Question

Pls who can solve this asap​

Answer 1

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)