What is the value of x according to the given lengths of angle bisectors?

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1 Answer
Feb 7, 2018

(a) Perimeter of the triangle ABC

P = (a + b + c) = 80.68 + 72 + 76.34 color(blue)(= 229.02)

Explanation:

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(a)
b_1 / b_2 = (a_1 + a_2) / (c_1 + c_2) Eqn (1)

a_1 / a_2 = (c_1 + c_2) / (b_1 + b_2) Eqn (2)

c_1 / c_2 = (b_1 + b_2) / (a_1 + a_2) Eqn (3)

Multiplying Eqn (1) by (3),

(b_1/ b_2) * (c_1/c_2) = (cancel(a_1 + a_2) / (c_1 + c_2)) / ( (b_1 + b_2) / cancel(a_1 + a_2))

(c_2 * b_1) / (c_1 * b_2) = (b_1 + b_2) / (c_1 + c_2)

(c_2 * b_1) * (c_1 + c_2) = (b_1 + b_2) * (c_1 * b_2)

c_2^2 b_1 + c_2 c_1 B-1 = b_1 b_2 c_1 + b_2^2 c_1

Substituting values of b_1, b_2, c_1 in the above equation,

32c_2^2 + 32*36*c_2 = 32*40*36 + 40^2 * 36

cancel(32)^1c_2^2 + cancel(1152)^36 c_2 = cancel(46080)^1280 + cancel(57600)^1800

c_2^2 + 36c_2 - 3080 = 0

c_2 = (-36 +- sqrt(36^2 + 4*3080))/2 = (-36 + 116..69)/2 = 40.34 rounded to two decimals and leaving the negative value.

Considering Equation (3),

c_1 / c_2 = (b_1 + b_2) / (a_1 + a_2)

36 / 40.34 = (40 + 32) / a where a = (a_1 + a_2)

a = (72 * 40.34) / 36 = 80.68

Considering Eqn (2),

(a - a_2) / a_2 = (c_1 + c_2) / (b_1 + b_2) = 76.34 / 72

#80.68 - a_2 = (76.34/72) * a_2

148.34a_2 = 80.68 * 72

a_2 = (80.68 * 72) / 148.34 = 39.16

a_1 = 80.68 - a_2 = 80.68 - 39.16 = 41.52

Perimeter of the triangle ABC

P = (a + b + c) = 80.68 + 72 + 76.34 color(blue)(= 229.02)

Similarly, we can find P for Question (b)