Question Number 208468 by Tawa11 last updated on 16/Jun/24
Answered by A5T last updated on 17/Jun/24
![((OD×OC)/(AO×OB))=(1/k^2 )=(9/(25))⇒k=(5/3) ((AB)/(DC))=(5/3); AB=5x,DC=3x [ABCD]=(((8x)h)/2)=4xh [OBC]=[OAD]=((5xh)/2)−25=((3xh)/2)−9⇒xh=16 ⇒[ABCD]=4×16=64cm^2](https://www.tinkutara.com/question/Q208479.png)
$$\frac{{OD}×{OC}}{{AO}×{OB}}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{25}}\Rightarrow{k}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{{AB}}{{DC}}=\frac{\mathrm{5}}{\mathrm{3}};\:{AB}=\mathrm{5}{x},{DC}=\mathrm{3}{x} \\ $$$$\left[{ABCD}\right]=\frac{\left(\mathrm{8}{x}\right){h}}{\mathrm{2}}=\mathrm{4}{xh} \\ $$$$\left[{OBC}\right]=\left[{OAD}\right]=\frac{\mathrm{5}{xh}}{\mathrm{2}}−\mathrm{25}=\frac{\mathrm{3}{xh}}{\mathrm{2}}−\mathrm{9}\Rightarrow{xh}=\mathrm{16} \\ $$$$\Rightarrow\left[{ABCD}\right]=\mathrm{4}×\mathrm{16}=\mathrm{64}{cm}^{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by som(math1967) last updated on 17/Jun/24
$${Ar}\bigtriangleup{ABD}={Ar}\bigtriangleup{ABC} \\ $$$$\Rightarrow\bigtriangleup{AOD}+\bigtriangleup{AOB}=\bigtriangleup{BOC}+\bigtriangleup{AOB} \\ $$$$\therefore\bigtriangleup{AOD}=\bigtriangleup{BOC}={x}\:\left({let}\right) \\ $$$$\frac{\bigtriangleup{AOD}}{\bigtriangleup{AOB}}=\frac{{OD}}{{OB}} \\ $$$$\Rightarrow\frac{{x}}{\mathrm{25}}=\frac{{OD}}{{OB}} \\ $$$${again}\:\frac{\bigtriangleup{DOC}}{\bigtriangleup{BOC}}=\frac{{OD}}{{OB}} \\ $$$$\Rightarrow\frac{\mathrm{9}}{{x}}=\frac{{OD}}{{OB}} \\ $$$$\therefore\frac{{x}}{\mathrm{25}}=\frac{\mathrm{9}}{{x}}\:\Rightarrow{x}=\sqrt{\mathrm{9}×\mathrm{25}}=\mathrm{15}{cm}^{\mathrm{2}} \\ $$$${Area}\:{of}\:{ABCD}=\mathrm{9}+\mathrm{15}+\mathrm{25}+\mathrm{15} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{64}{cm}^{\mathrm{2}} \\ $$
Commented by Tawa11 last updated on 17/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$