Please Solve this, it would be extremely helpful for me.

In the figure given below,

AB ll EF ll CD. If AB = 22.5 cm,

EP = 7.5 cm, PC =15 cm and

DC = 27 cm. Calculate:

(i) EF (ii) AC

Step-by-step explanation:

1) ΔCPD & ΔEPF

∠CPD = ∠EPF   { Vertically opposite angles}

∠CDP = ∠PFE {CD║EF, FD is transversal, Alternate interior angles are equal}

ΔCPD ≈ΔEPF  {AA criteria for similarity }

Cross multiply

EF * 15 = 27 * 7.5

EF = 27 * 0.5

EF = 13.5 cm

ii) EF // AB, so Triangles ACB & ECF are similar triangles

AC = 37.5 cm

1 is c

2 is a

3 is a

step-by-step explanation:

step-by-step explanation:

given that ab is divided into equal parts as

ac, cd, de, ef, fg, gh, hi, ib.

from this we find that f is the midpoint of ab.

so using mid point formula we find

f = (3,7)

similarly considering fb, we get

h = mid point of fb = ((6,6) and g, mid point of fh is

g = (9/2, 13/2)

i is mid point of hb

so i =(15/2,11/2)