Question

l, m and n are three equidistant parallel lines. AD, PQ and GH are three transversal, If BC = 2 cm and LM = 2.5 cm and AD || PQ, find MS and MN

Answer 1

Answer:

Step-by-step explanation:

Let l, m and n be three parallel lines intersected by two transversals p and q such that l, m and n cut off equal intercepts AB and BC on p i.e. AB = BC.

To show:  l, m and n cut off equal intercepts DE and EF on q also, i.e. DE = EF.

Construction:  Join AF intersecting m at G.

So, the trapezium ACFD is divided into two triangles: ΔACF and ΔAFD.

It is given that AB = BC

⇒ B is the mid point of AC

Now in ΔACF, B is the mid point of AC and BG || CF (as m || n )

∴ By mid point theorem, G is the mid point of AF.

Now in ΔAFD, G is the mid point of AF and GE || AD (as l || m)

⇒ E is the mid point of DF (by mid point theorem)

⇒ DE = EF

Answer 2

Answer:

l ,m and n are three equidistant parallel lines. Line AD,PQ and GH are three transversals.

                BC = RM = 2cm opposite sides of a║gm BCMR>

                BC = CE and RM = MS = 2 cm.

                                                                (By equal intercept theorem)

Then,          LM = MN =  2.5 cm

                                                         (By equal intercept theorem)

Hence,         MS = 2cm

                    MN = 2.5cm