A triangle ABC is drawn to circumscribe a circle of radius 3cm such that the segment BD and DC into which BC is divided by the point of contact D are of length 9cm and 3cm
Answers
Step-by-step explanation:
HELLO DEAR,
IN ∆ ABC,
CE = CD = 3cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point C)
BF = BD = 9cm (Tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point B)
AE = AF = x (Again, tangents drawn from an exterior point to a circle are equal. Here, tangent is drawn from exterior point A)
Now, AB = AF + FB
AB = (x + 9)cm
BC = BD + CD
BC = (9 + 3) = 12cm
AC = AE + EC
AC = (x + 3)cm
semi-perimeter of triangle ABC = sum of all sides)/2
s = (x + 9 + 12 + x + 3)/2
s = (2x + 24)/2
s = (x + 12)cm
using Heron's formula for calculating area of ∆ ABC,
area ∆ ABC = \sf{\sqrt{s(s - a)(s - b)(s - c)}}
s(s−a)(s−b)(s−c)
\sf{\Rightarrow \sqrt{(x + 12)(x + 12 - x - 9)(x + 12 - 12)(x + 12 - x - 3)}}⇒
(x+12)(x+12−x−9)(x+12−12)(x+12−x−3)
\sf{\Rightarrow \sqrt{(x + 12)(3)(x)(9)}}⇒
(x+12)(3)(x)(9)
\sf{\Rightarrow \sqrt{(x + 12)*27x}}⇒
(x+12)∗27x
\sf{\Rightarrow \triangle ABC = \sqrt{27x^2 + 324x}}⇒△ABC=
27x
2
+324x
-------------( 1 )
Also,
OF = OE = OD = 3CM (radius of circle)
NOW,
area ∆ABC = area∆AOC + area∆AOB + area∆BOC
∆ABC = (1/2 × OE × AC) + (1/2 × OF × AB) + (1/2 × OD × BC)
∆ABC = 1/2 × 3{x + 3 + x + 9 + 12}
∆ABC = 1/2 × 3{2x + 24}
∆ABC = (3x + 36)------( 2 )
from-----------( 1 ) & ---------------( 2 )
\sf{\therefore , \Rightarrow \sqrt{27x^2 + 324x} = 3x + 36}∴,⇒
27x
2
+324x
=3x+36
on squaring both side;
\sf{\Rightarrow 27x^2 + 324x = 9x^2 + 1296 + 216x}⇒27x
2
+324x=9x
2
+1296+216x
\sf{\Rightarrow 27x^2 - 9x^2 + 324x - 216x - 1296 = 0}⇒27x
2
−9x
2
+324x−216x−1296=0
\sf{\Rightarrow 18x^2 + 108 - 1296 = 0}⇒18x
2
+108−1296=0
\sf{\Rightarrow x^2 + 6x - 72 = 0}⇒x
2
+6x−72=0
\sf{\Rightarrow x^2 + 12x - 6x - 72 = 0}⇒x
2
+12x−6x−72=0
\sf{\Rightarrow x(x + 12) - 6(x + 12) = 0}⇒x(x+12)−6(x+12)=0
\sf{\Rightarrow (x - 6)(x + 12) = 0}⇒(x−6)(x+12)=0
x - 6 = 0 , x + 12 = 0
x = 6 , x = -12{neglect} because, side of triangle can't be negative
hence, AB = (x + 9)cm = (6 + 9)cm = 15cm
AC = (x + 3)cm = (6 + 3)cm = 9cm.
