Question

In the figure below, ABDC, EFHG, and ASHY are all squares; AB=1, EF=$, and AY=5.

What is the area of quadrilateral DYES?

Answer 1

16.97

Explanation:

you need to find the length of DY (the hypotenuse of one of the triangles), and to do that, you need to find the length of CY, so subtract AB (which is 1) from AY (which is 5) and you are left with 4. then use the Pythagorean theorem to solve for DY.
`1^(2)` +
`4^(2)` =
`17^(2)`.
`√(17)` = 4.12 (rounded to the nearest hundredth). The last step is to solve for area, or length times height. The quadrilateral has equal side lengths, so the equation is

4.12 × 4.12 which equals 16.97

Answer 2

The area of quadrilateral DYES is 3.5 square units.

To find the area of quadrilateral DYES, we can break it into two parts: triangle DYE and square DEFG.

1. Area of square DEFG:

Since EFHG is a square and EF = 1, the area of the square is
`\( EF^2 = 1^2 = 1 \)`.

2. Area of triangle DYE:

The area of a triangle is given by
`\( (1)/(2) * \text{base} * \text{height} \)`. Here, DE can serve as the base, and AY as the height since triangle DYE is a right-angled triangle with angle DYE being 90 degrees (because DEFG is a square, and thus DE is perpendicular to FG, and extension YG is also perpendicular to DE).

The length of DE is equal to the side of the square DEFG, which is 1. The length of AY is given as 5.

So the area of triangle DYE is:

`\[ \text{Area}_(\triangle DYE) = (1)/(2) * DE * AY = (1)/(2) * 1 * 5 = 2.5 \]`

3. Area of quadrilateral DYES:

Quadrilateral DYES is composed of the entire square DEFG and triangle DYE.

The area of DYES is therefore:

`Area $_(D Y E S)=$ Area $_(\square D E F G)+$ Area $_(\triangle D Y E)$`

`\[ \text{Area}_(DYES) = 1 + 2.5 = 3.5 \]`

So, The answer is 3.5.

The complete question is here:

In the figure below, ABDC, EFHG, and ASHY are all squares; AB=1, EF=1, and AY=5.

What is the area of quadrilateral DYES?