1.

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.​

Answer»

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.

 


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