Each side of square ABCD is 8 cm. Circle E is drawn through A and D so that it is tangent to BC. What is the radius of this circle?
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2006-12-15 17:01:19 · 4 answers · asked by Sarina 2 in Science & Mathematics ➔ Mathematics
Suppose Mid pt of BC is M. triangle AEM is an isosceles triangle
where AE=EM.
AM=sqrt(4^2 + 8^2)=4sqrt5
Angle AME = tan^-1(4/8) =26.565
Angle AEM = 180-2(26.565)=126.87
Using Sine rule:
4sqrt5/sin126.87=r/sin26.565
r=5
2006-12-15 18:28:28 · answer #1 · answered by Maths Rocks 4 · 0⤊ 0⤋
Radius = 8 cm / 2 / sin 60
2006-12-16 03:19:54 · answer #2 · answered by Renaud 3 · 0⤊ 0⤋
r^2 = 4^2 + (8 - r)^2
r^2 = 4^2 + 64 - 16r + r^2
16r = 16 + 64 = 80
r = 5
8 - 5 = 3
3^2 + 4^2 = 5^2
2006-12-15 17:34:41 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
(Sqrt (r^2 -16)) + r = 8, solve for r
r^2 -16 = r^2 - 16r + 64
16r = 80
r = 5
2006-12-15 17:14:55 · answer #4 · answered by Scythian1950 7 · 1⤊ 0⤋