Question

puzzle #3
1 can you find the trapezoid’s perimeter?
$9x + 4$, $7x + 8$, $12x - 2$ (and one side with a mark, assumed equal to a marked side)
2 can you find the square’s perimeter?
$4x - 8$ (side of square)
3 can you find the pentagon’s perimeter?
$2x - 4$ (a side, pentagon has equal sides marked)
4 can you find the rhombus’s perimeter?
$7x + 2$ (a side, rhombus has equal sides)
answer choices
a: $10x - 20$ b: $14x + 4$ c: $16x - 32$
d: $28x + 10$ e: $16x - 16$ f: $35x + 18$
g: $10x + 16$ h: $28x + 8$ i: $28x + 2$
type the 4 - letter code into the answer box. all caps, no spaces.
your answer
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Explanation:

Problem 1: Trapezoid Perimeter

The trapezoid has two equal sides (marked with blue ticks) and two other sides. The sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the equal side to \(7x + 8\) (so also \(7x + 8\))? Wait, no, the blue ticks: looking at the trapezoid, the two non - parallel sides? Wait, no, the labels: one side is \(7x + 8\), one is \(9x + 4\), one is \(12x - 2\), and the other side (with blue tick same as \(7x + 8\)'s side? Wait, no, the trapezoid has sides: \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\)? Wait, no, the blue ticks: maybe the two legs are equal? Wait, the trapezoid's sides: let's list all four sides. From the diagram, the sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\) (because of the blue tick). Wait, no, maybe the two bases? Wait, no, let's calculate the perimeter by adding all four sides.

So perimeter \(P=(7x + 8)+(9x + 4)+(12x - 2)+(7x + 8)\)? Wait, no, maybe I misread. Wait, the trapezoid: one side is \(7x + 8\), one is \(9x + 4\), one is \(12x - 2\), and the other side (with blue tick) is equal to \(7x + 8\)? Wait, no, the blue ticks: maybe the two non - parallel sides are equal. Wait, let's check the answer choices. Let's compute the sum:

\(7x+8 + 9x + 4+12x - 2+7x + 8\)? Wait, no, maybe the other side is \(7x + 8\)? Wait, no, let's add the four sides:

\((7x+8)+(9x + 4)+(12x - 2)+(7x + 8)\)

Combine like terms:

\(7x+9x + 12x+7x+8 + 4-2 + 8\)

\((7x+9x+12x+7x)+(8 + 4-2 + 8)\)

\(35x+18\)? No, that's option F. Wait, no, maybe I misread the sides. Wait, the trapezoid: the sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side with blue tick: maybe the side equal to \(7x + 8\) is not, maybe the other side is \(7x + 8\)? Wait, no, let's re - examine.

Wait, the trapezoid: one side is \(7x + 8\), one is \(9x + 4\), one is \(12x - 2\), and the other side (with blue tick) is \(7x + 8\)? Wait, no, maybe the sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\). Wait, but when we add them:

\(7x+8+9x + 4+12x - 2+7x + 8=35x + 18\), which is option F. But let's check the other problems.

Problem 2: Square Perimeter

A square has four equal sides. Each side is \(4x - 8\). So perimeter \(P = 4\times(4x - 8)=16x-32\), which is option C.

Problem 3: Pentagon Perimeter

The pentagon has five equal sides (blue ticks), each side is \(2x - 4\). So perimeter \(P = 5\times(2x - 4)=10x-20\), which is option A.

Problem 4: Rhombus Perimeter

A rhombus has four equal sides. Each side is \(7x + 2\). So perimeter \(P = 4\times(7x + 2)=28x+8\), which is option H.

Wait, let's re - do each problem:

Problem 1: Trapezoid Perimeter

Wait, maybe I misread the trapezoid's sides. Let's look again: the trapezoid has sides \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side with blue tick: maybe the side is \(7x + 8\)? Wait, no, the blue ticks: maybe the two legs are equal. Wait, the trapezoid's sides: let's list them correctly. The sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\) (because of the blue tick). So adding them:

\(7x+8+9x + 4+12x - 2+7x + 8\)

\(7x+9x+12x+7x=35x\)

\(8 + 4-2 + 8=18\)

So \(35x + 18\) (option F). But let's check the other problems.

Problem 2: Square Perimeter

Square has side \(4x - 8\). Perimeter of square is \(4\times\) side. So \(4\times(4x - 8)=16x-32\) (option C).

Problem 3: Pentagon Perimeter

Pentagon has 5 equal sides, each side \(2x - 4\). Perimeter \(=5\times(2x - 4)=10x-20\) (option A).

Problem 4: Rhombus Perimeter

Rhombus has 4 equal sides, each side \(7x + 2\). Perimeter \(=4\times(7x + 2)=28x + 8\) (option H).

So the four - letter code is FCAH? Wait, no, wait:

Wait, problem 1: trapezoid. Wait, maybe I made a mistake. Let's re - calculate trapezoid perimeter.

Wait, the trapezoid: the sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side with blue tick: maybe the side is \(7x + 8\)? No, maybe the other side is \(7x + 8\) is wrong. Wait, maybe the trapezoid's sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\) is not, maybe the side is \(7x + 8\) is a leg, and the other leg is equal. Wait, let's add the four sides again:

\(7x+8+9x + 4+12x - 2+7x + 8\)

\(7x+9x=16x\), \(16x+12x = 28x\), \(28x+7x=35x\)

\(8 + 4=12\), \(12-2 = 10\), \(10 + 8=18\)

So \(35x + 18\) (option F).

Problem 2: square, side \(4x - 8\), perimeter \(4*(4x - 8)=16x-32\) (option C).

Problem 3: pentagon, 5 sides, each \(2x - 4\), perimeter \(5*(2x - 4)=10x-20\) (option A).

Problem 4: rhombus, 4 sides, each \(7x + 2\), perimeter \(4*(7x + 2)=28x + 8\) (option H).

So the code is FCAH? Wait, no, the order is problem 1, problem 2, problem 3, problem 4. So problem 1 answer is F, problem 2 is C, problem 3 is A, problem 4 is H. So the code is FCAH? Wait, but let's check problem 1 again.

Wait, maybe the trapezoid's sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side with blue tick is \(7x + 8\). Wait, but maybe I misread the trapezoid's sides. Let's look at the answer choices. Option F is \(35x + 18\), which is what we got.

Problem 2: square, side \(4x - 8\), perimeter \(4*(4x - 8)=16x-32\) (option C).

Problem 3: pentagon, 5 sides, each \(2x - 4\), perimeter \(5*(2x - 4)=10x-20\) (option A).

Problem 4: rhombus, 4 sides, each \(7x + 2\), perimeter \(4*(7x + 2)=28x + 8\) (option H).

So the four - letter code is FCAH. Wait, but let's check problem 1 again. Maybe the trapezoid's sides are \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side equal to \(7x + 8\) is not, maybe the side is \(7x + 8\) is a leg, and the other leg is equal. Wait, another way: maybe the trapezoid has sides \(7x + 8\), \(9x + 4\), \(12x - 2\), and the side with blue tick is \(7x + 8\). So adding them:

\(7x+8+9x + 4+12x - 2+7x + 8 = 35x+18\) (option F).

Yes, so the four answers are:

1. Trapezoid: F (\(35x + 18\))

2. Square: C (\(16x - 32\))

3. Pentagon: A (\(10x - 20\))

4. Rhombus: H (\(28x + 8\))

So the 4 - letter code is FCAH.

Answer:

FCAH