Perimeter of a Trapezoid: Comparison between 2 of the same shape with an identical perimeter

Examples with solutions for Perimeter of a Trapezoid: Comparison between 2 of the same shape with an identical perimeter

Exercise #1

How long is B2C2 given that the perimeters of the two trapezoids are equal?

555666777XXX555777XXXA1A1A1B1B1B1D1D1D1C1C1C1A2A2A2B2B2B2C2C2C2D2D2D2

Video Solution

Step-by-Step Solution

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

  • Step 1: Calculate the perimeter of Trapezoid 1.
  • Step 2: Set the perimeters of the trapezoids equal to each other.
  • Step 3: Solve the equation for the unknown length B2C2 B2C2 .

Now, let's work through each step:
Step 1: The perimeter of Trapezoid 1 is given by the sum: P1=5+6+7+X=18+X P_1 = 5 + 6 + 7 + X = 18 + X
Step 2: Assume the perimeter of Trapezoid 2 is the same: P2=5+7+X+B2C2=12+X+B2C2 P_2 = 5 + 7 + X + B2C2 = 12 + X + B2C2
Step 3: Equate P1 P_1 and P2 P_2 : 18+X=12+X+B2C2 18 + X = 12 + X + B2C2 Subtract X X from both sides: 18=12+B2C2 18 = 12 + B2C2 Finally, solve for B2C2 B2C2 : B2C2=1812=6 B2C2 = 18 - 12 = 6

Therefore, the solution to the problem is B2C2=6 B2C2 = 6 .

Answer

6

Exercise #2

What are the perimeters of the trapezoid?

Are their areas identical?

4775BA21173

Video Solution

Step-by-Step Solution

To find the solution, we begin by calculating the perimeters of each trapezoid:

For trapezoid AA, the sides are 44, 77, 77, and 55. Therefore, the perimeter pApA is:

pA=4+7+7+5=23 pA = 4 + 7 + 7 + 5 = 23

For trapezoid BB, the sides are 22, 77, 1111, and 33. Thus, the perimeter pBpB is:

pB=2+7+11+3=23 pB = 2 + 7 + 11 + 3 = 23

Both trapezoids have identical perimeters of 2323. However, their areas cannot be determined to be identical without information about the heights of the trapezoids. Since these are crucial for the area calculation, the equivalence of their areas cannot be concluded from available data.

Therefore, the perimeter is pA=pB=23 pA = pB = 23 , but their areas are not necessarily identical.

Answer

pA=pB=23 pA=pB=23 , not necessarily

Exercise #3

The two trapezoids below are isosceles.

What is the size of C2D2 if the trapezoids are equal?

2X2X2X4443X3X3X4443X3X3X666666A1A1A1B1B1B1D1D1D1C1C1C1A2A2A2B2B2B2D2D2D2C2C2C2

Video Solution

Step-by-Step Solution

To solve this problem, let's calculate the perimeters of both trapezoids:

For the first trapezoid A1B1C1D1 A_1B_1C_1D_1 :

  • Top base: A1B1=2X A_1B_1 = 2X
  • Bottom base: C1D1=3X C_1D_1 = 3X
  • Two equal side extensions, each of length 4.

The perimeter P1 P_1 of the first trapezoid is:
P1=2X+3X+4+4=5X+8 P_1 = 2X + 3X + 4 + 4 = 5X + 8 .

For the second trapezoid A2B2C2D2 A_2B_2C_2D_2 :

  • Top base: A2B2=3X A_2B_2 = 3X
  • Bottom base: C2D2=? C_2D_2 = ?
  • Two equal side extensions, each of length 6.

The perimeter P2 P_2 of the second trapezoid is:
P2=3X+C2D2+6+6=3X+C2D2+12 P_2 = 3X + C_2D_2 + 6 + 6 = 3X + C_2D_2 + 12 .

Since the trapezoids are equal, their perimeters are the same:
5X+8=3X+C2D2+12 5X + 8 = 3X + C_2D_2 + 12 .

Solving for C2D2 C_2D_2 :

C2D2=5X+83X12 C_2D_2 = 5X + 8 - 3X - 12

C2D2=2X4 C_2D_2 = 2X - 4

Thus, the size of C2D2 C_2D_2 if the trapezoids are equal is 2X4 2X - 4 .

Answer

2X-4

Exercise #4

What can be said about the two trapezoids in the diagram?

x+517y10x12

Video Solution

Step-by-Step Solution

We calculate the perimeter of the left trapezoid:

P=10+12+x+y P=10+12+x+y

P=22+x+y P=22+x+y

We calculate the perimeterof the right trapezoid:

P=x+5+17+y3+2y3 P=x+5+17+\frac{y}{3}+\frac{2y}{3}

P=x+22+2y+y3 P=x+22+\frac{2y+y}{3}

P=x+22+3y3 P=x+22+\frac{3y}{3}

P=x+22+y P=x+22+y

It can be seen that the two perimeters are identical to each other.

Answer

Their perimeters are identical.

Exercise #5

Given the trapezoids in the drawing.

Are they the same trapezoid?

CCCxx+1610677

Video Solution

Step-by-Step Solution

We calculate the perimeter of the left trapezoid:

P=6+10+7+52x+5 P=6+10+7+\frac{5}{2}x+5

P=28+52x P=28+\frac{5}{2}x

We calculate the perimeter of the right trapezoid:

P=7+x+x+16+x2+5 P=7+x+x+16+\frac{x}{2}+5

P=212x+28 P=2\frac{1}{2}x+28

P=52x+28 P=\frac{5}{2}x+28

The perimeters of the two trapezoids are equal to each other.

Answer

No, but their perimeter is identical.

Exercise #6

Calculate X given that the perimeters of the two trapezoids are equal.

5X5X5X2X-22X-22X-26X6X6X2X+12X+12X+14X4X4X3X3X3XA1A1A1B1B1B1D1D1D1C1C1C1A2A2A2B2B2B2D2D2D2C2C2C2X-14X+1

Video Solution

Step-by-Step Solution

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

  • Step 1: Calculate the perimeter of each trapezoid
  • Step 2: Set the perimeters equal to each other
  • Step 3: Solve for X X

Let's work through the steps:

Step 1: Calculate the perimeter of each trapezoid.

For the first trapezoid, the perimeter is:

5X+(2X2)+(X1)+6X=14X3 5X + (2X - 2) + (X - 1) + 6X = 14X - 3

For the second trapezoid, the perimeter is:

(2X+1)+4X+(4X+1)+3X=13X+2 (2X + 1) + 4X + (4X + 1) + 3X = 13X + 2

Step 2: Set the two perimeter expressions equal:

14X3=13X+2 14X - 3 = 13X + 2

Step 3: Solve for X X .

Subtract 13X 13X from both sides:

X3=2 X - 3 = 2

Add 3 to both sides:

X=5 X = 5

Therefore, the solution to the problem is X=5 X = 5 .

Answer

5 5

Exercise #7

What is the perimeter of each trapezoid?

Are their areas equal?

43362275

Video Solution

Answer

a 16, b 16, yes.

Exercise #8

Two trapezoids are shown below.

Are their perimeters the same?

x3x152x+1052x

Video Solution

Answer

Yes, equal to 412x+15 4\frac{1}{2}x+15