Question Number 223301 by cemoosky last updated on 21/Jul/25
Answered by som(math1967) last updated on 21/Jul/25
![Ar of PQRK=3×△PSK area of△PSK=(1/2)×15sin 30×15cos 30 =(1/2)×((225(√3))/4)cm^2 area of PQRK=3×(1/2)×((225(√3))/4) =((675(√3))/8)cm^2 [If PQRS is rectangle]](https://www.tinkutara.com/question/Q223302.png)
$$\:\:\:{Ar}\:{of}\:{PQRK}=\mathrm{3}×\bigtriangleup{PSK} \\ $$$$\:{area}\:{of}\bigtriangleup{PSK}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{15sin}\:\mathrm{30}×\mathrm{15cos}\:\mathrm{30} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{225}\sqrt{\mathrm{3}}}{\mathrm{4}}{cm}^{\mathrm{2}} \\ $$$$\:{area}\:{of}\:{PQRK}=\mathrm{3}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{225}\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$=\frac{\mathrm{675}\sqrt{\mathrm{3}}}{\mathrm{8}}{cm}^{\mathrm{2}} \:\left[{If}\:{PQRS}\:{is}\:{rectangle}\right] \\ $$$$\:\: \\ $$
Answered by fantastic last updated on 21/Jul/25
$${PS}=\mathrm{15sin}\:\mathrm{30}^{\mathrm{0}} =\mathrm{15}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${SK}=\mathrm{15cos}\:\mathrm{30}^{\mathrm{0}} =\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${Trisngle}\:{area}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{15}}{\mathrm{2}}=\frac{\mathrm{225}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$${rectangle}\:{area}=\frac{\mathrm{15}}{\mathrm{2}}×\frac{\mathrm{15}\sqrt{\mathrm{3}}}{\mathrm{2}}×\mathrm{2}=\frac{\mathrm{225}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${shaded}\:{area}=\mathrm{225}\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\right)=\frac{\mathrm{675}\sqrt{\mathrm{3}}}{\mathrm{8}}{sq}.\:{units}\checkmark\checkmark \\ $$