Question

The length of the segment AD is 28 cm. The distance between the midpoints of segments AB and CD is 16 cm. Find the length of the segment BC .

Answer 1

Answer:

BC=4 cm

Step-by-step explanation:

Please refer to the image attached to this. AD is a line segment which 28 cm long. E and F are the mid point of AB and CD which are separated by a distance of 16 cm

If we say that AE=EB=x and CF=FD=y

it clear from the figure that

2x+BC+2y=28

2(x+y)+BC=28

EF=16

Hence

x+16+y=28

x+y=12

There for

2(x+y)+BC=28

2*12+BC=28

24+BC=28

BC=4cm

Answer 2

The length of the segment BC is for the specified situation is evaluated to be of 4 cm

What is the midpoint of a line segment?

The midpoint of a line segment is the point which is in the mid of its endpoints. That means, the distance of either of the two ends of the end points of that line segment from its midpoint is same.

For this case, referring to the figure attached below, and from the data specified, we have:

M = midpoint of AB,

N = midpoint of CD, thus, we get:

$|AM| = |MB|\\\\|NC| = |ND|$

Where |AM| means length of line segment AM (its a notation we use generally), and so for other line segments too.

It is specified that:

The distance between the midpoints of segments AB and CD is 16 cm.

Or,

$|MN| = 16 \: cm$

Since we're given that: |AD| = 28 cm, thus, as all points are on same line, we get:

$|AD| = |AM| + |MB| + |BC| + |NC| + |ND|$

Also, we have:

$|MN| = |MB| + |BC| + |NC|$

Thus, we get:

$|AD| = |AM| + |MB| + |BC| + |NC| + |ND|\\|AD| = |AM| + |MN| + |ND|\\28 = |AM| + |ND| + 16\\\\|AM| + |ND| = 12 \: \rm cm$

Since

$|AM| = |MB|\\and\\|NC| = |ND|$

Therefore, we get:

$|AM| + |ND| = 12\\|MB| + |NC| = 12$

Since $|MN| = |MB| + |BC| + |NC|$, we get:

$|MN| = |MB| + |BC| + |NC|\\16 = |MB| + |NC| + |BC|\\16 = 12 + |BC|\\\\|BC| = 4 \: \rm cm$

Thus, the length of the segment BC is for the specified situation is evaluated to be of 4 cm

Learn more about addition here: