Question

in triangle abc points d and e lies on ac and ab such that bc parallel to de ad=x cm bd=(x-2)cm ae=(x+2)cm and ec=(x-1)cm find ac​

Answer 1

Given, DE || BC

:. AD/BD=AE/EC

(Basic Proportionality Theorem)

:. x/x-2 = x+2/x-1

By cross multiplying, we get

x(x-1) =(x-2)(x+2)

x^2-x =x^2-2^2

x^2-x =x^2 - 4

x^2-x^2-x= -4

-x = -4

:. x = 4

AC = AE+EC

AC = (x+2)+(x-1)

AC = x+2+x-1

AC = 4+2+4-1 (since, x=4)

:. AC = 9cm

..... Here's your answer.....