Thread: geometry triangles

Given: AB=20, BC=7, CA=15, triangle DAB is ~ to triangle DCA.

FIND DC.

Yea, show all work, no fancy stuff, you probably can do it with a few proportions then substitution, you will need to find y first to find x...
once again, show all work.
thanks!

Originally Posted by gamecube10074

Given: AB=20, BC=7, CA=15, triangle DAB is ~ to triangle DCA.

FIND DC.

Yea, show all work, no fancy stuff, you probably can do it with a few proportions then substitution, you will need to find y first to find x...
once again, show all work.
thanks!

There isn't enough information to finish this problem. The key is that we have no idea what angle DAC is supposed to be because both x and y are unknown. We may construct any number of such triangles with different angles DAC.

-Dan

Originally Posted by topsquark

There isn't enough information to finish this problem. The key is that we have no idea what angle DAC is supposed to be because both x and y are unknown. We may construct any number of such triangles with different angles DAC.

-Dan

wrong, there is enough, solve for Y first, make a few proportions, and see a similarity and plug it in.

edit: several people have solved at my school, but they refuse to tell me x.x

The first step is to determine which angles are corresponding since they are similar. To do that we will travel back 2300 years to Euclid.

Proposition 16 from Book the First:
In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.

1)Look at triangle DCA and DBA. Note they share a common angle at D. That is the first pair of corresponding angles.

2)Gaze upon angle <ACD, it is equal to either <ABC or <BAD, because we are told the triangles are similar. Note by Proposition 16 <ACD is greater than <ABC. Thus, <ACD=<BAD.

3)Hence, the remainging angles <DAC and <ABC are equal.

4)By similarity:
AC/DC=AB/AD
AC/AD=AB/DB

5)Thus,
15/x=20/y
15/y=20/(x+15)

I recommend to take reciprocals of both sides,
x/15=y/20
y/15=(x+15)/20
Multiply both sides in both equations by 60,
4x=3y
4y=3(x+15)=3x+45
Thus,
3y-4x=0
4y-3x=45
Solve it yourself.

Originally Posted by gamecube10074

Given: AB=20, BC=7, CA=15, triangle DAB is ~ to triangle DCA.

FIND DC.

Yea, show all work, no fancy stuff, you probably can do it with a few proportions then substitution, you will need to find y first to find x...
once again, show all work.
thanks!

The figure, as shown, doesn't help. As shown, the (CA=15) is longer than the (AB=20). Confusing.

Sketch the figure approximately closer to true scale.

I assume you mean by the "worm" that (triangle DAB) is "proportional" to (triangle DCA). So if they are proportional, then the two triangles are similar.

The key or trick is to superimpose one on the other, so that we can see that they are similar.
So, move triangle DCA on top of triangle DAB such that "DC = x" is along DA, and "CA = 15" is parallel to AB.

By proportion,
x/15 = y/20 <-----the x here is along DA.
y = 20x/15 = (4/3)x ------------(1)

By proportion again,
(x+7)/20 = y/15 <----the y here is along DB.
y = (15/20)(x+7)
y = (3/4)(x+7) -----------------(2)

y from (1) = y from (2),
(4/3)x = (3/4)(x+7)
Clear the fractions, multiply both sides by 3*4,
16x = 9(x+7)
16x = 9x +63
16x -9x = 63
7x = 63
x = 63/7 = 9 ---------------answer.

And so, say, in (1),
y = (4/3)9 = 12 -----------------answer.

Check,
In (2),
12 =? (3/4)(9+7)
12 =? (3/4)(16)
12 =? 12
Yes, so, OK.

Originally Posted by topsquark

There isn't enough information to finish this problem. The key is that we have no idea what angle DAC is supposed to be because both x and y are unknown. We may construct any number of such triangles with different angles DAC.

-Dan

I missed the proportionality statement in the problem.

-Dan