Question

The perimeter of a rhombus is 40 cm and the measure of an angle is 60°, then the area of it is-
(1) 100 √3 cm² (2) 75 √3 cm² (3) 180 √3 cm² (4) 50 √3 cm²

Answer 1

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Answer 2

Answer:

50ROOT(3)

Step-by-step explanation:

DRAW A RHOMBUS ABCD AND DIAGONALS AC AND DB INTERSECT ON POINT O AND ANGLE A = 60 DEG

IN TRI AOD AND TRI AOB

AO = AO -- COMMON

AD = AB  -- SIDES OF RHOMBUS

OD = OB -- EQUAL HALVES OF DIAGONAL BD

TRI AOD CONGRUENT TO TRI AOB

ANGLE BAO = ANGLE DAO BYU CPCT

ANGLE BAO + ANGLE DAO = ANGLE A

2ANGLE DAO = 60DEG

ANGLE DAO = 30 DEG

IN RIGHT TRI AOD(ANGLE O 90-- DIAGONALS OF RHOMBUS ARE PERPENDICULAR BISCETORS OF EACH OTHER)

SIN 30 = PERPENDICULAR/HYPOTENUSE

1/2 = OD/AD

OD = 5CM

2(OD) = 10CM

DB = 10CM

COS 30 = BASE /HYPOTENUSE

ROOT(3)/2 = AO/AD

5ROOT(3) = AO

2(AO) = AC

10ROOT(3) = AC

AREA OF RHOMBUS ABCD = 1/2 * DIAGONAL1 * DIAGONAL2

= 1/2 * 10 * 10ROOT(3)

= 50ROOT(3) CM^2