rhombus ABCD has diagonals that intersect at E. DF is drawn perpendicular to DC and AD at D. With AB = 10cm, <BCD = 60 degrees and FD = 4cm, Find the length of FE

The diagonals in a rhombus bisect each other at right angles.

Make use of that information in your sketch.

sqrt of 41 cm

To find the length of FE, we can use the properties of the diagonals of a rhombus.

1. In a rhombus, the diagonals are perpendicular bisectors of each other. This means that the line segment DE is bisected at point E.

2. We know that FD = 4 cm, so the length of ED would be 2 times FD, which is 2 * 4 cm = 8 cm.

3. Since DE is bisected at point E, the length of each half of DE would be half of ED, which is 8 cm / 2 = 4 cm.

4. Now, let's consider the triangle FED. We have a right triangle with FD = 4 cm and DE = 4 cm (half of ED).

5. Using the Pythagorean theorem, we can find the length of EF. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

EF^2 = FD^2 + DE^2
EF^2 = (4 cm)^2 + (4 cm)^2
EF^2 = 16 cm^2 + 16 cm^2
EF^2 = 32 cm^2

6. Taking the square root of both sides, we find:

EF = sqrt(32 cm^2)
EF = 4 * sqrt(2) cm (rounded to two decimal places)

Therefore, the length of FE is approximately 4 * sqrt(2) cm.

To find the length of FE, we can use the properties of a rhombus and trigonometry.

First, let's draw the diagram described:

```
A
/ \
/ \
/ \
/ \
/ \
/ \
D-------------C
F
```

Given:
- AB = 10 cm (side length of the rhombus)
- <BCD = 60 degrees (angle at vertex C)
- FD = 4 cm

Now, let's analyze the properties of a rhombus:

1. All sides of a rhombus are equal in length. Since AB = 10 cm, BC = CD = DA = 10 cm.

2. The diagonals of a rhombus bisect each other at right angles. So, DE = EC and BE = EA.

3. The diagonals of a rhombus divide each other into segments with equal lengths. Hence, DE = EC = BE = EA = x (let's assume this length as x).

Now, let's use the properties of triangles to find the length of x:

In triangle FDC, we have:
FD = 4 cm
DC = 10 cm (from property 1)

Since FD is perpendicular to DC (FD is drawn perpendicular to DC and AD at D), we have a right triangle. Let's find the length of FC:

Using the Pythagorean theorem:
FC^2 = FD^2 + DC^2
FC^2 = 4^2 + 10^2
FC^2 = 16 + 100
FC^2 = 116
FC = √116
FC ≈ 10.77 cm

Now, let's find the lengths of BF and FA:

By applying the properties of a rhombus, we know that BF = FA = x.

Since BF = FA = x, we can form a right triangle BFA:

Using the Pythagorean theorem:
BF^2 = AB^2 - AF^2
x^2 = 10^2 - x^2
2x^2 = 100
x^2 = 50
x = √50
x ≈ 7.07 cm

Now we know that DE = EC = BE = EA = x = √50 ≈ 7.07 cm.

Since DE = EC, triangle DEC is an isosceles triangle with base DE.

The angle DEC = 180 - <BCD = 180 - 60 = 120 degrees (using the property <BCD = 60 degrees).

Now, let's use the law of cosines to find the length of FE:

In triangle DEC, we have:
DE = EC ≈ 7.07 cm
DC = 10 cm
<DEC = 120 degrees

Using the law of cosines:
DC^2 = DE^2 + EC^2 - 2 * DE * EC * cos(<DEC)
10^2 = 7.07^2 + 7.07^2 - 2 * 7.07 * 7.07 * cos(120)
100 = 50 + 50 - 2 * 50 * cos(120)
100 = 100 - 100 * cos(120)
100 = 100 + 100 * cos(60) (cos(120) = -cos(60))
100 = 100 + 100 * (1/2) (cos(60) = 1/2)
100 = 100 + 50
100 = 150

The equation 100 = 150 is not true, which means there is no solution for the given parameters. Please verify the provided values and angles for accuracy.