plz sol question number 12
Answers
Let the ΔABC be drawn to circumscribe a circle with centre O and radius 4 cm.
Since lengths of two tangents drawn from an external point to circle are equal. Therefore, BF = BD = 8 cm, CE = CD = 6 cm and AF = AE = x cm.
Then the sides of the triangle are x, (x + 6) and (x + 8) cm.
∴ 2s = 14 + (x + 6) + (x + 8)
=> s = 14 + x.
s - a = 14 + x - 14 = x
s - b = 14 + x - x - 6 = 8
s - c = 14 + x - x - 8 = 6
∴ Area of ΔABC = √s(s - a)(s - b)(s - c)
= √(14 + x)(x)(8)(6)
= √48(x^2 + 14x)
∴ Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB
√48(x^2 + 14x) = (1/2) * BC * OD + (1/2) * CA * OE + (1/2) * AB * OF
√48(x^2 + 14x) = (1/2) * 14 * 4 + (1/2) * (x + 6) * 4 + (1/2) * (x + 8) * 4
√48(x^2 + 14x) = 2(28 + 2x)
√48(x^2 + 14x) = 4(14 + x)
Squaring on both sides, we get
48(x^2 + 14x) = 16(4 + x)^2
3(x^2 + 14x) = 196 + 28x + x^2
x = 7 or x = -14
But x cannot be negative
x = 7
Hence, the sides AB and AC are 15 cm and 13 cm respectively.
Hope it help!
