In figure a triangle ABC is drawn to circumscribe a circle of radius 2 cm such that the tangents BD
And DC into which BC is divided by the point of contact Dare the lengths 4 cm and 3 cm. If area of
∆ABC = 21 cm2
, then find the lengths of sides AB and AC
Answers
ANSWER
In the given figure, ΔABC circumscribed the circle with centre O.
Radius OD=3cm
BD=6cm,DC=9cm
Area of ΔABC=54cm
2
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=
2
a+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
54
2
=54x(x+15)
x
2
+15x−54=0
Solve this quadratic equation and find the value of x.
x
2
+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
Answer :-
AB=x+6=3+6=9cm
AC=x+9=3+9=12cm
Step-by-step explanation:
Answer
In the given figure, ΔABC circumscribed the circle with centre O.
Radius OD=3cm
BD=6cm,DC=9cm
Area of ΔABC=54cm
2
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=
2
a+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
54
2
=54x(x+15)
x
2
+15x−54=0
Solve this quadratic equation and find the value of x.
x
2
+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
Answer :-
AB=x+6=3+6=9cm
AC=x+9=3+9=12cm
Please see the attachment.
