Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

the length of a rectangle is 3cm more than twice the width. The area of the rectangle is 90cm^2. Find the dimensions of the rectangle.

Sagot :

logo88
A=l*w
if the length is 3cm more than the width let the length be x+3
and let the width be x
since the area=90cm^2
90=(x+3) *x
90=x(x+3)
90=x²+3x
x²+3x-90=0
x=(-3+ or - √369) /2
x= 8.105 and -11.105
but the dimensions must be positive so they are 8.105 and 11.105 (they are rounded to 3 decimal places, if you don't want them rounded look at them in the
calculator. 
Width= w 
Length= 2w+3
(Length)(Width)= Area

w(2w+3)
2w^2+3w= 90
2w^2+3w-90= 0
(2w+15)(w-6)=0
This can be broken down into two different equations:
2w+15=0 ---------------> w= -15/2
w-6=0 -------------------> w= 6

Since w cannot be negative, the width is 6 cm, and the length is 15 cm. 

This can also be solved using the quadratic formula:
-b±√[b^2-4(a)(c)]
            2a

Start with 2w^2+3w-90= 0
a= 2
b= 3
c= -90
-3±√[9-4(2)(-90)]
           4
-3±√(729)
      4
-3±27
   4
Therefore, the two answers are 6, and -30/4, AKA -15/2.
Once again, 6 is the only one that works because it is positive, so when plugged in, the length is 15, and the width is 6.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.