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A magician’s stage has a rectangle trapdoor which is (x+ 1/2) by 2x ft. The outside rectangle length is (x+ 16) ft.

Part A. The total area (in square feet) of the stage can be represented by x2 + 27x + 176. Write an expression for the width of the stage.

Part B: Write an expression for the perimeter of the stage.

Part C: The area of the trapdoor is 10 square feet. Find the value of x.

Part D: The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.

Respuesta :

Answer:

Part A:- [tex]x +11 ft[/tex]     Part B:- [tex](4x+54)ft[/tex]   part C:- [tex]x= 2ft[/tex]  

Part D:- Yes it satisfy his requirement

Step-by-step explanation:

Given that,

length and breadth of Trapdoor are [tex](x+ \frac{1}{2} ) by 2x[/tex] ft.

The Outside length of Rectangle is [tex](x+16 )[/tex] ft.

Part A:- Total area (in square feet) of the stage can be represented by  

             [tex]x^{2} + 27x+ 176[/tex]. Write an expression for the width of stage.

So,        Area of rectangle =  [tex]length \times breadth[/tex]

                 [tex]x^{2} + 27x+ 176 = (x+16) \times breadth[/tex]

                     [tex]Breadth = \frac{x^{2} + 27x+ 176}{x+16}[/tex]

                     [tex]Breadth = x+11[/tex] ft.                       ................(1)

Part B:- Write an expression for the Perimeter of the stage.

Here,   Perimeter of Rectangle = [tex]2(L+B)[/tex]

                                                    = [tex]2(x+16 + x +11)[/tex]

                                                    = [tex](4x + 54) ft[/tex]

Part C:- The area of trapdoor is [tex]10 ft^{2}[/tex]. Find the value of x.

So,                        Area of trapdoor = [tex]10 ft^{2}[/tex]

                              [tex](x +\frac{1}{2}) \times 2x =10 ft^{2}[/tex]

                               [tex]2x^{2} +x - 10 = 0[/tex]

                                  [tex]x = \frac{-1-9}{4}[/tex] , [tex]\frac{-1+9}{4}[/tex]

                                  [tex]x = \frac{-5}{2}[/tex] , [tex]2[/tex]

Hence value of x is [tex]2 ft[/tex]             (Neglecting the negative value because                      

                                                   length cannot be negative).

Part D:-  The magician wishes to have the area of the stage be at least 20 times the area of the trapdoor. Does this stage satisfy his requirement? Explain.

So,   Length of trapdoor is [tex](2+\frac{1}{2} ) = \frac{5}{2} ft[/tex]

        breadth of trapdoor is [tex]2\times 2 = 4ft[/tex]

Now, outside length of Rectangle is = [tex]18ft[/tex]

And outside breadth of rectangle is = [tex]13ft[/tex]

Here, Area of trapdoor = [tex]\frac{5}{2} \times 4 = 10ft^{2}[/tex]              ...................(2)

         Area of rectangle = [tex]18 \times 13= 234 ft^{2}[/tex]        ....................(3)

Thus comparing Equation (1) & (2) we found that

Area of rectangular stage is 20 times greater than the area of trapdoor.

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