A triangle ABC is drawn to circumscribe a circle of radius 2 cm. Such that the segments BD and DC into which BC is divided by a point of contact D are of lengths 4cm and 3 cm respectively. If the area of triangle ABC is equal to 21 cm sq. Find the lengths of sides AB and AC.
Answers
Answer:
AB = 7.5 cm
AC = 6.5 cm
Step-by-step explanation:
Circle with center at O is inside a ΔABC.
Given that radius of circle = 2 cm = OD = OE = OF.
BD = 4 cm, CD = 3 cm
Since BD, BE are tangents drawn from same point B external to the circle, BD = BE = 4 cm
Similarly CD = CF = 3 cm
Similarly AE = AF = x cm.
BC = 7, AB = x+4, AC = x+3
Area of the ΔABC = √(S(S – AB)(S – BC)(S – AC)
S = (AB + BC + CA) / 2 = (7 +x+4+x+3)/2 = x + 7
Area of triangle = √((x +7)(X+7-x-4)(x+7-x-3)(x+7-7))
= √((x+7)(3)(4)x)
=√(12x(x+7) = 21
12x(x+7) = 21*21
4x(x+7) = 21*7
4x² +28x – 147 = 0
Solving we get x = 7/2 or -21/2. (Ignore the negative value)
X = 7/2.
AB = 7/2 + 4 = 7.5 cm
AC = 7/2 + 3 = 6.5 cm
Step-by-step explanation:
tangent from same ext points are equal in length
therefore;
BD =BF=4cm
CD=CE=3cm
let AF=AE=x cm
AR(ABC) = AR( AOB) + AR(BOC) + AR(AOC)
21 = 1/2 [ OF * (x+4)] + 1/2(OD*7) + 1/2[ OE *(x+3)]
21 = 1/2* r ( x+4 + 7 + x+3 ) [ as OF=OD=OE=r]
21 = 1/2*2 ( 14+2x)
21 = 14 + 2x
2x= 21-14
2x =7
x = 7/2 = 3.5 cm
AB =x+4
=3.5 +4
=7.5 cm
AC =x+3
= 3.5+3
= 6.5 cm
