Question

Guys pls give solution for
( Class 10th ncert text book )
Qno12 EX 10.2

Answer 1

$\bold\mathcal{SOLUTION:}$

In ΔABC,

Length of two tangents drawn from the same point to the circle are equal,

∴ CF = CD = 6cm

∴ BE = BD = 8cm

∴ AE = AF = x

We observed that,

AB = AE + EB = x + 8

BC = BD + DC = 8 + 6 = 14

CA = CF + FA = 6 + x

Now semi perimeter of triangle (s) is,

⇒ 2s = AB + BC + CA

= x + 8 + 14 + 6 + x

= 28 + 2x

⇒s = 14 + x

Area of ΔABC = √ s (s - a)(s - b)(s - c)

= √ (14 + x) (14 + x - 14)(14 + x - x - 6)(14 + x - x - 8)

= √ (14 + x) (x)(8)(6)

= √ (14 + x) 48 x ... (i)

also, Area of ΔABC = 2×area of (ΔAOF + ΔCOD + ΔDOB)

= 2×[(1/2×OF×AF) + (1/2×CD×OD) + (1/2×DB×OD)]

= 2×1/2 (4x + 24 + 32) = 56 + 4x ... (ii)

Equating equation (i) and (ii) we get....

√ (14 + x) 48 x = 56 + 4x

Squaring both sides...

48x (14 + x) = (56 + 4x)2

⇒ 48x = [4(14 + x)]2/(14 + x)

⇒ 48x = 16 (14 + x)

⇒ 48x = 224 + 16x

⇒ 32x = 224

⇒ x = 7 cm

Hence, AB = x + 8 = 7 + 8 = 15 cm

CA = 6 + x = 6 + 7 = 13 cm

$\boxed{CA = 13\;cm}$

$\mathbb{THANK\;\; YOU}$