A blackboard has an area of 28 feet squared and a perimeter of 22 ft. What are the dimensions of the board?
A. 2 ft by 11 ft
B. 2 ft by 14 ft
C. 4 ft by 7 ft
D. 7 ft by 7 ft

Respuesta :

The dimensions are 4 ft. by 7 ft. 4*7 is 28, the area, and 4+4+7+7 is 22, the perimeter. 

Answer:

Option (c) is correct.

Dimension of board is 4 ft by 7 ft.

Step-by-step explanation:

Given : A blackboard has an area of 28 feet squared and a perimeter of 22 ft.

We have to find the dimension of blackboard and choose the correct option from given options.

Since blackboard is in shape of rectangle thus,

Area of blackboard is 28 feet²

That is, Area of rectangle =  28 feet²

We know Area of rectangle = length × Breadth

substitute, we get,

28 = length × Breadth ......(1)

Also, Perimeter of blackboard is 22 feet.

That is, Perimeter of rectangle =  22 feet

We know, Perimeter of rectangle = 2 (length + Breadth)

substitute, we get,

22 = 2 (length + Breadth)  

11 = (length + breadth)   ........(2)

Let length be l and breadth be b

Solving (1) and (2) ,

from (2)

b = 11 - l  ..(3)

put (3)  in (1) , we get,

28 = L (11 - L)

Multiply, we get,

[tex]28=11L-L^2[/tex]

[tex]L^2-11L+28=0[/tex]

Solving above quadratic equation using middle term split method, we get,

[tex]L^2-4L-7L+28=0[/tex]

Solving, e get,

[tex]L(L-4)-7(L-4)=0[/tex]

[tex](L-7)(L-4)=0[/tex]

Using zero product property, we get,

[tex](L-7)=0[/tex]  and [tex](L-4)=0[/tex]

We get ,

[tex]L=7[/tex]  and [tex]L=4[/tex]

Thus, length can be 4ft or 7 ft

Substitute, L in (3) , we get,

b = 11 - 4 = 7 ft

b = 11 - 7 = 4 ft

Thus, Dimension of board is 4 ft by 7 ft.

Option (c) is correct.