Q. 234.0( 4 Votes )

In fig. 4,

Answer :


First, we join OB, OF, OA, OE, and OC.


D, E, and F are the tangent points.


As we know that Tangents to a circle
from a point are equal.


So, BF= BD = 4 cm (given)


CE= CD= 3 cm (given)


AF=AE= x cm (assume)


We are also given that area of Δ ABC=21cm2


Now,


area(Δ ABC) = area(Δ BOC) + area(Δ COA) + area(Δ AOB)


area (Δ BOC) = 1/2 × Base(BC) × Height(OD)
= 1/2 × 7 × 2=7cm2


area(Δ COA) = 1/2 × Base(AC) × Height(OE)
= 1/2 × (3 + x) × 2= 3 + x


area(Δ AOB) = 1/2 × Base(AB) × Height(OF)
= 1/2 × (4 + x) × 2= 4 + x


Putting the values in the equation;
area(
Δ ABC) = area(Δ BOC) + area(Δ COA) + area(Δ AOB)


21=7 + 3 + x + 4 + x


21=14 + 2x


7=2x


x=3.5cm


Length of AB= 4 + x = 4 + 3.5= 7.5cm


And, Length of AC = 3 + x= 3 + 3.5=6.5cm

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