In the given figure, find the length of Ab and AC.
Please tell right and full answer..
Answers
Step-by-step explanation:
Given :-
A circle inscribed in a triangle ABC ,
BD = 4 cm and CD = 3 cm and the radius of the circle is 2 cm
To find :-
Find AB and AC ?
Solution :-
Given that :
Radius of the circle = 2 cm
Join OE and OF
OD = OE = OF = 2 cm
BD = 4 cm
DC = 3 cm
BC = BD + DC
=> BC = 4 + 3 = 7 cm
BC = 7 cm
From the figure,
B is the exterior points to the circle and BD and BF are tangents to the circle from the point B
We know that
"The tangents drwan from the point exterior to the centre of the circle are equal".
=> BD = BF
=>BF = 4 cm
and
Similarly
C is the exterior points to the circle and CD and CE are tangents to the circle from the point C
=>CD = CE
=> CE = 3 cm
and
A is the exterior points to the circle and AF and AE are tangents to the circle from the point A
=> AE = AF
Let AE = AF = x cm
Now ,
Now the three sides of the triangle ABC
AB = AF+BF = x+4 cm
BC = 7 cm
AC = AE+EC = x+3 cm
Finding Area ( ∆ ABC) :-
Area of a triangle by Heron's formula
=√[S(S-a)(S-b)(S-c)] sq.units
S = (a+b+c)/2 units
Let a = x+4 cm
Let b = x+3 cm
Let c = 7 cm
=> S = (x+4+x+3+7)/2
=> S = (2x+14)/2
=> S = 2(x+7)/2
=> S = x+7 cm
Now
Area of ∆ABC =
√[(x+7)(x+7-x-4)(x+7-x-3)(x+7-7)] cm²
=> √[(x+7)(3)(4)(x)]
=> √(x+7)(12x)]
=>Area of ∆ABC =√(12x²+84x) cm²-----(1)
Now ,
Finding Ar( ∆AOB) :-
We know that
Area of a triangle = (1/2) base × height
In ∆ AOB , Base = AB = x+4 cm
height = OF = 2 cm
Area of ∆ AOB = (1/2)×AB×OF
=> (1/2)×(x+4)×2 cm²
=>Area of ∆ AOB = (x+4) cm²-----------(2)
Finding Ar( ∆AOC) :-
In ∆ AOC , Base = AC= x+3 cm
height = OE = 2 cm
Area of ∆ AOC = (1/2)×AC×OE
=> (1/2)×(x+3)×2 cm²
=>Area of ∆ AOC = (x+3) cm²-----------(3)
Finding Ar( ∆BOC) :-
In ∆ BOC , Base = BC= 7cm
height = OD = 2 cm
Area of ∆ BOC = (1/2)×BC×OD
=> (1/2)×7×2 cm²
=>Area of ∆ AOB = 7 cm²---------------(4)
We know that
Area of ∆ ABC = Ar(∆AOB)+Ar(∆AOC)+Ar(∆BOC)
From (1),(2),(3)&(4)
=> √(12x²+84x) = (x+4)+(x+3)+7
=> √(12x²+84x) = x+4+x+3+7
=> √(12x²+84x) = 2x+14
=> √4(3x²+21x) = 2(x+7)
=> 2√(3x²+21x) = 2(x+7)
=> √(3x²+21x) = x+7
On squaring both sides then
=>[√(3x²+21x)]² = (x+7)²
=> 3x²+21x = x²+2(x)(7)+7²
=> 3x²+21x=x²+14x+49
=> 3x²+21x-x²-14x-49 = 0
=> 2x²+7x-49 = 0
=> 2x²+14x-7x-49 = 0
=> 2x(x+7)-7(x+7) = 0
=> (x+7)(2x-7) = 0
=> x+7 = 0 or 2x-7 = 0
=> x = -7 or x = 7/2
X can't be negative.
So, x = 7/2 cm
=> x = 3.5 cm
Now AE = AF = 3.5 cm
Now , AB = AF+BF = 3.5+4 = 7.5 cm
AC = AE+EC = 3.5+3 = 6.5 cm
Answer:-
The length of AB = 7.5 cm
The length of AC = 6.5 cm
Used formulae:-
- "The tangents drwan from the point exterior to the centre of the circle are equal".
- The angle between the radius and the tangent drawn from the exterior to the circle at the point of contact is 90°
- Area of a triangle by Heron's formula ∆ = √[S(S-a)(S-b)(S-c) sq.units
- S = (a+b+c)/2 units
- Area of a triangle = (1/2) base × height sq.units
