ABCD is a trapezoid. PQRS and MLKJ are two rhombusDiagonal of PQRS are

1.	ABCD is a trapezoid. PQRS and MLKJ are two rhombusDiagonal of PQRS are 6 cm and 8 cm. One of the angle ofMLKJ is 120 degree and the diagonal bisecting that anglemeasures 15cm. Side of PQRS - AB, side of MLKJ - CDFind XY (median of trapezoid)
Answer» 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

ABCD is a trapezoid. PQRS and MLKJ are two rhombusDiagonal of PQRS are 6 cm and 8 cm. One of the angle ofMLKJ is 120 degree and the diagonal bisecting that anglemeasures 15cm. Side of PQRS - AB, side of MLKJ - CDFind XY (median of trapezoid)

Answer»

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

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ABCD is a trapezoid. PQRS and MLKJ are two rhombusDiagonal of PQRS are 6 cm and 8 cm. One of the angle ofMLKJ is 120 degree and the diagonal bisecting that anglemeasures 15cm. Side of PQRS - AB, side of MLKJ - CDFind XY (median of trapezoid)​

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ABCD is a trapezoid. PQRS and MLKJ are two rhombusDiagonal of PQRS are 6 cm and 8 cm. One of the angle ofMLKJ is 120 degree and the diagonal bisecting that anglemeasures 15cm. Side of PQRS - AB, side of MLKJ - CDFind XY (median of trapezoid)