Question

,OABC is a rhombus whose three vertices A,B,C lie on a circle of radius 10 cm and centre O. Find the area of the rhombus . [ Take √3 = 1.732]

Answer 1

Clearly , OA = OB = OC = 10 cm . Let OB and AC intersect at P.


Since the diagonals of a rhombus bisect each other at right angles , we have OP = 5 cm and /_ OPC = 90°.



Now , AC = 2CP and

CP = √(OC)² - ( OP)²



CP = ✓(10)² - (5)²



CP = √75

CP = 5√3 cm.


Therefore,


AC = ( 2 × 5√3 ) cm = 10√3 cm.


=> ( 10 × 1.732) cm


=> 17.31 cm.




Therefore,


Ar(Rhombus OABC) = 1/2 × OB × AC



=> (1/2 × 10 × 17.32 ) cm².


=> 86.6 cm².

Answer 2

Hey mate ^_^

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Answer:
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Given OABC is a rhombus whose three vertices A, B, C lies on circle with centre O and radius 10 cm.

Suppose the diagonals of the Rhombus OABC intersects at S. 

Radius of circle (r) = 10 cm

∴ OA = OB = OC = 10 cm

Diagonals of rhombus bisect each other at 90o

∴ OS = SB = OB / 2 = 10/2 = 5 cm

And,

SC = SA 

In a right angle triangle OCS,

OC^2 = OS^2 + SC^2

⇒ (10)^2 = 5^2 + SC^2

⇒ SC^2 = 100 – 25

⇒ SC^2 = 75 

⇒ SC = 5√3 cm 

∴ AC = 2 × SC = 2 × 5√3 = 10√3 cm 

Area of Rhombus = ½ × d1 × d2 

= ½ × 10 × 10√3 

= 50√3 cm^2 

∴ Area of Rhombus = 50√3 cm^2

#Be Brainly❤️