Solution:

Length of the total area will be = length of the building + width of walkway from 2 sides =

Width of the total area will be = width of the building + width of walkway from 2 sides =

Now Total area(around and including the building) = Length of the total area Width of the total area

Now solving for both equation we get.

Now, We get 2 values for x which -128 and 3 but the width of the building can't be in negative value hence we will consider value of x be 3.

∴ Width of the walkway is 3 ft.

The walkway is 3 feet wide around the building

Step-by-step explanation:

Given as :

The height of building  = 200 feet tall

The shape of the base of building is rectangular

And The Length of building = 150 feet

The width of building = 100 feet

∵ The building has a walkway which x feet wide surrounding all four sides

The total area of building = 16,536 feet²

Let the walkway be x feet wide

So, The length of building including pathway = ( 150 + 2 x ) feet

And The width of building including pathway = ( 100 + 2 x ) feet

∴ The total area of building =  ( 150 + 2 x ) feet × ( 100 + 2 x ) feet

Or,   ( 150 + 2 x ) feet × ( 100 + 2 x ) feet = 16,536 feet²

Or,  15000 + 300 x + 200 x + 4 x² = 16,536

Or, 4 x² + 500 x - 1536 = 0

Or,  x² + 125 x - 384 = 0

Or,   x² + 128 x - 3 x - 384 = 0

Or,  x ( x + 128 ) - 3 ( x + 128 ) = 0

∴  ( x + 128 ) ( x - 3 ) = 0

I.e x = 3 , - 125

Here we consider x= 3 value

So The width of building including pathway = ( 100 + 2 ×3 ) feet

                                                                        = 106 feet

And The length of building including pathway = ( 150 + 2 ×3 ) feet

                                                                             = 156 feet

Hence The walkway is 3 feet wide around the building . Answer

it's both, now can you answer my most recent question?

so far i think you have the right thing going for you just remember soh cah toh

step-by-step explanation:

sine over hypotenuse

cosine   equals adjacent

tangent over hypotenuse