Mathematics Question and Solution
Let O be the centre of the circle.
Since AE = EB, therefore E is the midpoint of length AB.
AB = 10cm, therefore AE = EB = 10/2 = 5cm.
r will be;
Triangle EBC Area+Triangle EOC Area+Triangle AOE Area+Triangle AOD Area+Triangle DOC Area = Square ABCD Area.
The triangles height and base.
EBC, height=5cm, base=10cm.
EOC, height=rcm, base=√(5²+10²)=√(125)=5√(5)cm.
AOE, height=(10-r)cm, base=5cm.
AOD, height=rcm, base=10cm.
DOC, height=rcm, base=10cm.
Therefore r is;
(5x10)/2+(5√(5)r)/2+(5(10-r))/2+(10r)/2+(10r)/2 = (10x10)
25+(5√(5)r)/2+(50-5r)/2+5r+5r = 100
50+5√(5)r+50-5r+10r+10r=200
100+5√(5)r+20r-5r=200
5√(5)r+15r=200-100
5√(5)r+15r=100
Dividing through by common factor 5.
√(5)r+3r=20
r is common at the left, factoring.
(√(5)+3)r=20
Divide through by r coefficient, (√(5)+3).
Therefore;
r = 20/(√(5)+3) cm.
r = 3.82 cm² to 2 decimal places.
