Question

All the vertices of a rhombus lie on a circle .Find the area of the rhombus if the area of of a circle is 1256cm sqare

Answer 1

$\large\bf\underline{Given:-}$

• All the vertices of a rhombus lie on a circle
• Area of circle = 1256cm²

$\large\bf\underline {To \: find:-}$

• Area of rhombus.

$\huge\bf\underline{Solution:-}$

we know that,

$\dag \large\boxed{\bf\:Area\:of\:circle= \rm \: \pi {r}^{2}}$

Given area of circle = 1256 cm²

$\dashrightarrow \rm \: \pi {r}^{2} = 1256 \\ \\ \dashrightarrow \rm \: {r}^{2} =1256 \times \frac{1}{ \pi} \\ \\\dashrightarrow \rm \: {r}^{2} = \frac{1256}{3.14} \\ \\ \dashrightarrow \rm \: {r}^{2} = 400 \\ \\ \dashrightarrow \rm \: r = \sqrt{400} \\ \\ \dashrightarrow \rm \:r = 20 \: cm$

So , we get the radius of the circle = 20 cm

Diameter of circle = 2 × Radius

Diameter of circle = 2 × 20

≫ Diameter of circle = 40 cm

So, it is given in the Question that all the vertices of rhombus lie on circle. So Diameter of circle is the diagonal of rhombus and diagonal of rhombus are equal.

So,

length of Diagonal of rhombus = 40cm.

Now,

$\large \dag \boxed{ \bf \: area \: of \: rhombus = \frac{1}{2} \times d _ 1 \times \: d _ 2 }$

Area of rhombus = 1/2 × 40 × 40

Area of rhombus = 20 × 40

Area of rhombus = 800cm²

hence,

»★ Area of rhombus = 800 cm²

Answer 2

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ area \ of \ rhombus \ is \ 800 \ cm^{2}}$

$\sf\orange{Given:}$

$\sf{\implies{All \ vertices \ of \ rhombus \ lie \ on \ a \ circle}}$

$\sf{\implies{Area \ of \ circle=1256 \ cm^{2}}}$

$\sf\pink{To \ find:}$

$\sf{Area \ of \ the \ rhombus. }$

$\sf\green{\underline{\underline{Solution:}}}$

$\boxed{\sf{Area \ of \ circle=\pi\times \ r^{2}}}$

$\sf{\therefore{1256=\frac{22}{7}\times \ r^{2}}}$

$\sf{\therefore{r^{2}=\frac{1256\times7}{22}}}$

$\sf{\therefore{r^{2}=400(approx)}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides.}$

$\sf{\implies{r=20 \ cm}}$

$\sf{\implies{\therefore{Diameter=2\times20=40 \ cm}}}$

$\sf{According \ to \ given \ condition}$

$\sf{Quadrilateral \ is \ rhombus}$

$\sf{Also \ it's \ cyclic \ quadrilateral, \ so \ opposite}$

$\sf{angles \ will \ be \ supplementary. }$

$\sf{\therefore{The \ rhombus \ is \ square.}}$

$\sf{Here, \ Diagonal \ of \ square=Diameter}$

$\sf{\therefore{Diagonal=40 \ cm}}$

$\boxed{\sf{Diagonal \ of \ square=Side\times\sqrt2}}$

$\sf{\therefore{Side \ of \ square=\frac{40}{\sqrt2}}}$

$\sf{\therefore{Side \ of \ square=20\sqrt2 \ cm}}$

$\boxed{\sf{Area \ of \ square=Side^{2}}}$

$\sf{\therefore{Area \ of \ square=(20\sqrt2)^{2}}}$

$\sf{\therefore{Area \ of \ square=800 \ cm^{2}}}$

$\sf{But, \ here \ rhombus \ is \ square. }$

$\sf{\therefore{Area \ of \ rhombus=800 \ cm^{2}}}$

$\sf\purple{\tt{\therefore{The \ area \ of \ rhombus \ is \ 800 \ cm^{2}}}}$