Question

A new building is being constructed at UT Austin. It is 200 feet tall, and has a

Answer 1

Answer:

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

Answer 2

Answer:

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 = $150 +2x$

Width of the total area will be = width of the building + width of walkway from 2 sides = $100 +2x$

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

$(150+2x)\times(100+2x)=16536\\15000+300x+200x+4x^2=16536\\500x+4x^2=16536-15000\\4(125x+x^2)=1536\\125x+x^2= \frac{1536}{4}=384$

$x^2+125x-384=0\\x^2-3x+128x-384=0\\x(x-3)+128(x-3)=0\\(x-3)(x+128)=0$

Now solving for both equation we get.

$x+128=0\\x=-128\\x-3=0\\x=3$

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.