in the diagram above, ABCD is a square. the circles centred at E and F just touch the sides of the square and each other at G. the circle centred at E has a radius of 4cm and the circle centred at F has a radius of 6cm
a. show that the lengths DE and BF are 4√2 and 6√2 cmm respectively.
b. show that the length of the diagonal BD is 10 + 10√2cm
c. use this to show that the exact area of the square ABCD is 150 + 100√2 cm^2
d. show that the total area of the 2 circles is 52pi cm^2
e. find the percentage of the area of the square which is filled by circles.
+130516 a) From E, draw a perpendicular to AD and a perpendicular to DC. The length of each of these will be the radius of the circle centered at E. Then, DE is just the hypotenuse of a isoceles right triangle and its length is just 4√2 cm.
By similar reasoning, BF = 6√2cm
b) So, the length of the diagonal will be 4√2 + 4 + 6 + 6√2 = 10 + 10√2
c) The length of the side of the square will be [10 + 10√2] / √2.....and the area will be
([10 + 10√2] / √2)^2 = [100 + 200√2 + 200] / 2 = [300 + 200√2] / 2 =
[150 + 100√2]cm^2
d) The area of the two circles = pi(4^2 + 6^2] =pi[16 + 36] = 52pi cm^2
e) The % area occupied by the two circles will be [52pi / (150 + 100√2) ] = about 56% of the total area of the square
+130516 a) From E, draw a perpendicular to AD and a perpendicular to DC. The length of each of these will be the radius of the circle centered at E. Then, DE is just the hypotenuse of a isoceles right triangle and its length is just 4√2 cm.
By similar reasoning, BF = 6√2cm
b) So, the length of the diagonal will be 4√2 + 4 + 6 + 6√2 = 10 + 10√2
c) The length of the side of the square will be [10 + 10√2] / √2.....and the area will be
([10 + 10√2] / √2)^2 = [100 + 200√2 + 200] / 2 = [300 + 200√2] / 2 =
[150 + 100√2]cm^2
d) The area of the two circles = pi(4^2 + 6^2] =pi[16 + 36] = 52pi cm^2
e) The % area occupied by the two circles will be [52pi / (150 + 100√2) ] = about 56% of the total area of the square
