The full question is: Given that the side of the square has a length b−a, find the area of one of the four triangles and the area of the small inner square.
Give the area of one of the triangles followed by the area of the small inner square separated by a comma. Express your answers in terms of the variables a and b.
Answer: (b-a)^2, {(b-a)^2}/4
Step-by-step explanation:
One side of the big square = b-a, therefore,
(total area of big square =(b-a)^2
The four inner triangles are equal because they are divided by diagonal lines of a square. So area of one triangle will be: area of square/4. So,
Area of each of 4 inner triangles= ( {b-a)^2} /4
From the image attached, I've drawn the diagram to reflect what the question is saying. The sides of the small inner square are connected by mid points of inner triangle sides.
By symmetry, we can tell that the area of this small square would be a quarter of that of the big square seeing that the sides of this small square will be half of the side of the big square.
Therefore, area of small square =
{(b-a)^2}/4
Answer: [tex]\frac{ba}{2}[/tex],(b-a)^2
Step-by-step explanation:
Area of square: 4 * ([tex]\frac{1}{2} ab[/tex]) + (b-a)^2
= 2ab + ([tex]b^{2}[/tex] - 2ab + [tex]a^{2}[/tex])
= [tex]a^{2}[/tex] + [tex]b^{2}[/tex]
