Question

slove questions 12
please ​

Answer 1

Let

E be the point where circle touches AC

F be the point where circle touches AB

though we can prove it easily, we will take the following properry as proved

Two tangents can be drawn on a circle from an outer point of the circle. The length of the two tangents are equal.

so DC = CE = 6 cm

BD = FB = 8 cm

AE = AF = x (let)

now area of ∆ ABC = ar(∆AOB) + ar(∆AOC) + ar(∆BOC)

= (1/2)*AB*OF + (1/2)*AC*OE + )1/2)*BC*OD

= (1/2) [(x+8)4 + (x+6)4 + 14*4]

= (1/2) [4x + 32 + 4x + 24 + 56]

= (1/2) (8x + 112)

= 4x + 56..........(1)

again, area of ∆ABC using heron formula

s = (AB + AC + BC)/2

= (x + 8 + x + 6 + 14)/2

= (2x + 28)/2

= x + 14

s - a = s - BC = (x + 14 - 14) = x

s - b = s - AC = x + 14 - x - 8 = 6

s - c = s - AB = x + 14 - x - 6 = 8

area = √s(s-a)(s-b)(s-c)

= √(x + 14)*x*6*8......(2)

thus, (1) = (2)

√(x + 14)*x*6*8 = (4x + 56) = 4(x + 14)

squaring both side

48x(x+14) = 16(x+14)^2

3x(x+14) - (x + 14)^2 = 0

(x + 14) ( 3x - (x + 14)) = 0

(x + 14) ( 3x - x - 14) = 0

(x + 14) (2x - 14) = 0

this gives

x + 14 = 0 or x = -14 (this is not possible)

2x - 14 = 0

2x = 14

x = 7

thus

AC = x + 8 = 7 + 8 = 15 cm

AB = x + 6 = 7 + 6 = 13 cm