Question

find the area and perimeter.
1
trapezoid with top base (4x^2 + 4x + 1), bottom base (x^2 + 2x + 6), height (x^2 + x - 1), and the two non - parallel sides (8x^2 - 2) (both sides).
area:
perimeter:
2
right triangle with legs (3x + 4) and (2x^2 - x + 3), hypotenuse (5x^2 + x).
area:
perimeter:

Explanation:

Problem 1: Trapezoid (Area and Perimeter)

Area of Trapezoid

The formula for the area of a trapezoid is \( A = \frac{1}{2} \times (b_1 + b_2) \times h \), where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides (bases) and \( h \) is the height.

• \( b_1 = 4x^2 + 4x + 1 \)
• \( b_2 = x^2 + 2x + 6 \)
• \( h = x^2 + x - 1 \)

Step 1: Add the two bases

\( b_1 + b_2 = (4x^2 + 4x + 1) + (x^2 + 2x + 6) = 5x^2 + 6x + 7 \)

Step 2: Multiply by height and \( \frac{1}{2} \)

\( A = \frac{1}{2} \times (5x^2 + 6x + 7) \times (x^2 + x - 1) \)
First, multiply \( (5x^2 + 6x + 7)(x^2 + x - 1) \):
\[

$$\begin{align*} &5x^2(x^2 + x - 1) + 6x(x^2 + x - 1) + 7(x^2 + x - 1)\\ =& 5x^4 + 5x^3 - 5x^2 + 6x^3 + 6x^2 - 6x + 7x^2 + 7x - 7\\ =& 5x^4 + (5x^3 + 6x^3) + (-5x^2 + 6x^2 + 7x^2) + (-6x + 7x) - 7\\ =& 5x^4 + 11x^3 + 8x^2 + x - 7 \end{align*}$$

\]
Then multiply by \( \frac{1}{2} \):
\( A = \frac{1}{2}(5x^4 + 11x^3 + 8x^2 + x - 7) = \frac{5}{2}x^4 + \frac{11}{2}x^3 + 4x^2 + \frac{1}{2}x - \frac{7}{2} \)

Perimeter of Trapezoid

The perimeter is the sum of all four sides. The two non-parallel sides are both \( 8x^2 - 2 \) (assuming the diagram shows both non-parallel sides as \( 8x^2 - 2 \); if there's a typo and one is different, but based on the image, both non-parallel sides seem to be \( 8x^2 - 2 \)).

• Side 1: \( 4x^2 + 4x + 1 \)
• Side 2: \( 8x^2 - 2 \)
• Side 3: \( x^2 + 2x + 6 \)
• Side 4: \( 8x^2 - 2 \)

Step 3: Sum all sides

\( P = (4x^2 + 4x + 1) + (8x^2 - 2) + (x^2 + 2x + 6) + (8x^2 - 2) \)
Combine like terms:
\[

$$\begin{align*} &(4x^2 + 8x^2 + x^2 + 8x^2) + (4x + 2x) + (1 - 2 + 6 - 2)\\ =& 21x^2 + 6x + 3 \end{align*}$$

\]

Problem 2: Right Triangle (Area and Perimeter)

Area of Right Triangle

The formula for the area of a right triangle is \( A = \frac{1}{2} \times \text{base} \times \text{height} \).

• Base: \( 2x^2 - x + 3 \)
• Height: \( 3x + 4 \)

Step 1: Multiply base and height

\( \text{base} \times \text{height} = (2x^2 - x + 3)(3x + 4) \)
\[

$$\begin{align*} &2x^2(3x + 4) - x(3x + 4) + 3(3x + 4)\\ =& 6x^3 + 8x^2 - 3x^2 - 4x + 9x + 12\\ =& 6x^3 + 5x^2 + 5x + 12 \end{align*}$$

\]

Step 2: Multiply by \( \frac{1}{2} \)

\( A = \frac{1}{2}(6x^3 + 5x^2 + 5x + 12) = 3x^3 + \frac{5}{2}x^2 + \frac{5}{2}x + 6 \)

Perimeter of Right Triangle

The perimeter is the sum of all three sides. The sides are:

• Leg 1: \( 3x + 4 \)
• Leg 2: \( 2x^2 - x + 3 \)
• Hypotenuse: \( 5x^2 + x \)

Step 3: Sum all sides

\( P = (3x + 4) + (2x^2 - x + 3) + (5x^2 + x) \)
Combine like terms:
\[

$$\begin{align*} &(2x^2 + 5x^2) + (3x - x + x) + (4 + 3)\\ =& 7x^2 + 3x + 7 \end{align*}$$

\]

Final Answers

Problem 1:

• Area: \( \frac{5}{2}x^4 + \frac{11}{2}x^3 + 4x^2 + \frac{1}{2}x - \frac{7}{2} \)
• Perimeter: \( 21x^2 + 6x + 3 \)

Problem 2:

• Area: \( 3x^3 + \frac{5}{2}x^2 + \frac{5}{2}x + 6 \)
• Perimeter: \( 7x^2 + 3x + 7 \)

Answer:

Problem 1: Trapezoid (Area and Perimeter)

Area of Trapezoid

The formula for the area of a trapezoid is \( A = \frac{1}{2} \times (b_1 + b_2) \times h \), where \( b_1 \) and \( b_2 \) are the lengths of the two parallel sides (bases) and \( h \) is the height.

• \( b_1 = 4x^2 + 4x + 1 \)
• \( b_2 = x^2 + 2x + 6 \)
• \( h = x^2 + x - 1 \)

Step 1: Add the two bases

\( b_1 + b_2 = (4x^2 + 4x + 1) + (x^2 + 2x + 6) = 5x^2 + 6x + 7 \)

Step 2: Multiply by height and \( \frac{1}{2} \)

\( A = \frac{1}{2} \times (5x^2 + 6x + 7) \times (x^2 + x - 1) \)
First, multiply \( (5x^2 + 6x + 7)(x^2 + x - 1) \):
\[

$$\begin{align*} &5x^2(x^2 + x - 1) + 6x(x^2 + x - 1) + 7(x^2 + x - 1)\\ =& 5x^4 + 5x^3 - 5x^2 + 6x^3 + 6x^2 - 6x + 7x^2 + 7x - 7\\ =& 5x^4 + (5x^3 + 6x^3) + (-5x^2 + 6x^2 + 7x^2) + (-6x + 7x) - 7\\ =& 5x^4 + 11x^3 + 8x^2 + x - 7 \end{align*}$$

\]
Then multiply by \( \frac{1}{2} \):
\( A = \frac{1}{2}(5x^4 + 11x^3 + 8x^2 + x - 7) = \frac{5}{2}x^4 + \frac{11}{2}x^3 + 4x^2 + \frac{1}{2}x - \frac{7}{2} \)

Perimeter of Trapezoid

The perimeter is the sum of all four sides. The two non-parallel sides are both \( 8x^2 - 2 \) (assuming the diagram shows both non-parallel sides as \( 8x^2 - 2 \); if there's a typo and one is different, but based on the image, both non-parallel sides seem to be \( 8x^2 - 2 \)).

• Side 1: \( 4x^2 + 4x + 1 \)
• Side 2: \( 8x^2 - 2 \)
• Side 3: \( x^2 + 2x + 6 \)
• Side 4: \( 8x^2 - 2 \)

Step 3: Sum all sides

\( P = (4x^2 + 4x + 1) + (8x^2 - 2) + (x^2 + 2x + 6) + (8x^2 - 2) \)
Combine like terms:
\[

$$\begin{align*} &(4x^2 + 8x^2 + x^2 + 8x^2) + (4x + 2x) + (1 - 2 + 6 - 2)\\ =& 21x^2 + 6x + 3 \end{align*}$$

\]

Problem 2: Right Triangle (Area and Perimeter)

Area of Right Triangle

The formula for the area of a right triangle is \( A = \frac{1}{2} \times \text{base} \times \text{height} \).

• Base: \( 2x^2 - x + 3 \)
• Height: \( 3x + 4 \)

Step 1: Multiply base and height

\( \text{base} \times \text{height} = (2x^2 - x + 3)(3x + 4) \)
\[

$$\begin{align*} &2x^2(3x + 4) - x(3x + 4) + 3(3x + 4)\\ =& 6x^3 + 8x^2 - 3x^2 - 4x + 9x + 12\\ =& 6x^3 + 5x^2 + 5x + 12 \end{align*}$$

\]

Step 2: Multiply by \( \frac{1}{2} \)

\( A = \frac{1}{2}(6x^3 + 5x^2 + 5x + 12) = 3x^3 + \frac{5}{2}x^2 + \frac{5}{2}x + 6 \)

Perimeter of Right Triangle

The perimeter is the sum of all three sides. The sides are:

• Leg 1: \( 3x + 4 \)
• Leg 2: \( 2x^2 - x + 3 \)
• Hypotenuse: \( 5x^2 + x \)

Step 3: Sum all sides

\( P = (3x + 4) + (2x^2 - x + 3) + (5x^2 + x) \)
Combine like terms:
\[

$$\begin{align*} &(2x^2 + 5x^2) + (3x - x + x) + (4 + 3)\\ =& 7x^2 + 3x + 7 \end{align*}$$

\]

Final Answers

Problem 1:

• Area: \( \frac{5}{2}x^4 + \frac{11}{2}x^3 + 4x^2 + \frac{1}{2}x - \frac{7}{2} \)
• Perimeter: \( 21x^2 + 6x + 3 \)

Problem 2:

• Area: \( 3x^3 + \frac{5}{2}x^2 + \frac{5}{2}x + 6 \)
• Perimeter: \( 7x^2 + 3x + 7 \)