Question

OA BC is a rhombus whose 3 vertices A, Band C lie on a circle with centre O. if the area of the rhombus is 32√3cm square. find the radius of the circle
please tell fast​

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow OABC\:is\:a\:rhombus$

$\sf\dashrightarrow vertices\:of\:rhombus\:A,B\:and\:C\:lies\:on\:a\:circle\:with\:centre\:"O".$

$\sf\dashrightarrow area\:of\:rhombus\:is\:32 \sqrt{3}$

$\large\underline\bold{TO\:FIND,}$

$\sf\bold\star radius\:of\:the\:circle$

ACCORDING TO THE QUESTION,

$\sf\dashrightarrow area\:of\:rhombus= \dfrac{1}{2} \times d_1 \times d_2$

$\sf\large\star looking\:at\:the\:given\:figure,$

$\sf\star we\:don't\:know\:the\:value \:of\:any\:one\:diagonal.so,$

$\sf\dashrightarrow two\: equilateral\:triangles\:are\:formed\:in\:a\:rhombus\:by\:joining \:the\:diagonals\:of\:the\:rhombus$

$\sf\dashrightarrow \:radius\:of\:circle\:can\:be\:also\:find$

$\sf by\:using\:area\:of\:two\: equilateral\:triangles\:formed\:in\:the\:rhombus$

we KNOW,

✯AREA OF RHOMBUS = 32√3.........EQUATION¹

LOOKING AT THE GIVEN FIGURE,

"THE DIAGONAL OF RHOMBUS MAKES TWO EQUILATERAL TRIANGLES"

$\sf\large\dashrightarrow \triangle ABO and \triangle CBO$

THEREFORE,

BY GIVEN EQUATION AND STATEMENT,

WE GET,

AREA OF RHOMBUS= 2× (AREA OF EQUILATERAL TRIANGLES).........EQUATION²

$\sf\therefore area\:of\: equilateral\:triangle= \dfrac{\sqrt{3}}{4} \times a^2$

$\sf\therefore a=r$

$\sf\therefore area\:of\:rhombus=2 \times \dfrac{\sqrt{3}}{4} \times r^2$

NOTE:-

✯."DIAGONAL OF RHOMBUS IS EQUAL TO RADIUS OF THE CIRCLE".

✯FORMULA IN USE,

$\rm{\boxed{\bf{\star \:\: 2 \times \dfrac{\sqrt{3}}{4} \times r^2\:\:\star }}}$

$\large\underline\bold{SOLUTION,}$

WE KNOW,

AREA OF RHOMBUS = $\sf\dfrac{1}{2} \times d_1 \times d_2$

✯.FROM EQUATION (1)AND (2) WE GET,

area of rhombus = 32√3cm²

$\sf\implies 2 \times \dfrac{\sqrt{3}}{4} \times r^2=32\sqrt{3}$

$\sf\implies 2 \times \dfrac{\sqrt{3}}{4} \times r^2=32\sqrt{3}$

$\sf\implies \dfrac{\sqrt{3}}{2} \times r^2=32\sqrt{3}$

$\sf\implies \sqrt{3} \times r^2=2 \times 32\sqrt{3}$

$\sf\implies \sqrt{3} \times r^2=64 \sqrt{3}$

$\sf\implies \dfrac{\sqrt{3}}{\sqrt{3}} \times r^2=64$

$\sf\implies \dfrac{\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}} \times r^2=64$

$\sf\implies r^2=64$

$\sf\implies r= \sqrt{64}$

$\sf\implies r=8$

$\large{\boxed{\bf{ r=8cm}}}$

$\large\underline\bold{RADIUS\:OF\:THE\:CIRCLE\:IS\:8CM,}$

____________________________

Answer 2

$\huge\underline{\underline{\bold{\green{AnSwEr}}}}$

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow OABC\:is\:a\:rhombus$

$\sf\dashrightarrow vertices\:of\:rhombus\:A,B\:and\:C\:lies\:on\:a\:circle\:with\:centre\:"O".$

$\sf\dashrightarrow area\:of\:rhombus\:is\:32 \sqrt{3}$

$\large\underline\bold{TO\:FIND,}$

$\sf\bold\star radius\:of\:the\:circle$

ACCORDING TO THE QUESTION,

$\sf\dashrightarrow area\:of\:rhombus= \dfrac{1}{2} \times d_1 \times d_2$

$\sf\large\star looking\:at\:the\:given\:figure,$

$\sf\star we\:don't\:know\:the\:value \:of\:any\:one\:diagonal.so,$

$\sf\dashrightarrow two\: equilateral\:triangles\:are\:formed\:in\:a\:rhombus\:by\:joining \:the\:diagonals\:of\:the\:rhombus$

$\sf\dashrightarrow \:radius\:of\:circle\:can\:be\:also\:find$

$\sf by\:using\:area\:of\:two\: equilateral\:triangles\:formed\:in\:the\:rhombus$

we KNOW,

✯AREA OF RHOMBUS = 32√3.........EQUATION¹

LOOKING AT THE GIVEN FIGURE,

"THE DIAGONAL OF RHOMBUS MAKES TWO EQUILATERAL TRIANGLES"

$\sf\large\dashrightarrow \triangle ABO and \triangle CBO$

THEREFORE,

THEREFORE,BY GIVEN EQUATION AND STATEMENT,

THEREFORE,BY GIVEN EQUATION AND STATEMENT,WE GET,

AREA OF RHOMBUS= 2× (AREA OF EQUILATERAL TRIANGLES).........EQUATION²

$\sf\therefore area\:of\: equilateral\:triangle= \dfrac{\sqrt{3}}{4} \times a^2$

$\sf\therefore a=r$

$\sf\therefore area\:of\:rhombus=2 \times \dfrac{\sqrt{3}}{4} \times r^2$

NOTE:-

NOTE:-✯."DIAGONAL OF RHOMBUS IS EQUAL TO RADIUS OF THE CIRCLE".

NOTE:-✯."DIAGONAL OF RHOMBUS IS EQUAL TO RADIUS OF THE CIRCLE". ✯FORMULA IN USE,

$\rm{\boxed{\bf{\star \:\: 2 \times \dfrac{\sqrt{3}}{4} \times r^2\:\:\star }}}$

$\large\underline\bold{SOLUTION,}$

WE KNOW,

AREA OF RHOMBUS = $\sf\dfrac{1}{2} \times d_1 \times d_2$

✯.FROM EQUATION (1)AND (2) WE GET,

area of rhombus = 32√3cm²

$\sf\implies 2 \times \dfrac{\sqrt{3}}{4} \times r^2=32\sqrt{3}$

$\sf\implies 2 \times \dfrac{\sqrt{3}}{4} \times r^2=32\sqrt{3}$

$\sf\implies \dfrac{\sqrt{3}}{2} \times r^2=32\sqrt{3}$

$\sf\implies \sqrt{3} \times r^2=2 \times 32\sqrt{3}$

$\sf\implies \sqrt{3} \times r^2=64 \sqrt{3}$

$\sf\implies \dfrac{\sqrt{3}}{\sqrt{3}} \times r^2=64$

$\sf\implies \dfrac{\cancel{\sqrt{3}}}{\cancel{\sqrt{3}}} \times r^2=64$

$\sf\implies r^2=64$

$\sf\implies r= \sqrt{64}$

$\sf\implies r=8$

$\large{\boxed{\bf{ r=8cm}}}$

$\large\underline\bold{RADIUS\:OF\:THE\:CIRCLE\:IS\:8CM,}$