AB = CD = 10.
A = D = 45o.
h = 10*sin45 = 7.07 = Altitude
CF = h/sin45 = 7.07/sin45 =10= Diagonal
BC = EF = CF*cos45 = 10*cos45 = 7.07 =
Short base.
A circle with radius 3 is inscribed in a isosceles trapezoid with legs of 10. Find the length of the smaller base.
When I draw a diagram, calling the trapezoid ABCD with A and D at the bottom, I see that the length from where the altitude from B and C hits AC to D is 8 from the pythagorean theorem. But the 6 looks a lot bigger. Also, how to find the smaller base?!
2 answers
Correction:
AB = CD = 10.
h = Diameter = 2*3 = 6=Ht. or altitude.
Draw altitudes CE and BF
Draw diagonal CF which bisects BCE and
BFE. Therefore, CFE = 45o
tan45 = h/BC = 6/BC
BC = 6/tan45 = 6. = Shortest base.
AB = CD = 10.
h = Diameter = 2*3 = 6=Ht. or altitude.
Draw altitudes CE and BF
Draw diagonal CF which bisects BCE and
BFE. Therefore, CFE = 45o
tan45 = h/BC = 6/BC
BC = 6/tan45 = 6. = Shortest base.