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kakasveta [241]
3 years ago
9

What is the ratio of the area of the inner square to the area of the outer

Mathematics
1 answer:
Rama09 [41]3 years ago
8 0

Answer:

\frac{(a-b)^2+b^2}{a^{2}}

Step-by-step explanation:

see the attached figure to better understand the problem

we know that

The area of the square is

A=x^{2}

where

x is the length side of the square

step 1

Find the area of the outer square (A_o)

we have that

x=a\ units

substitute in the formula

A_o=a^{2}\ units^2

step 2

Find the area of the inner square (A_i)

we know that

The length side of the inner square is equal to the hypotenuse of a right triangle

so

Applying Pythagoras theorem

x^{2}=(a-b)^2+b^2

Remember that

A_i=x^2

so

A_i=(a-b)^2+b^2

step 3

Find the ratio of the area of the inner square to the area of the outer

ratio=\frac{A_i}{A_o}

substitute the values

ratio=\frac{(a-b)^2+b^2}{a^{2}}

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