Question Number 134396 by I want to learn more last updated on 03/Mar/21 Commented by mr W last updated on 03/Mar/21 $${P}+{Q}=\mathrm{2}×\mathrm{10}×\mathrm{sin}\:\frac{\mathrm{75}°}{\mathrm{2}}=\mathrm{12}.\mathrm{2}\:{unit} \\ $$ Commented…
Question Number 3304 by prakash jain last updated on 09/Dec/15 $$\mathrm{Bring}\:\mathrm{up}\:\mathrm{the}\:\mathrm{topic}/\mathrm{challenge}\:\mathrm{started}\:\mathrm{by}\:\mathrm{Filup}\:\mathrm{at}\:\mathrm{the}\:\mathrm{top}. \\ $$$$\mathrm{Shall}\:\mathrm{we}\:\mathrm{start}\:\mathrm{new}\:\mathrm{topic}\:\mathrm{at}\:\mathrm{the}\:\mathrm{beginning} \\ $$$$\mathrm{of}\:\mathrm{calendar}\:\mathrm{month}? \\ $$$$\mathrm{See}\:\mathrm{older}\:\mathrm{post}\:\mathrm{dt}\:\mathrm{24}.\mathrm{11}\:\mathrm{by}\:\mathrm{Filup} \\ $$ Commented by Filup last updated on…
Question Number 68831 by Maclaurin Stickker last updated on 15/Sep/19 $$\mathrm{The}\:\mathrm{square}\:{ABCD}\:\mathrm{has}\:\mathrm{side}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{distance}\:{AP}\:\:\mathrm{is}\:\:\frac{\mathrm{1}}{\mathrm{8}}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:{PMN}\:\mathrm{inscribed}\:\mathrm{in}\:\mathrm{the}\:\mathrm{square}. \\ $$ Commented by Maclaurin Stickker last updated…
Question Number 3280 by Filup last updated on 09/Dec/15 $$\mathrm{For}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{perpandicular} \\ $$$$\mathrm{height}\:{h}\:\mathrm{and}\:\mathrm{base}\:\mathrm{length}\:{b},\:\mathrm{the}\: \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}: \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}{hb} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:\mathrm{this}\:\mathrm{the}\:\mathrm{case}? \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{that}\:\mathrm{two}\:\mathrm{identicle}\:\mathrm{triangles} \\ $$$$\mathrm{can}\:\mathrm{construct}\:\mathrm{a}\:{rectangle},\:\mathrm{so}\:\mathrm{the}\:\mathrm{area} \\…
Question Number 3262 by Rasheed Soomro last updated on 08/Dec/15 $$\mathcal{C}{ould}\:\:^{\mathrm{3}} \sqrt{\mathrm{2}}\:\:{be}\:{drawn}\:{on}\:{numbered}\:{line}\:{with}\: \\ $$$${the}\:{help}\:\:{of}\:\:{ruler}\:{and}\:{compass}\:{only}? \\ $$ Commented by prakash jain last updated on 09/Dec/15 $$\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{using}\:\mathrm{a}\:\mathrm{ruler}\:\mathrm{and}\:\mathrm{compass}.…
Question Number 3249 by Rasheed Soomro last updated on 08/Dec/15 $$\mathcal{H}{ow}\:{could}\:\sqrt{\mathrm{5}}\:\:{be}\:{drawn}\:{on}\:{numbered}\:{line}\:{using} \\ $$$${scale}\:{and}\:{compass}\:{only}?\:\left({Exactly}\:\sqrt{\mathrm{5}}\:{not}\:{its}\:{decimal}\:{approximation}.\right) \\ $$ Answered by prakash jain last updated on 08/Dec/15 $$\mathrm{For}\:\sqrt{\mathrm{5}} \\…
Question Number 68666 by ahmadshah last updated on 14/Sep/19 Commented by mr W last updated on 14/Sep/19 Commented by mr W last updated on 14/Sep/19…
Question Number 68664 by Maclaurin Stickker last updated on 14/Sep/19 $$\mathrm{In}\:\mathrm{a}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{whose} \\ $$$$\mathrm{side}\:\mathrm{is}\:\boldsymbol{{a}},\:\mathrm{the}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}\:\mathrm{are}\:\mathrm{taken} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{side}\:{BC},\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{triangles} \\ $$$${ABM},\:{AMN}\:\mathrm{and}\:{ANC}\:\mathrm{have}\:\mathrm{the}\:\mathrm{same}\: \\ $$$$\mathrm{perimeter}.\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{distances}\:\mathrm{from} \\ $$$$\mathrm{vertex}\:{A}\:\mathrm{to}\:\mathrm{points}\:{M}\:\mathrm{and}\:{N}. \\ $$$$\left(\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{detail}}.\right) \\ $$…
Question Number 68629 by TawaTawa last updated on 14/Sep/19 Answered by $@ty@m123 last updated on 14/Sep/19 $$\angle{XYT}=\angle{XTP} \\ $$$$\Rightarrow\angle{XYT}=\mathrm{50}^{\mathrm{o}} \\ $$$$\mathrm{L}{et}\:\angle{YXT}=\angle{XTY}={x} \\ $$$$\Rightarrow{x}+{x}+\mathrm{50}^{\mathrm{o}} =\mathrm{180}^{\mathrm{o}} \\…
Question Number 68611 by TawaTawa last updated on 14/Sep/19 Answered by mind is power last updated on 15/Sep/19 $$\frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}\left({r}−\mathrm{2}\right)}−\frac{\mathrm{1}}{\mathrm{4}\left({r}+\mathrm{2}\right)} \\ $$$$\Rightarrow\sum_{{r}=\mathrm{3}} ^{\mathrm{100}} \frac{\mathrm{1}}{{r}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\Sigma\left(\frac{\mathrm{1}}{{r}−\mathrm{2}}−\frac{\mathrm{1}}{{r}+\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{99}}−\frac{\mathrm{1}}{\mathrm{100}}−\frac{\mathrm{1}}{\mathrm{101}}−\frac{\mathrm{1}}{\mathrm{102}}\right)…