Triangle

[123]

A right triangle ABC has area equal to \(\displaystyle 43.5cm^2\), \(\displaystyle AC=5.8cm\), \(\displaystyle AM=MB\).
MBD is a triangular area of \(\displaystyle 1/3\) triangular area ABC.
1) Show, that \(\displaystyle BD=\frac{2}{3}BC\)

2) Find BD lenght. (I get that BD=10) correct?

 

[TchrWill]

5.8(BC)/2 = 43.5 making BC = 15.

If AM = MB then CD = DB and MD = 5.8/2 = 2.9

Then, the area of triangle MDB = 2.9(7.5)/2 = 10.875, not 1/3 the area of ABC, or 14.5.

BD is 1/2 BC = 7.5.

What appears to be the problem is that MD is really perpendicular to AB.

Then, leave the old MD line where it is but just reidentifying it as ME.

Then, the triangle MED leads us to 2.9((DE)/2 14.5 - 10.875 = 3.625 making DE = 2.5 and DB = 10 or 2/3 CB.

 

[soroban]

Hello, 123!

\(\displaystyle \text{A right triangle }ABC \text{ has area equal to }43.5\text{ cm}^2,\;AC=5.8\text{ cm},\;AM=MB\).

\(\displaystyle MBD\text{ is a triangular area of }\tfrac{1}{3}\text{ the area of }\Delta ABC.\)

\(\displaystyle \text{1) Show that: }\:BD\,=\,\tfrac{2}{3}BC\)

\(\displaystyle \text{2) Find BD length.\;\;(I get that BD = 10, correct?)}\) . Yes!

              C
              o
             *    *
            *         *
           *              *    D
      5.8 *               /   o
         *              h/  *     *
        *               / *           *
       *               /*                 *
    A o   *   *   *   o   *   *   *   *   *   o B
                      M

\(\displaystyle \text{From }M\text{ draw a perpendicular }h\text{ to }BC\!:\;h = 2.9\)


\(\displaystyle \text{The area of }\Delta ABC \:=\:\tfrac{1}{2}(BC)(5.8) \:=\:43.5 \quad\Rightarrow\quad \boxed{BC \,=\,15}\)


\(\displaystyle \text{The area of }\Delta MBC = \tfrac{1}{3}(\Delta ABC) \:=\:\tfrac{1}{3}(42,5) \:=\:14.5\)

\(\displaystyle \text{The area of }\Delta MBC \:=\:\tfrac{1}{2}(BD)(h) \:=\:\tfrac{1}{2}(BD)(2.9) \:=\:14.5 \quad\Rightarrow\quad \boxed{BD \:=\:10}\)


\(\displaystyle \text{And we're done!}\)