+14 In quadrilateral \(BCED\), sides \(\overline{BD}\) and \(\overline{CE}\) are extended past B and C, respectively, to meet at point A.
If:
BD = 18
BC = 8
CE = 2
AC = 7
and AB = 3...
Then what is DE?
Thanks guys 
+129820 
Using the Law of Cosines twice :
[BC^2 - BA^2 - AC^2] / [ -2 BA * AC] = cos DAE
[8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 7 ] = cos DAE = -1/7
And
DE^2 = DA^2 + AE^2 - [2* DA * AE] cos ( DAE)
DE^2 = 21^2 + 9^2 - [ 2* 21 * 9] (-1/7)
DE^2 = 522 + 54
DE^2 = 576
DE = sqrt 576 = 24
+1752 To determine the length of \(DE\) in quadrilateral \(BCED\), we can use the properties of similar triangles and the concept of similar triangles formed by intersecting lines.
Given:
- \(BD = 18\)
- \(BC = 8\)
- \(CE = 2\)
- \(AC = 7\)
- \(AB = 3\)
We need to find \(DE\).
### Step-by-Step Solution:
1. **Identify triangles**:
- Triangles \(ABD\) and \(ACE\) share a common vertex \(A\) and extend through points \(D\) and \(E\) on \(BD\) and \(CE\), respectively.
2. **Use the intercept theorem (also known as Thales' theorem)**:
According to the intercept theorem, if two lines are intersected by a pair of parallel lines, the segments intercepted on one line are proportional to the corresponding segments intercepted on the other line.
3. **Similar triangles**:
Triangles \(ABD\) and \(ACE\) are similar by the AA criterion (angle-angle similarity) because they share angle \(A\) and both have right angles.
4. **Set up the proportionality**:
\[
\frac{BD}{DE} = \frac{AB}{AC}
\]
5. **Plug in the known values**:
\[
\frac{18}{DE} = \frac{3}{7}
\]
6. **Solve for \(DE\)**:
\[
DE = 18 \cdot \frac{7}{3} = 18 \cdot \frac{7}{3} = 6 \cdot 7 = 42
\]
Thus, the length of \(DE\) is \(\boxed{42}\).
+1786 Alright. I'm not the best with these problems, but let me give it a shot,
First, let's use the Law of Cosines. According to the law, we have
\(BC^2 = AB^2 + AC^2 - 2(AB* AC) cos ( BAC)\)
Now, we know a lot of the information given already, so we want to find cos BAC. Thus, we have
\(8^2=3^2+7^2-2(3*7) \cdot cos(BAC)\)
Now, isolating BAC, we have that
\(64=16 \cdot cos(BAC)\\ cos(BAC)=4\)
Now, why not apply Law of Cosines again? Let's do it! :)
This time, let's use the equation
\(DE^2 = AE^2 + AD^2 - 2(AE * AD) cos (BAC) \)
We figured out and know every term in the right side of the equation, so plugging in all the information, we have
\(DE^2 = 9^2+21^2-2(9*21)*4\)
mmm...let's check something out.
This simplifies to \(DE^2=-990 \), which defintely IS NOT possible.
Now, disclaimer though....it's kinda late at night, and my brain is fried. my solution is probably faulty.
...
+129820
Using the Law of Cosines twice :
[BC^2 - BA^2 - AC^2] / [ -2 BA * AC] = cos DAE
[8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 7 ] = cos DAE = -1/7
And
DE^2 = DA^2 + AE^2 - [2* DA * AE] cos ( DAE)
DE^2 = 21^2 + 9^2 - [ 2* 21 * 9] (-1/7)
DE^2 = 522 + 54
DE^2 = 576
DE = sqrt 576 = 24
