Geometry help

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I am having trouble with a problem and need some help. I have a triangle that has one right angle. If the traingle sides are AB, BC, and AC, the side of AB =6, BC=10 and AC =8. angle A is a right angle. If a line from point D is drawn perpendicular to BC to angle A I need to find the height of line AD. I know only the one side of the new triangle formed ABD and that is AB is still 6. What do I do?
 
You have two right triangles
ABD and ACD
AD²+CD²=8²
AD²+(10-CD)²=6²
Solve for CD then use that to find AD.
 
Those little ones are Right Triangles, too, right? Where's Pythagoras when you need him?

6^2 = AD^2 + BD^2
8^2 = AD^2 + DC^2
BD + DC = 10

That's enough information. Solve away...
 
Ok I get all that, but when I solve for CD I get CD=AD-2.84 and still don't know where to go from here, because I still can't get a numeric answer?
 
Expand the (10-CD)² then subtract the other equation. You get a linear in CD only.
 
when I just try to plug in numbers to see what works, if I put AD in at 4.8 then BD is 3.6 and CD is 6.4 and it all works, but I don't know how to do it other than guess until I get the right number.
 
Hello, Becky99!

The triangle sides are AB = 6, BC = 10, and AC = 8.
If a line from A is drawn perpendicular to BC at poinT d.
Find the length of AD.
Click to expand...
Code: 
            A
            *
           /: \
          / :   \
        6/  :     \8
        /   :h      \
       /    :         \
      *-----+-----------*
      B     D           C
      : - - -  10 - - - :
Note that: \(\displaystyle \,\Delta BAC\,\sim\,\Delta ADC\)
\(\displaystyle \;\;\)(They are right triangles that share angle C.)

From "proportional parts", we have: \(\displaystyle \frac{h}{8}\,=\,\frac{6}{10}\;\;\Rightarrow\;\;h\,=\,4.8\)
 
Becky99 said:
Ok I get all that, but when I solve for CD I get CD=AD-2.84 and still don't know where to go from here, because I still can't get a numeric answer?
Click to expand...
You're complicating it; no need to use AD:

8^2 = AD^2 + CD^2 : AD^2 = 64 - CD^2 [1]
6^2 = AD^2 + BD^2 : AD^2 = 36 - BD^2 [2]

from [1] and [2]: 64 - CD^2 = 36 - BD^2
plus you have:
BD + CD = 10 : CD = 10 - BC
 
oh, ok thanks, I didn't know about proportional parts-is that a formula?
 
Could you tell me how to find the area of a concave kite? I know that a convex kite formula is (D1 X D2)/2---is it the same for a concave kite?
 
Hello, Becky99!

Could you tell me how to find the area of a concave kite?
I know that a convex kite formula is: \(\displaystyle \,\frac{D_1\,\times\,D_2}{2}\)
Is it the same for a concave kite? \(\displaystyle \;\) . . . yes!
Click to expand...
Code: 
      P           R
      *...........*
       * *  Q  * *
        *   *   *
         *  :  *
          * : *
           *:*
            *
            S
The concave kite is \(\displaystyle PQRS\).

The diagonals are: \(\displaystyle \,D_1\,=\,PR,\;D_2\,=\,QS\).

Yes . . . the area is still: \(\displaystyle \,A\;=\;\frac{D_1\,\times\,D_2}{2}\)
 
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