Question

Ar of ABC=54cm square
find AB and AC​

Answer 1

$\huge\red{ANSWER}$

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= a+b+c / 2

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

54²=54x(x+15)

x ²+15x−54=0

Solve this quadratic equation and find the value of x.

x²+18x−3x−54=0

x(x+18)−3(x+18)=0

(x−3)(x+18)=0

Either x=3 or x=−18

But x cannot be negative.

So, x=3

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Answer :-

AB=x+6=3+6=9cm

AC=x+9=3+9=12cm

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