Category: Mensuration

Question Number 193199 by mnjuly1970 last updated on 07/Jun/23 $$ \\ $$$$\:\:\:{a}_{\mathrm{1}} \:,\:{a}_{\mathrm{2}} \:,…,{a}_{{n}} \:{are}\:\:{mutually}\:{distinct} \\ $$$$\:\:{and}\:{is}\:{a}\:\:\:\:{am}\:\:{sequence}\:. \\ $$$$\:\:\:{if}\:{a}_{\:\mathrm{1}} \:+{a}_{\:\mathrm{2}} \:+…+{a}_{{n}} \:={A} \\ $$$$\:\:\:{and}\:\: \\…

Question Number 193149 by Mingma last updated on 04/Jun/23 Answered by ajfour last updated on 05/Jun/23 $$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{p}^{\mathrm{2}} ={A} \\ $$$${p}={k}\sqrt{{A}}\:\:\:\:{where}\:\:\:{k}^{\mathrm{2}} =\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$${q}={k}\sqrt{{B}} \\ $$$${B}=\mathrm{9}{A}…

Question Number 193161 by Mingma last updated on 05/Jun/23 Answered by a.lgnaoui last updated on 05/Jun/23 $$\mathrm{Triangle}\:\mathrm{ABC}\:\:\mathrm{equilaterale}\: \\ $$$$\boldsymbol{\mathrm{MN}}=\boldsymbol{\mathrm{NP}}=\boldsymbol{\mathrm{MP}} \\ $$$$\measuredangle\mathrm{BAH}=\mathrm{60}=\measuredangle\mathrm{DAH} \\ $$$$\Rightarrow\:\measuredangle\mathrm{ADH}=\mathrm{30}°\:\:\:\measuredangle\mathrm{AMN}=\mathrm{90}−\mathrm{30}=\mathrm{60} \\ $$$$\mathrm{MNP}\:\:\:\mathrm{Triangle}\:\mathrm{equilaterale}…

Question Number 64778 by aliesam last updated on 21/Jul/19 Commented by som(math1967) last updated on 21/Jul/19 $${join}\:{O},{Q}\:\:{O},{N}\:{O},{M}\:\:{M},{Q}\:{N},{Q} \\ $$$${now}\:{ON}={NQ}={OQ}={OM}={MQ} \\ $$$$\therefore\angle{NOM}=\mathrm{60}+\mathrm{60}=\mathrm{120}° \\ $$$$\angle{MLN}=\frac{\mathrm{1}}{\mathrm{2}}\angle{NOM}=\mathrm{60}° \\ $$$$\angle{NKM}=\angle{NOM}=\mathrm{120}°\:\:\left[{subtend}\:{on}\:{same}\:{segment}\right]…

Question Number 193069 by Mingma last updated on 03/Jun/23 Answered by ajfour last updated on 03/Jun/23 Commented by ajfour last updated on 03/Jun/23 $${a}=\mathrm{17cos}\:\theta \\…