# Geometric problems:the area of a rectangular field is 78 sq meters and its perimeter is 38 meters.

1
by geluh

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by geluh

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We let the length be l and the width be w

The area of the rectangle is computed as lw so:

78 = lw

The perimeter is computed as 2(l+w) so:

38 = 2(l+w)

19 = l + w

Method 1: Guess and Check

We look for two numbers with a product of 78 and a sum of 19.

78 + 1 = 79 X

39 + 2 = 41 X

13 + 6 = 19 √

Therefore the dimensions of the rectangle are 13 by 6 m.

Method 2: Quadratic Formula and Vieta's Formula

By Vieta's formula:

αβ = 78 = c

α + β = 19 = -b

*take note that we let a=1

The quadratic equation would be:

t² - 19t + 78

Then we do the quadratic formula

t =__19 ± √(19² - 4(78))__ = __19 ± √49__ = __19 ____± 7__ = 13 or 6

2 2 2

78/13 = 6 and 78/6 = 13

Therefore the dimensions are 13 by 6 m.

The area of the rectangle is computed as lw so:

78 = lw

The perimeter is computed as 2(l+w) so:

38 = 2(l+w)

19 = l + w

Method 1: Guess and Check

We look for two numbers with a product of 78 and a sum of 19.

78 + 1 = 79 X

39 + 2 = 41 X

13 + 6 = 19 √

Therefore the dimensions of the rectangle are 13 by 6 m.

Method 2: Quadratic Formula and Vieta's Formula

By Vieta's formula:

αβ = 78 = c

α + β = 19 = -b

*take note that we let a=1

The quadratic equation would be:

t² - 19t + 78

Then we do the quadratic formula

t =

2 2 2

78/13 = 6 and 78/6 = 13

Therefore the dimensions are 13 by 6 m.