Question casey’s den is 10feet longer than it is wide. If the den’s area is 119 square feet, what are the dimensions of the room?

Answer: The width is 7ft and the length is 17 ft Step-by-step explanation: Let w represent the width of the room and L represents the length of the den. Given that: caseys den is 10feet longer than it is wide Then; w + 10 = L —- (1) The area of the den = 119 ft² Area = w × L 119 = w × L —- (2) From equation (1); let w = L – 10. Now we can replace the value of w into equation (2) i.e. (L – 10) × L = 119 L² – 10L – 119 = 0 (L + 7) ( L – 17) = 0 L + 7 = 0 or L – 17 = 0 L = – 7 ft or L = 17 ft Recall that: w = L – 10 w = -7 – 10 or w = 17 – 10 w = -17 or w = 7 ft The area of the den can now be: Area = -17 ft × – 7ft or 17ft × 7 ft Area = 119 ft² or 119 ft² Since it is ideal to use the positive integer, Then the width is 7ft and the length is 17 ft Log in to Reply

Answer:The width is 7ft and the length is 17 ft

Step-by-step explanation:Let w represent the width of the room and L represents the length of the den.

Given that:

caseys den is 10feet longer than it is wide

Then;

w + 10 = L —- (1)

The area of the den = 119 ft²

Area = w × L

119 = w × L —- (2)

From equation (1);

let w = L – 10.

Now we can replace the value of w into equation (2)

i.e.

(L – 10) × L = 119

L² – 10L – 119 = 0

(L + 7) ( L – 17) = 0

L + 7 = 0 or L – 17 = 0

L = – 7 ft or L = 17 ft

Recall that:

w = L – 10

w = -7 – 10 or w = 17 – 10

w = -17 or w = 7 ft

The area of the den can now be:

Area = -17 ft × – 7ft or 17ft × 7 ft

Area = 119 ft² or 119 ft²

Since it is ideal to use the positive integer,

Then the width is 7ft and the length is 17 ft