The area of a hall is 806 sq. ft. Its length is 5 ft greater than its width. Find the dimensions of the hall. Enter the results in the answer box. Show work that leads to the final answer and clearly state the final answer.

width is = 26ft

length is = 31ft

Step-by-step explanation:

The area of hall is 806 sq. ft.

length is 5 ft greater than its width.

L = 5+b

L*b = 806

(5+b)*b= 806

5b +b² = 806

B²+5B-806= 0

Solving quadratically

B² +31B -26B -806= 0

B(B+31)-26(B+31)= 0

(B-26)(B+31)= 0

B= 26 or -31

So the value of the width must be positive so the width is = 26ft

If width is 26, length is 26+5 = 31ft

d. bc = 12 and ef = 16

step-by-step explanation:

we are given that δabc : δdef such that .

also, it is given that

so, from the options, we need that .

a.

b.

c.  

d.

as, we can see that only option d gives the same ratio as i.e. .

hence, the possible lengths are bc = 12 and ef = 16.

a

explanation: