Question

What is the area of the triangle in the diagram?

Answer 1

Answer:

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

which agrees with the first choice (A) in your list of possible answers

Step-by-step explanation:

Notice that the area of the outer square is given by the square of it side $(a)$.Then the outer area is given by: $a^2$

Now, the inner square's area we calculate by subtracting from the outer area four equal right angle triangles (those shaded in grey in the figure provided). Each of such right angle triangles have an area given by the general formula of a triangle's area:

$Area_{triangle}= \frac{base\,*\,height}{2} =\frac{(a-b)\,b}{2} =\frac{ab-b^2}{2}$

Therefore the internal square's area is:

$a^2-4\frac{ab-b^2}{2} =a^2-2ab+2b^2$

And then the quotient of this are over the larger square area becomes:

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