Geometry Triangle Missing Side
- Thread starter kwikpwn
- Start date
- Joined
- Apr 12, 2005
- Messages
- 11,339
It's also not clear what is marked 52 and 48.
Plus, you still haven't shown any work at all. What's your plan?
Note: Once you solve the little problem Dennis gave you (call the result 'h'), the length of the giant hypotenuse might be 52 + h + x
Plus, you still haven't shown any work at all. What's your plan?
Note: Once you solve the little problem Dennis gave you (call the result 'h'), the length of the giant hypotenuse might be 52 + h + x
Can you please solve for X and tell me what you did? Thanks!
Assuming that the 52 and 48 apply to the small triangle at the lower left of the large triangle
Lets label what we know.
Triangle ABC, A upper right, B lower left and C lower right.
Small lower left triangle, D upper right and E lower right.
BD = 52 and BE = 48
Semi-circle of radius R with center G on BC, tangent to AB at F and DE at E.
AF = AC = X
Draw line DG which bisects angle EDF making DE = DF
DE = sqrt(52^2 - 48^2) = DF = 20
BF = 52 + 20 = 72
(72 + X)/52 = X/20
I'll leave the rest for you.
Assuming that the 52 and 48 apply to the small triangle at the lower left of the large triangle
Lets label what we know.
Triangle ABC, A upper right, B lower left and C lower right.
Small lower left triangle, D upper right and E lower right.
BD = 52 and BE = 48
Semi-circle of radius R with center G on BC, tangent to AB at F and DE at E.
AF = AC = X
Draw line DG which bisects angle EDF making DE = DF
DE = sqrt(52^2 - 48^2) = DF = 20
BF = 52 + 20 = 72
(72 + X)/52 = X/20
I'll leave the rest for you.
HA! After drawing all this and assuming smaller right triangle = 20-48-52 and labelling
as per TchrWill (thanks Will!), I see it is still easier using similar triangles:
Triangle BDE is similar to triangle BFG (both right triangles, sharing angleDBE:
DE/BD = FG/BG ; 20 / 52 = r / (r + 48) ; r = 30
Triangle BDE is of course similar to triangle ABC:
DE / BE = AC / BC ; 20 / 48 = x / (2r + 48) ; you gor r: solve for x
You'll end up with triangle ABC = 45-108-117
as per TchrWill (thanks Will!), I see it is still easier using similar triangles:
Triangle BDE is similar to triangle BFG (both right triangles, sharing angleDBE:
DE/BD = FG/BG ; 20 / 52 = r / (r + 48) ; r = 30
Triangle BDE is of course similar to triangle ABC:
DE / BE = AC / BC ; 20 / 48 = x / (2r + 48) ; you gor r: solve for x
You'll end up with triangle ABC = 45-108-117