Question

ABCD is a trapezoid. PQRS and MLKJ are two rhombus
Diagonal of PQRS are 6 cm and 8 cm. One of the angle of
MLKJ is 120 degree and the diagonal bisecting that angle
measures 15cm. Side of PQRS - AB, side of MLKJ - CD
Find XY (median of trapezoid)​

Answer 1

Answer:

10 cm

Step-by-step explanation:

For rhombus ABCd

Side of ABCD = 12 62+82−−−−−−√ = 12 36+64−−−−−−√

= 12 100−−−√ = 102 = 5cm

For rhombus PQRS,

Here, PQR = 120°

PQO = 60°

POQ = 90°

(diagnoals bisect each other at 90°)

OPQ = 180° - (60° + 90°) = 30°

SQ = 15cm

OQ = 152 = 7.5cm

(diagonals of rhombus bisect each other)

sin 30° = OQpQ

12 = 7.5PQ => PQ = 15 cm

Side of rhombus PQRS = 15 cm

Now, ML = 5cm, JK = 15 cm

So median = 5+152 = 10 cm