Question

please solve question

Class 10

Circles

Answer 1

Hey there !!


▶ Given :-

→ Radius of a circle with centre O = 10 cm .

→ OABC is a rhombus .


▶ To find :-

→ Area of the rhombus .


▶ Solution :-

AO = OC = OB = 10 cm [ As radius of the circle].

And,

OA = AB = BC = CO = 10 cm [ As sides of rhombus are equal ] .

Since, we get 2 equilateral triangle of sides 10 cm each .


Therefore, Area of rhombus = 2 × Area of equilateral triangle .

$= 2 \times \frac{ \sqrt{3} {a}^{2} }{4} . \\ \\ = 2 \times \frac{ \sqrt{3} \times {(10)}^{2} }{4} . \\ \\ = 2 \times \frac{ \sqrt{3} \times 100}{4} . \\ \\ = 2 \times \sqrt{3} \times 25. \\ \\ \boxed{ \pink{ = 50 \sqrt{3} {cm}^{2} .}} \\ \\ or \\ \\ \boxed{ \red{ = 86.60 {cm}^{2} .}}$


[ a = side ] .


✔✔ Hence, it is solved ✅✅.



THANKS


#BeBrainly.

Answer 2

hey

here is answer

given= radius of circle= 10Cm

and OABC is a rhombus.

we know diagonal of rhombus divides it into two equilateral triangles.

and sides of triangle are equal

so the side of triangles=10cm each

now area of equilateral triangle=

$\frac{ \sqrt{3} }{4} {a}^{2}$
now

there are two triangles

so area=

$2 \times \frac{ \sqrt{3} }{4} {10}^{2}$

$2 \times \frac{ \sqrt{3} }{4} \times 100$

$\not2 \times \frac{ \sqrt{3} }{ \not4 \: 2} 100$

$\frac{ \sqrt{3} }{ \not2} \times \not{100}$

$\bf{50 \sqrt{3}}$ cm^2


hence

answer is


$\huge{50{\sqrt{3}}$cm^2
hope it helps

thanks