Question Number 4437 by Rasheed Soomro last updated on 25/Jan/16
$$\mathrm{An}\:\boldsymbol{\mathrm{ellipse}}\:\mathrm{having}\:\boldsymbol{\mathrm{semi}}-\boldsymbol{\mathrm{major}}\:\boldsymbol{\mathrm{axis}}\: \\ $$$$\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\boldsymbol{\mathrm{semi}}-\boldsymbol{\mathrm{minor}}\:\boldsymbol{\mathrm{axis}}\:\boldsymbol{\mathrm{length}}\:\boldsymbol{\mathrm{b}} \\ $$$$\mathrm{and}\:\mathrm{a}\:\boldsymbol{\mathrm{circle}}\:\mathrm{having}\:\boldsymbol{\mathrm{radius}}\:\boldsymbol{\mathrm{r}}\:\mathrm{have}\:\mathrm{equal} \\ $$$$\boldsymbol{\mathrm{area}}. \\ $$$$\mathrm{Express}\:\boldsymbol{\mathrm{r}}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\boldsymbol{\mathrm{a}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}. \\ $$$$ \\ $$
Answered by Yozzii last updated on 25/Jan/16
$${Area}\:{of}\:{ellipse},\:{A}_{\mathrm{1}} =\pi{ab}. \\ $$$${Area}\:{of}\:{circle},\:{A}_{\mathrm{2}} =\pi{r}^{\mathrm{2}} . \\ $$$${A}_{\mathrm{1}} ={A}_{\mathrm{2}} \Rightarrow{r}=\sqrt{{ab}}. \\ $$