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Category: Geometry

Question-167011

Question Number 167011 by mnjuly1970 last updated on 04/Mar/22 Answered by TheSupreme last updated on 04/Mar/22 $${Talete}\:{theorem}\:{respect}\:{point}\:{A}\:{far}\:{away}\:{x}\:{from}\:{center}\: \\ $$$${of}\:{circle}\:{of}\:{radius}\:\mathrm{4} \\ $$$${let}'{s}\:{y}={PQ} \\ $$$${x}:\mathrm{4}=\left({x}+\mathrm{4}\right):{y}=\left({x}+\mathrm{12}\right):\mathrm{8} \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{{x}+\mathrm{12}}{\mathrm{8}}\rightarrow{x}=\mathrm{12}…

new-idea-and-solution-to-questions-35178-amp-35195-triangle-ABC-a-BC-b-CA-c-AB-CAB-ABC-BCA-d-a-b-c-a-b-c-a-b-c-a-b-c-put-it-as-this-A-0-0-B-c-0-C-

Question Number 35871 by MJS last updated on 25/May/18 $$\mathrm{new}\:\mathrm{idea}\:\left(\mathrm{and}\:\mathrm{solution}\right)\:\mathrm{to}\:\mathrm{questions}\:\mathrm{35178}\:\&\:\:\mathrm{35195} \\ $$$$ \\ $$$$\mathrm{triangle}: \\ $$$${ABC};\:{a}={BC},\:{b}={CA},\:{c}={AB}; \\ $$$$\alpha=\angle{CAB},\:\beta=\angle{ABC},\:\gamma=\angle{BCA} \\ $$$${d}=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$ \\ $$$$\mathrm{put}\:\mathrm{it}\:\mathrm{as}\:\mathrm{this}: \\…

Question-35798

Question Number 35798 by ajfour last updated on 23/May/18 Commented by ajfour last updated on 23/May/18 $$\bigtriangleup{ABC}\:{and}\:\bigtriangleup{CDE}\:{are}\:{congruent}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{quadrilateral} \\ $$$${APQC}\:{in}\:{terms}\:{of}\:{a},\:{b},\:{and}\:{c}\:;\:{the} \\ $$$${sides}\:{of}\:\bigtriangleup{ABC}\:. \\ $$…

Question-166861

Question Number 166861 by Tawa11 last updated on 01/Mar/22 Commented by cortano1 last updated on 01/Mar/22 $$\Rightarrow\mathrm{3}.\mathrm{x}=\mathrm{6}.\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{4} \\ $$$$\Rightarrow\mathrm{r}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:=\sqrt{\mathrm{49}+\mathrm{16}}=\sqrt{\mathrm{65}} \\ $$…