Answered

Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields 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 appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.