Pls who can solve this asap
1 answer:
Answer:
the radius 1 < r < 3, so r = 1.39
Step-by-step explanation:
notice that you have 3 cm radius for the big circle
r radius for small
and you have this rectangle with dimension 6 cm by 8cm
so if you can imagine the radius segments moving into vertical and horizontal position
there are two boxes... 6cm by 6cm and 2r by 2r, but 6cm + 2r is not 8cm
diagonal of rectangle = root(8*8 + 6*6) = 10 cm
diagonals intersect at ((0 + 8)/ 2, (0 + 6) / 2)= (4, 3) but center of big circle at (3, 3)
It appears that the point (4, 3) is on the diagonal but does not go through the centers of the circles. but I do see
slope of diagonal is 3/4 so if rise is 2r, then the run.
(2r/x) = 3/4
(4/3)* 2r = x
equation of diagonal : y = (3/4)x + 0 if the bottom left is (0,0)
equation of circle r : (x - (8-r))^2 + (y - (6 - r))^2 = r*r
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Use a trig function to find the length of this line segment from the center of the big circle to the intersection of the small circle.
The triangle has lengths 1 cm , 3cm and a segment called x
x = line segment from the center of the big circle to the intersection of the small circle.
the angle between 1cm and x is 180 - arctan (3/4) = 180 - 36.8698 deg.
=143.1302 degrees
trig formula:
c^2 = a^2 + b^2 - 2ab*cos C
c^2 = 1^2 + 3^2 - 2*1*3*cos (81.8698 )
c^2 = 10 - 6 cos (81.8698)
c^2 = 9.15146173
c = 3.0251383 units
ok so we know the diagonal is 10
the distance from the intersection to the corner is 10 - 5 - 3.0251383
= 1.9748616 units
we can say 1.9748 = root(2) * r
so that r = 1.9748 / root(2)
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