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

Sagot :

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.