In triangle DEF, DE=5, DF=6, and EF=7. Circles are drawn centered at D, E, and F, so that the circles are pairwise externally tangent. Find the sum of the areas of the three circles.
Chords AB and CD of a circle are perpendicular, and meet at P. If AP = 3, BP = 4, CP = 6, and DP = 2, then find the diameter of the circle.
Peter has built a gazebo, whose shape is a regular heptagon, with a side length of 1 unit. He has also built a pathway around the gazebo, of constant width 1 unit, as shown below. (Every point on the ground that is within 1 unit of the gazebo and outside the gazebo is covered by the pathway.) Find the area of the pathway.
Trapezoid ABCD is inscribed in a semicircle, as shown below. The length of base CD is 7 and the lengths of legs AD and BC are 3. Find the radius of the semicircle.
2) Let O be the center of the circle, and draw OA. Drop the perpendicular from O to AB, and you'll hit the midpoint of chord AB. You can build right triangles for the other chord, which will eventually give you an answer of sqrt(51).
4) Drop pependiculars from C and D to AB, and you can work with the right triangles. By Pythagoras, r^2 - (3/2)^2 = 7^2 - (r - 7/2)^2. Therefore, r = 13/2.
1. Let the radius of te circle with center D = x
Let the radius of the circle with center E = y
Let the radius of the circle with the center F = z
So we have these equations
x + y = 5
x + z = 6
y + z = 7
Subtract thesecond equation from the first and we have that y - z = -1
Add this to equation to the third equation and we have that
2y = 6
y = 3
x = 2
z = 4
So...the total area of the circles are pi ( 2^2 + 3^2 +4^2) = pi ( 29) = 29 pi units^2 ≈
Draw DE and CF perpendicular to AB
So AE = BF = sqrt (3^2 - H^2) where H is the height of the trapezoid
And AB = 2R....so....
2R =AE + BF + 7
2R = 2*AE + 7
2R = 2 sqrt ( 9 - H^2) + 7
R = sqrt (9- H^2) + 3/2
R - 3/2 = sqrt (9 - H^2) square both sides
R^2 - 3R + 9/4 = 9 - H^2 rearrange as
H^2 = 3R - R^2 + 6.75
Let the midpoint of AB = O ...... connect OD and draw OM perpendicular to DC
And OM = H and DM = 3.5 and OD is the radius of the semi-circle = R
So....triangle ODM is right and we have that
OD^2 - OM^2 = DM^2 ....so.....
R^2 - H^2 = 3.5^2
R^2 - (3R - R^2 + 6.75) = 3.5^2 rearrange as
R^2 + R^2 - 3R -6.75 = 12.25 subtract 12.25 from each side
2R^2 - 3R - 19 = 0
Using the quad formula we have that
R = 3 ± √ [ 9 + 4* 2 * 19 ] 3 ± √ [ 161]
___________________ = ___________
Take the positive answer and we get that
R = 3 + √161
_________ ≈ 3.92
And the diameter ≈ 3 * 3.92 = 7.84