Question

The length of one pair of opposite sides of a square is reduced by 10% and that the other pair is increased by 10%. Compare the area of the new rectangle with the area of the original square.​

Answer 1

Answer:

99:100

Step-by-step explanation:

let x be the length

10%of x =x/10

comparing area

11/40x*9/10x : x*x

=99:100

hope it helps

Answer 2

Answer:

$\huge\underline\bold {Question:}$

The length of one pair of opposite sides of a square is reduced by 10% and that other pair is increased by 10%. Compare the area of the original square.

$\huge\underline\bold {Answer:}$

Let the side of the square be a.

Therefore, increased length

= a + 10% of a = 11a/10

Decreased length

= a – 10% of a = 9a/10

Area of original square = a^2

Area of the new rectangle

$= \frac{11a}{10} \times \frac{9a}{10} = \frac{99a {}^{2} }{100}$

Difference of the two areas

$= {a}^{2} - \frac{99a {}^{2} }{100} = \frac{a {}^{2} }{100}$

=> The area of the new rectangle is 1% less than the area of original square.