the radius of the circle of a triangle is 4 cm and the segment into which one side is divided by the point of contact are 6 cm and 8 cm. determine the other two sides of the triangle.
Share
the radius of the circle of a triangle is 4 cm and the segment into which one side is divided by the point of contact are 6 cm and 8 cm. determine the other two sides of the triangle.
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
15 cm and 13 cm
Step-by-step explanation:
Let there is a circle having centre O touches the sides AB and AC of the triangle at point E and F respectively.
Let the length of the line segment AE is x.
Now in ΔABC,
CF = CD = 6 (tangents on the circle from point C)
BE = BD = 6 (tangents on the circle from point B)
AE = AF = x (tangents on the circle from point A)
Now AB = AE + EB
=> AB = x + 8
BC = BD + DC
=> BC = 8+6 = 14
CA = CF + FA
=> CA = 6 + x
Now
s = (AB + BC + CA )/2
=> s = (x + 8 + 14 + 6 +x)/2
=> s = (2x + 28)/2
=> s = x + 14
Area of the ΔABC = √{s*(s-a)*(s-b)*(s-c)}
= √[(14+x)*{(14+x)-14}*{(14+x)-(6+x)}*{(14+x)-(8+x)}]
= √[(14+x)*x*8*6]
= √[(14+x)*x*2*4*2*3]
=> Area of the ΔABC = 4√[3x(14+x)] .............................1
Now area of ΔOBC = (1/2)*OD*BC
= (1/2)*4*14
= 56/2
= 28
Area of ΔOBC = (1/2)*OF*AC
= (1/2)*4*(6+x)
= 2(6+x)
= 12 + 2x
Area of ΔOAB = (1/2)*OE*AB
= (1/2)*4*(8+x)
= 2(8+x)
= 16 + 2x
Now Area of the ΔABC = Area of ΔOBC + Area of ΔOBC + Area of ΔOAB
=> 4√[3x(14+x)] = 28 + 12 + 2x + 16 + 2x
=> 4√[3x(14+x)] = 56 + 4x
=> 4√[3x(14+x)] = 4(14 + x)
=> √[3x(14+x)] = 14 + x
On squaring bothe side, we get
3x(14 + x) = (14 + x)2
=> 3x = 14 + x (14 + x = 0 => x = -14 is not possible)
=> 3x - x = 14
=> 2x = 14
=> x = 14/2
=> x = 7
Hense
AB = x + 8
=> AB = 7+8
=> AB = 15
AC = 6 + x
=> AC = 6 + 7
=> AC = 13
So value of AB is 15 cm and value of AC is 13 cm
Answer:
Step-by-step explanation: