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?!
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.
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.
To find the length of the smaller base of the isosceles trapezoid, let's consider the properties of the circle that is inscribed in it.
First, note that the center of the circle is the point where the perpendicular bisectors of the two non-parallel sides of the trapezoid meet. Let's call this point O.
Since the circle is inscribed in the trapezoid, the distance from point O to each of the sides of the trapezoid is equal to the radius of the circle, which is 3.
Let's label the four vertices of the trapezoid as A, B, C, and D, where A and D are the bases of the trapezoid.
To find the length of the smaller base, we need to determine the distance between points B and C along the line AD. Let's call this distance x.
Now, let's draw a line segment from point B to point O. This line segment is the perpendicular bisector of side AD.
Since the trapezoid is isosceles, the length of the line segment from O to the midpoint of AD is equal to the length of the altitude from B to AD. Let's call this length h.
From your description, you correctly found h to be 8 using the Pythagorean theorem.
Now, let's consider the right triangle formed by points O, B, and C. The hypotenuse of this triangle is the distance between points B and C, which is x.
The other two sides of this triangle are the radius of the circle (3) and the altitude from B to AD (8).
Using the Pythagorean theorem, we have:
x^2 = (3^2) + (8^2)
x^2 = 9 + 64
x^2 = 73
Taking the square root of both sides, we find:
x = √73
So, the length of the smaller base of the isosceles trapezoid is approximately equal to √73.