ABCD is a square of side 8, Mis the centre of the circle taking AD as diameter, E is a point on the side AB such that CE is tangent to the circle. Find the area of the triangle CBE.
Answers
Answer:
i)
AB, BC and AC are tangents to the circle at E, D and F.
BD=30cm,DC=7cm,∠BAC=90
From the theorem stated,
BE=BD=30cm
Also FC=DC=7cm
Let AE=AF=x …. (1)
Then AB=BE+AE=(30+x)
AC=AF+FC=(7+x)
BC=BD+DC=30+7=37cm
Consider right trianlge ABC, by Pythagoras theorem we have
BC
2
=AB
2
+AC
2
(37)
2
=(30+x)
2
+(7+x)
2
1369=900+60x+x
2
+49+14x+x
2
2x
2
+74x+949–1369=0
2x
2
+74x–420=0
x
2
+37x–210=0
x
2
+42x–5x–210=0
x(x+42)–5(x+42)=0
(x–5)(x+42)=0
(x–5)=0or(x+42)=0
x=5orx=–42
x=5[Since x cannot be negative]
∴AF=5cm[From (1)]
ThereforeAB=30+x=30+5=35cm
(ii)
AC=7+x=7+5=12cm
Let ‘O’ be the centre of the circle and ‘r’ the radius of the circle.
Join point O, F; points O, D and points O, E.
From the figure,
2
1
×AC×AB=
2
1
×AB×OE+
2
1
×BC×OD+
2
1
×AC×OC
AC×AB=AB×OE+BC×OD+AC×OC
12×35=35×r+37×r+12×r
420=84r
∴r=5
Thus the radius of the circle is 5 cm.
Answer:
i don't know sorry ....