MIDSEGMENT THEOREM OF A TRAPEZOID

Example 1 :

In the figure, YZ is the midsegment of trapezoid TWRV.  Determine the value of x.

Solution :

Length of line segment YZ = length of (TW + VR)/2

8 = (14.8 + x)/2

8(2) = 14.8 + x

16 = 14.8 + x

x = 16 - 14.8

x = 1.2

So, length of line segment VR is 1.2

Example 2 :

Solve for x.

Solution :

2x + 0.75 = (2x - 5.5 + x + 16)/2

2x + 0.75 = (3x + 10.5)/2

2(2x + 0.75) = 3x + 10.5

4x + 1.5 = 3x + 10.5

Subtracting 3x and 1.5 on both sides.

4x - 3x = 10.5 - 1.5

x = 9

Example 3 :

Find the length of TU.

Solution :

TU = (LM + QP)/2

2x + 4 = (2x - 4 + 3x + 2)/2

2x + 4 = (5x - 2)/2

2(2x + 4) = 5x - 2

4x + 8 = 5x - 2

4x - 5x = -2 - 8

-x = -10

x = 10

TU = 2x + 4

TU = 2(10) + 4

TU = 20 + 4

TU = 24

Example 4 :

CD is the midsegment of trapezoid WXYZ.

a. What is the value of x?

b. What is XY?

c. What is WZ?

Solution :

CD = (WZ + XY)/2

22 = (x + 3 + 4x + 1)/2

44 = 5x + 4

40 = 5x 

x = 40/5

x = 8

a) The value of x is 8.

Example 5 :

Find the length of the midsegment of the trapezoid with the given vertices.

S(0, 0) T(2, 7) U(6, 10) and V(8, 6)

Solution :

T(2, 7) and U(6, 10)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Length of TU = √(6 - 2)² + (10 - 7)²

= √4² + 3²

= √(16 + 9)

= √25

= 5

S(0, 0) and V(8, 6)

Length of SV = √(8 - 0)² + (6 - 0)²

= √8² + 6²

= √(64 + 36)

= √100

= 10

Length of midsegment = (1/2) (TU + SV)

= (1/2)(5 + 10)

= 15/2

 = 7.5 units

Example 6 :

In the diagram, NP = 8 inches, and LR = 20 inches. What is the diameter of the bottom layer of the cake?

Solution :

Given that, NP = 8 inches and LR = 20 inches.

MQ = (1/2)(NP + LR)

= (1/2) (8 + 20)

= (1/2) (28)

MQ = 14

Using midsegment theorem, we get

LR = (1/2) (MQ + KS)

20 = (1/2)(14 + KS)

20(2) = 14 + KS

40 = 14 + KS

KS = 40 - 14

= 26

So, the required diameter of the bottom layer of the cake is 26 inches.

Example 7 :

The bases of a trapezoid lie on the lines y = 2x + 7 and y = 2x − 5. Write the equation of the line that contains the midsegment of the trapezoid.

Solution :

Length of midsegment = (1/2) (2x + 7 + 2x - 5)

= (1/2)(4x + 2)

Factoring 2, we get

= (1/2)2(2x + 1)

= 2x + 1

So, the equation of the midsegment is 2x + 1.

Example 8 :

In the figure above, EO is the midsegment of trapezoid TRAP and RP intersect EO at point Z. If RA = 15 and EO = 18, what is the length of TP?

Solution :

RA = 15 and EO = 18

EO = (1/2) (RA + TP)

18 = (1/2)(15 + TP)

18(2) = 15 + TP

36 = 15 + TP

36 - 15 = TP

TP = 21

Example 9 :

You are building a plant stand with three equally-spaced circular shelves. The diagram shows a vertical cross section of the plant stand. What is the diameter of the middle shelf?

Solution :

x = (1/2)(6 + 15)

x = 1/2 (21)

x = 21/2

= 10.5

So, the diameter of the middle shelf is 10.5 inches.

Example 10 :

Find the length of the midsegment EF in trapezoid ABCD and verify the distance between EF using distance between two points formula.

Solution :

A(-5, 2), B(-1, 2) C(1, 0) and D(1, -4)

Distance between two points = √(x₂ - x₁)² + (y₂ - y₁)²

Length of BC = √(1 + 1)² + (0 - 2)²

= √2² + (-2)²

= √(4 + 4)

= √8

= 2√2

Length of AD = √(-5 - 1)² + (2 + 4)²

= √(-6)² + 6²

= √(36 + 36)

= √72

= √(2 x 2 x 2 x 3 x 3)

= 6√2

Length of BC = √(1 + 1)² + (0 - 2)²

= √2² + (-2)²

= √(4 + 4)

= √8

= 2√2

Using midsegment theorem,

= (1/2)(BC + AD)

= (1/2) (2√2 + 6√2)

= (1/2) (8√2)

= 4√2

E(-3, 2) and F(1, -2)

Length of EF = √(-3 - 1)² + (2 + 2)²

= √(-4)² + 4²

= √(16 + 16)

= √32

= 4√2

In both ways, we will get the same answer.