Math Problem - Help?

Question

I need help with a math problem that has to do with Special Parts of Triangles & Congruent Triangles in Geometry.

The triangle for the problem is in this link: http://i12.tinypic.com/5ykkfas.jpg

DE=x+6
BC=7x-6
BC=____

I know BC=19.2 I just don't know how they got that and how to work out the problem!

Please help fast, thanks.

Answer 1

triangles EAD and CAB are similar. Hence
(x+6)/(7x-6) = 1/2......(corresponding sides of similar triangles are proportional.)
from this we get x = 18/5.
substitute x in BC = 7x-6, you get BC = 19.2.

Answer 2

From the construction of the triangle, it appears as if D and E are the midpoints of AB and AC respectively.

We have AC/BC = sin B = AE/DE

AC/BC = AE/DE

BC = AC x DE / AE

AC is 2AE

So, BC = 2DE

7x - 6 = 2(x + 6)

7x - 6 = 2x + 12

7x - 2x = 12 + 6 = 18

5x = 18

x = 18/5 = 3.6

So, BC = 7 (3.6) - 6 = 25.2 - 6 = 19.2

So, your answer is right and the above is a fairly simple method of arriving at it.

Answer 3

Hi Cheryl,

Well, BC must be double DE.
(since the sides are double).

So 7x-6 = 2*(x+6)
So 5x=18
So x=3.6

Since BC is 7x-6,
BC= 25.2 - 6 = 19.2
.

Answer 4

DA=DB, therefore AB=2DA
EC=AE, therefore AC=2AE
In triangle ADE
DE^2=DA^2+AE^2
(x+6)^2=DA^2+AE^2

In triangle ABC
BC^2=AB^2+AC^2
(7x-6)^2=(2DA)^2+(2AE)^2
(7x-6)^2=4DA^2+4AE^2
(7x-6)^2=4(DA^2+AE^2)
(7x-6)^2=4(x+6)^2
(7x-6)^2=[2(x+6)]^2
Taking square root on both sides
7x-6=2(x+6)
7x-6=2x+12
7x-2x=12+6
5x=18
x=18/5=3.6

Therefore, BC=7x-6
=7*3.6-6
=25.2-6
=19.2