Category: Geometry

Question Number 63981 by ajfour last updated on 11/Jul/19 Answered by mr W last updated on 11/Jul/19 $$\left[\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\sqrt{\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{r}\right)^{\mathrm{2}} −\left({b}−{r}\right)^{\mathrm{2}}…

Question Number 63750 by Tawa1 last updated on 08/Jul/19 Commented by Prithwish sen last updated on 08/Jul/19 $$\mathrm{perimeter}\mathrm{of}\mathrm{circle}\mathrm{A}=\frac{2}{3}\pi \mathrm{r}\mathrm{unit}$$$$\mathrm{perimeter}\mathrm{of}\mathrm{circle}\mathrm{B}=2\pi \mathrm{r}\mathrm{unit}$$$$\mathrm{Circle}\mathrm{A}$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{covers}\:\frac{\mathrm{2}}{\mathrm{3}}\pi\mathrm{r}\:\mathrm{in}\:\mathrm{1}\:\mathrm{time} \…

Question Number 63663 by Tawa1 last updated on 07/Jul/19 Answered by mr W last updated on 07/Jul/19 $$\mathrm{\angle }ADB=360-(180-6-30)-(180-6-24)=66°$$$$\frac{AD}{DC}=\frac{\sin 30°}{\sin 6°}$$$$\frac{BD}{DC}=\frac{\sin 24°}{\sin 6°}$$$$\Rightarrow\frac{{AD}}{{BD}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{24}°} \…

Question Number 129105 by I want to learn more last updated on 12/Jan/21 $${\int }_0^{\frac{\pi }{4}}\sqrt{\tan \mathrm{x}(1-\tan \mathrm{x})}\mathrm{dx}$$ Answered by Lordose last updated on 13/Jan/21…

Question Number 129096 by Gulnoza last updated on 12/Jan/21 Answered by talminator2856791 last updated on 15/Jan/21 $$2\left(\frac{300-75\pi }{4}\right)$$ Commented by Gulnoza last updated on…