Find perimeter of BAED, lengths of segments, values of 4 num

[monkeyrock227]

I have a few review problems for a test that I'm taking tomorrow since I was sick when the review was handed out. I am a student that understands things after they are explained once. I have a great memory, but my teacher doesn't explain things very well. So here are a few problems I am having trouble with please help.

ED || AB, BE bisects <ABC
1: Find the perimeter of BAED. Round to the nearest thousandth. (connect A to B, B to E, B to D and C, A to E and C, and D to E) (A to C measures 32, A to B is 17, and B to D is 8)

                A .
                             
                            .E

B.                  D .                 .C

(E is between A and C, by the way, if that's not clear above.)

2. The sides of a triangle are 10, 8, 15. Find the length of the 2 segments cut off by the angle bisector of the largest angle. Give the answers as fractions in lowest terms.

3. Four numbers total 417. The ratio of the 1st to the 2nd is 3:5. The ratio of the 2nd to the 3rd is 4:5, and the 3rd to the 4th is 2:3. What are the numbers rounded to the nearest thousandth.

 

[soroban]

Hello, monkeyrock227!

3. Four numbers total 417. .The ratio of the 1st to the 2nd is 3:5.
The ratio of the 2nd to the 3rd is 4:5, and the 3rd to the 4th is 2:3.
What are the numbers, rounded to the nearest thousandth?

\(\displaystyle \text{Let the four numbers be: }\:a,b,c,d\)

\(\displaystyle \text{We have: }\:a + b + c + d \:=\:417\;\;[1]\)

\(\displaystyle \text{We also have: }\:\begin{array}{cccccccc}\frac{a}{b} \:=\:\frac{3}{5} & \Rightarrow & b \:=\:\frac{5}{3}a & [2] \\ \\[-3mm]\frac{b}{c}\:=\:\frac{4}{5} & \Rightarrow & c \:=\:\frac{5}{4}b & [3] \\ \\[-3mm] \frac{c}{d}\:=\:\frac{2}{3} & \Rightarrow & d \:=\:\frac{3}{2}c & [4] \end{array}\)

\(\displaystyle \text{Substitute [2] into [3]: }\:c \:=\:\frac{5}{4}\left(\frac{5}{3}a\right) \quad\Rightarrow\quad c \:=\:\frac{25}{12}a\;\;[5]\)

\(\displaystyle \text{Substiute [5] into [4]: }\:d \:=\:\frac{3}{2}\left(\frac{25}{12}a\right) \quad\Rightarrow\quad d \:=\:\frac{75}{24}a\;\;[6]\)


\(\displaystyle \text{Substitute [2], [5], [6] into [1]: }\:a + \frac{5}{3}a + \frac{25}{12}a + \frac{75}{24}a \:=\:417\)

\(\displaystyle \text{Multiply by 24: }\:24a + 40a + 50a + 75a \:=\:10,\!008 \quad\Rightarrow\quad 189a \:=\:10,\!008\)


\(\displaystyle \text{Therefore : }\:\boxed{\begin{array}{ccccc}a &&&=& 52.952 \\ \\[-3mm] b &=&\frac{5}{3}a &=& 88.253 \\ \\[-3mm] c&=&\frac{25}{12}a &=& 110.317 \\ \\[-3mm] d &=& \frac{75}{24}a &=& 165.475 \end{array}}\)

 

[soroban]

Hello again, monkeyrock227!

2. The sides of a triangle are 10, 8, 15.
Find the length of the 2 segments cut off by the angle bisector of the largest angle.
Give the answers as fractions in lowest terms.

We're expected to know this theorem:
. . The angle bisector of a triangle divides the opposite side
. . into two segments proportional to the other two sides.

            B
            *
           * * *
       8  *   *   *     10
         *     *      *
        *       *         *
       *         *            *
    A *   *   *   *   *   *   *   * C
      :    8a     D      10a      :
      : - - - - - - 15- - - - - - :

\(\displaystyle BD\text{ is the bisector of }\angle B.\)

\(\displaystyle D\text{ divides }AC\text{ in the ratio }8:10\)

. . \(\displaystyle \text{Hence: }AD = 8a,\;DC = 10a\)


\(\displaystyle \text{We have: }\:8a + 10a \:=\:15 \quad\Rightarrow\quad 18a \:=\:15 \quad\Rightarrow\quad a \:=\:\frac{5}{6}\)


\(\displaystyle \text{Therefore: }\;\begin{array}{ccccccc}AD &=&8\left(\frac{5}{6}\right) &=& \frac{20}{3} \\ \\[-3mm] DC &=& 10\left(\frac{5}{6}\right) &=& \frac{25}{3} \end{array}\)