Area of the Square: Worded problems

We will refer to two separate areas: the area of the square with side y and the total area of the four squares with sides x,

We'll use the formula for the area of a square with side b:

$S=b^2$and therefore when applying it to the problem, we get that the area of the square with side y in the drawing is:

$S_1=y^2$Next, we'll calculate the area of the square with side x in the drawing:

$S_2=x^2$and to get the total area of the four squares in the drawing, we'll multiply this area by 4:

$4S_2=4x^2$Therefore, the area of the required figure in the problem, which includes the area of the square with side y and the area of the four squares with side x is:

$S_1+4S_2=y^2+4x^2$Therefore, the correct answer is A.

First, recall the formula for calculating square area:

The area of a square (where all sides are equal and all angles are $90\degree$) with a side length of $a$(length units - u)

, is given by the formula:

$\boxed{ S_{\textcolor{red}{\boxed{}}}=a^2}$(square units - sq.u),

Let's proceed to solve the problem:

First, let's mark the square's vertices with letters: $ABCD$

Next, considering the given data (that the square's side length is: $x+1$ cm), apply the above square area formula in order to express the area of the given square using its side length-$AB=BC=CD=DA= x +1$ (cm):

$S_{\textcolor{red}{\boxed{}}}=AB^2\\ \downarrow\\ S_{\textcolor{red}{\boxed{}}}=(x+1)^2$(sq.cm)

Continue to simplify the algebraic expression that we obtained for the square's area. This can be achieved by using the shortened multiplication formula for squaring a binomial:

$(c+d)^2=c^2+2cd+d^2$ Therefore, we'll apply this formula to our square area expression:

$S_{\textcolor{red}{\boxed{}}}=(x+1)^2 \\ \downarrow\\ \boxed{S_{\textcolor{red}{\boxed{}}}=x^2+2x+1}$(sq.cm)

The correct answer is answer D.

To solve this problem, we'll follow these steps:

• Step 1: Identify the areas for both square and rectangle using given dimensions.
• Step 2: Equate the two areas as the problem states they are equal.
• Step 3: Solve the resulting equation for $x$.
• Step 4: Check that the condition $x > 1$ is satisfied and select the correct answer choice.

Now, let's work through the solution:

Step 1: The area of the square with side length $x$ is given by:

$x^2$.

For the rectangle, where one side is extended by 3 cm and an adjacent side is shortened by 1 cm, we have:

Original length and width of the rectangle are $x + 3$ and $x - 1$, respectively.

The area of the rectangle becomes:

$(x + 3)(x - 1)$.

Step 2: As per the problem, these two areas are equal:

$x^2 = (x + 3)(x - 1)$.

Step 3: Expanding the right-hand side of the equation:

$(x + 3)(x - 1) = x^2 + 3x - x - 3 = x^2 + 2x - 3$.

Now, equate and simplify:

$x^2 = x^2 + 2x - 3$.

Subtract $x^2$ from both sides:

$0 = 2x - 3$.

Adding 3 to both sides gives:

$3 = 2x$.

Divide both sides by 2 to solve for $x$:

$x = \frac{3}{2}$.

Step 4: We check the condition $x > 1$ and find $x = \frac{3}{2} \approx 1.5$ which satisfies it.

Therefore, the side length of the square is $x = \frac{3}{2}$ cm.

First, recall the formula for calculating the area of a rectangle:

The area of a rectangle (which has two pairs of equal opposite sides and all angles are $90\degree$) with sides of length $a,\hspace{4pt} b$ units, is given by the formula:

$\boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=a\cdot b }$(square units)

Proceed to solve the problem:

Calculate the area of the rectangle whose vertices we'll mark with letters $EFGH$ (drawing)

$$It is given in the problem that one side of the rectangle is obtained by extending one side of the square with side length $x +1$ (cm) by 1 cm, and the second side of the rectangle is obtained by shortening the adjacent side of the given square by 1 cm:

Therefore, the lengths of the rectangle's sides are:

$EF=HG=(x+1)+1\\ \downarrow\\ \boxed{ EF=HG=x+2}\\ \hspace{2pt}\\ \\ EH=FG=(x+1)-1\\ \downarrow\\ \boxed{ EH=FG=x }$ (cm)

Apply the above formula to calculate the area of the rectangle that was formed from the square in the way described in the problem:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=EF\cdot EH\\ \downarrow\\ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x+2)x$ (sq cm)

Continue to simplify the expression that we obtained for the rectangle's area, using the distributive property:

$(m+n)d=md+nd$ Therefore, applying the distributive property, we obtain the following area for the rectangle:

$S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=(x+2)x \\ \boxed{ S_{\textcolor{blue}{\boxed{\hspace{6pt}}}}=x^2+2x}$(sq cm)

The correct answer is answer B.

To determine which shape has a larger area, we need to compare the areas of the square and the rectangle:

• Step 1: Calculate the area of the original square:

The side length of the square is $X$, so its area is given by:

• Step 2: Calculate the area of the rectangle:

The dimensions of the rectangle are $X + 3$ cm and $X - 3$ cm. Thus, its area is:

Using the difference of squares formula, we find:

• Step 3: Compare the areas:

We compute the difference between the square's area and the rectangle's area:

Since 9 is positive, the area of the square is larger than the area of the rectangle.

Therefore, the square has a larger area than the rectangle.

To solve this problem, follow these steps:

• Step 1: Set up the equation based on the given information. The side of the square is given by $x + 1$, and the area formula for a square is $(\text{side})^2$. Thus, we set up the equation as $(x + 1)^2 = 36$.
• Step 2: Solve for $x$. Take the square root of both sides of the equation:

$(x + 1) = \pm \sqrt{36}$
$(x + 1) = \pm 6$

• Since $x > 0$ is a condition, only the positive value of $x + 1$ is valid, so $x + 1 = 6$.
• Solve for $x$ by subtracting 1 from both sides:

$x = 6 - 1$
$x = 5$

Therefore, the solution to the problem is $x = 5$.