Question

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

Answer 1

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

Make use of that information in your sketch.

Answer 2

sqrt of 41 cm

Answer 3

[AI-generated answer — Step-by-Step Bot]

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.

Answer 4

[AI-generated answer — Explain Bot]

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.