"Weird" Right Triangle problem

[Infantry]

"Weird" for me anyways, I don't really get the solution and the textbook doesn't explain it really well IMO. Anyway, here it goes...

​< link to objectionable page removed >

The textbook says I should multiply BD x DC = 8^2 -> DC = 16, then

AC^2= CDxCB -> AC^2=16 x 20 ; AC = 17.89

Could someone walk me through the solution?

 

[Deleted member 4993]

"Weird" for me anyways, I don't really get the solution and the textbook doesn't explain it really well IMO. Anyway, here it goes...

​<link removed>

The textbook says I should multiply BD x DC = 8^2 -> DC = 16, then

AC^2= CDxCB -> AC^2=16 x 20 ; AC = 17.89

Could someone walk me through the solution?

Most of us are not willing to go to another website to retrieve problems. Please copy and paste (or type it up) and post here.

 

[HallsofIvy]

There are three right triangles here- the big one and the two right triangles that altitude divides it into. Use the fact that all three triangles are similar and the Pythagorean theorem.

 

[Steven G]

"Weird" for me anyways, I don't really get the solution and the textbook doesn't explain it really well IMO. Anyway, here it goes...

<link removed>

The textbook says I should multiply BD x DC = 8^2 -> DC = 16, then

AC^2= CDxCB -> AC^2=16 x 20 ; AC = 17.89

Could someone walk me through the solution?

In right triangle ABC we have angle B + angle C =90. In the smaller top left triangle small angle A + angle B = 90 since triangle ABD is a right angle. In the bottom right Right Triangle we have small angle A + angle C = 90. So we have these two smaller right triangle are similar right triangles. Side marked 4 in the left triangle is similar (proportional) the the side marked 8 in the right side triangle. So the ratio from the left to the right is 1 to 2. So we can set Y to be AB, 2Y to AC. We want to find 2Y. We know from the left triangle that 4^2 +8^2 =80 = y^2. So Y =sqrt(80) = sqrt(16)*sqrt(5)= 4 sqrt(5). Then 2Y=8sqrt(5)