1.

IND ABC, AR-3cm, AC=9cm thenfind A(TRIANGLE ABR) :A(TRIANGLE ABC)​

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Answer:

In the given figure, ΔABC CIRCUMSCRIBED the circle with centre O.

Radius OD=3cm

BD=6cm,DC=9cm

Area of ΔABC=54cm2

To find :  LENGTH of AB and AC.

AF and EA are tangents to the circle at point A.

Let AF=EA=x

BD and BF are tangents to the circle at point B.

BD=BF=6cm

CD and CE are tangents to the circle at point C.

CD=CE=9cm

Now, NEW sides of the triangle are:

AB=AF+FB=x+6cm

AC=AE+EC=x+9cm

BC=BD+DC=6+9=15cm

Now, using Heron's formula:

Area of triangle ABC=s(s−a)(s−b)(s−c)

Where S=2a+b+c

S=1/2(x+6+x+9+15)=x+15

Area of ABC=(x+15)(x+15−(x+6))(x+15−(x−9))(x+15−15) 

Or

54=(x+15)(9)(6)(x)

SQUARING both sides, we have

Step-by-step explanation:

hopes its helps


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