Answer:
If the side of the square is increased by 50 percent, then 125% in the area.
Step-by-step explanation:
Let side of square =x
Area [tex]= Side^2 = x^2[/tex]
After 50% increase in side
side length =x+50% of x [tex]=3/2 x[/tex]
Area[tex]=( \frac{3x}{2})^2 = \frac{9}{4}x^2[/tex]
percentage increase in are =
[tex]= \frac{A final -A initial}{ A initial} \times 100\\=\frac{\frac{9}{4}x^2 -x^2}{x^2} \times 100\\ = \frac{5x^2}{4x^2} \times 100[/tex]
= 125%
So, If the side of the square is increased by 50 percent, then 125% in the area.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Verified answer
Answer:
If the side of the square is increased by 50 percent, then 125% in the area.
Step-by-step explanation:
Let side of square =x
Area [tex]= Side^2 = x^2[/tex]
After 50% increase in side
side length =x+50% of x [tex]=3/2 x[/tex]
Area[tex]=( \frac{3x}{2})^2 = \frac{9}{4}x^2[/tex]
percentage increase in are =
[tex]= \frac{A final -A initial}{ A initial} \times 100\\=\frac{\frac{9}{4}x^2 -x^2}{x^2} \times 100\\ = \frac{5x^2}{4x^2} \times 100[/tex]
= 125%
So, If the side of the square is increased by 50 percent, then 125% in the area.