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Category: Geometry

Question-62542

Question Number 62542 by mr W last updated on 22/Jun/19 Commented by mr W last updated on 22/Jun/19 $${the}\:{distances}\:{of}\:{a}\:{point}\:{M}\:{to}\:{the} \\ $$$${vertexes}\:{of}\:{a}\:{triangle}\:{are}\:{p},{q},{r}. \\ $$$${find}\:{the}\:{side}\:{lengthes}\:{and}\:{hence}\:{the} \\ $$$${perimeter}\:{of}\:{the}\:{triangle}\:{with}\:{the}…

Question-62456

Question Number 62456 by ajfour last updated on 21/Jun/19 Commented by ajfour last updated on 21/Jun/19 $${If}\:{the}\:{regular}\:{tetrahedron}\:{with} \\ $$$${edge}\:{length}\:{unity}\:{inscribes}\:{a} \\ $$$${hemisphere}\:{such}\:{that}\:{three}\:{faces} \\ $$$${are}\:{tangent}\:{to}\:{the}\:{spherical} \\ $$$${surface}\:{while}\:{fourth}\:{serves}\:{to}…

Question-62448

Question Number 62448 by mr W last updated on 21/Jun/19 Commented by mr W last updated on 21/Jun/19 $${The}\:{distances}\:{of}\:{a}\:{point}\:{M}\:{to}\:{the} \\ $$$${vertexes}\:{of}\:{a}\:{triangle}\:{are}\:{p},{q},{r}. \\ $$$$\left({assume}\:{p}\geqslant{q}\geqslant{r}\right) \\ $$$${Find}\:{the}\:{side}\:{lengthes}\:{and}\:{thus}\:{the}…

Prove-that-if-the-lengths-of-a-triangle-form-an-arithmetic-progression-then-the-centre-of-incircle-and-the-centroid-of-triangle-lie-on-a-line-parallel-to-the-side-of-middle-length-of-the-triangle-

Question Number 62380 by ajfour last updated on 20/Jun/19 $${Prove}\:{that}\:{if}\:{the}\:{lengths}\:{of}\:{a}\: \\ $$$${triangle}\:{form}\:{an}\:{arithmetic} \\ $$$${progression},\:{then}\:{the}\:{centre}\:{of} \\ $$$${incircle}\:{and}\:{the}\:{centroid}\:{of} \\ $$$${triangle}\:{lie}\:{on}\:{a}\:{line}\:{parallel}\:{to} \\ $$$${the}\:{side}\:{of}\:{middle}\:{length}\:{of}\:{the} \\ $$$${triangle}. \\ $$ Answered…

Question-62363

Question Number 62363 by Tawa1 last updated on 20/Jun/19 Answered by MJS last updated on 20/Jun/19 $$\mathrm{big}\:\mathrm{circle}\:\left(\mathrm{let}\:{R}=\mathrm{1}\right) \\ $$$${x}=\pm\sqrt{\mathrm{1}−{y}^{\mathrm{2}} } \\ $$$$\mathrm{lines}\:\mathrm{of}\:\mathrm{tbe}\:\mathrm{triangle} \\ $$$${x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{y} \\…