Respuesta :
Complete Question:
Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.
Correct Answer:
A) [tex](10-x)(12-x)=15[/tex]
B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
Step-by-step explanation:
a. Write an equation to represent the area of the reduced image.
Let the reduced dimensions is by x , So the new dimensions are
[tex]length=10-x\\breadth=12-x[/tex]
According to question , Area of new image is :
⇒ [tex]Area = \frac{1}{8}Length(breadth)[/tex]
⇒ [tex]Area = \frac{1}{8}(10)(12)[/tex]
⇒ [tex]Area = 15[/tex]
So the equation will be :
⇒ [tex](10-x)(12-x)=15[/tex]
b. Find the dimensions of the reduced image
Let's solve : [tex](10-x)(12-x)=15[/tex]
⇒ [tex]120-10x-12x+x^2=15[/tex]
⇒ [tex]120-22x+x^2=15[/tex]
⇒ [tex]x^2-22x+105=0[/tex]
By Quadratic formula :
⇒ [tex]x = \frac{-b \pm \sqrt{b^2-4ac} }{2a}[/tex]
⇒ [tex]x = \frac{22 \pm8 }{2}[/tex]
⇒ [tex]x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}[/tex]
⇒ [tex]x = 15 , x =7[/tex]
x = 15 is rejected ! as 15 > 10 ! Side can't be negative
⇒ [tex]x =7[/tex]
Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch
