Question

The width of a field is 5m less than its length. If the area is 204 m2, find the dimmensions of the field

Answer 1

Answer:

The width of a rectangle is 5 meters less than its length. The area is 84 square meters. Find the dimensions of the rectangle and the perimeter of the rectangle.

First it helps to draw a picture so draw a rectangle

Remember that area of a rectangle =Length times width and the perimeter is P= 2 times the length plus 2 times the width

So A=LW and P=2L+2W

84= area

L=length

W= length -5

Step-by-step explanation:

L(L-5)=84 distribute the L and you have L2 -5L=84 now subtract 84 from both sides and get L2 �5L-84 next factor to get (L+7)(L-12) next set both to =0 (L+7)=0 (L+12)=0 solve for L

L=-7 and L=12 you can�t have a negative length so -7 is out. So L=12 now plug in to the original area equation

84=LW 84= 12w now divide both sides by 12 to isolate W, this gives you W=7

L=12 and W= 7

For perimeter use 12 for the length and 7 for width Remember that P= 2L+2W

P=2(12) +2(7) =38

Answer 2

Answer:

Answer:

\large{\red{\bold{\underline{✫\: Given :-}}}}

✫Given:−

↦ The width of a field is 5 m less than its length.

↦ Area is 204 m²

\large{\red{\bold{\underline{✫\: Find\: Out :-}}}}

✫FindOut:−

↦ What is the dimension of the field.

\large{\red{\bold{\underline{✫\: Formula\: Required :-}}}}

✫FormulaRequired:−

\large{\boxed{\underline{\underline{\bf{✮\: A\: =\: L\: \times W\: ✮}}}}}

✮A=L×W✮

【 where, A = Area, L = Length, W = Width 】

\large{\red{\bold{\underline{✫\: Solution :-}}}}

✫Solution:−

Let, the length be x m and the width will be (x - 5) m

According to the question, by substitute the values we get,

➳ 204 = x(x - 5)

➳ 204 = x² - 5x

➳ x² - 5x - 204 = 0

By doing middle term we get,

➳ x² - (17 - 12)x - 204 = 0

➳ x² - 17x + 12x - 204 = 0

➳ x(x - 17) + 12(x - 17) = 0

➳ x - 17 = 0 ; x + 12 = 0

➤ x = 17 ; x = - 12

We can't take length as negative so we take x = 17 as length.

Now, we have to find the width,

↬ (x - 5)

↬ (17 - 5)

➙ 12 m

Hence, width will be 12 m

Then, we get length = 17 m and breadth = 12 m

Hence, the dimension will be,

↪ Length × Width

↪ 17 m × 12 m

➠ 17 × 12 m²

∴ The dimensions of the field is 17 × 12 m² .