Question

Find x.AC=10√3.BD=5​

Answer 1

Answer:

In ∆ ADC

AD^2 = AC^2 - DC^2

AD^2 = (10√3)^2 - x^2

AD^2 = 300 - x^2..............(1)

now, in ∆ABC

AB^2 = BC^2 - AC^2

(5 + x)^2 - (10√3)^2

AB^2 = (5+x)^2 - 300..........(2)

again in ∆ABD

AD^2 = AB^2 - BD^2

AD^2 = (5+x)^2 - 300 + 5^2

AD^2 = (5+x)^2 -300 + 25...........(3)

from equation (1) and (3)

(5+x)^2 - 300 - 25 = 300 - x^2

25 + x^2 + 10x - 325 = 300 - x^2

2x^2 + 10x - 600 = 0

x^2 + 5x - 300 = 0

x^2 + 20x - 15x - 300 = 0

x(x + 20) - 15(x + 20) = 0

(x + 20) (x - 15) = 0

x = 15 or -20

x= 15 (ignoring -ve value)