Question

The perimeter of a rectangular field is 60 meters. Its area is 200 square meters. Find its dimensions.

Answer 1

Answer:

THIS IS THE REPRENTATION :I(LENGTH ),W(WIDTH)

2IW= 200

I,II USE THE SUBSTITUTION METHOD SINCE :I=30-W WE CAN SUBSTITUTE THE LENGTH TO 30

-W TO IW=200

(30-W)W=200

W2 + 30W=200---------------TRANSPOSE -w2

+30w

w2-30w +200 = 0------------------- USE FACTORING

(W-20)(w-10 =0----------------- THEN EQUATE EACH FACTOR TO ZERO

W- 20=0 W-10 =0

W=20 W=10

THEN SUBSTITUTE THE VALUE OF w TO ANY OF THE EQUATIONS TO GET THE VALUE OF I

Iw =200 Iw=200

I(20) =200 I(10)

I(20)/20= 200/20 I(10)/10

=200/10

I=10

SO the answer to the problem are:

width = 20 then lenght =10 or

width = 10 then lenght =20

Answer 2

Answer:

20 m by 10 m

Step-by-step explanation:

let w be width and l be length , then

2(l + w) = 60 ( divide both sides by 2 )

l + w = 30 ( subtract w from both sides )

l = 30 - w → (1)

lw = 200 → (2)

Substitute l = 30 - w into (2)

w(30 - w) = 200 ← distribute parenthesis on left side

30w - w² = 200 ( subtract 200 from both sides )

30w - w² - 200 = 0 ( multiply through by - 1 )

w² - 30w + 200 = 0 ← in standard form

(w - 10)(w - 20) = 0 ← in factored form

Equate each factor to zero and solve for w

w - 10 = 0 ⇒ w = 10

w - 20 = 0 ⇒ w = 20

Substitute these values into (1)

l = 30 - 10 = 20

l = 30 - 20 = 10

dimensions of field is 20 m by 10 m