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Category: Mensuration

Question-185978

Question Number 185978 by Rupesh123 last updated on 30/Jan/23 Answered by ajfour last updated on 30/Jan/23 $${x}=?=\mathrm{2}\left(\mathrm{2sin}\:\theta\right) \\ $$$${Y}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} \right)\left(\mathrm{2}\theta\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\right)\left(\mathrm{2sin}\:\mathrm{2}\theta\right) \\ $$$${G}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}^{\mathrm{2}} \right)\left(\frac{\pi}{\mathrm{2}}−\mathrm{2}\theta\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2cos}\:\mathrm{2}\theta\right)\left(\mathrm{2sin}\:\mathrm{2}\theta\right)…

Question-54869

Question Number 54869 by peter frank last updated on 13/Feb/19 Answered by $@ty@m last updated on 13/Feb/19 $${Given}\:\beta=\frac{\mathrm{1}}{\alpha}\:\Rightarrow\alpha\beta=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{10}}{\mathrm{3}{k}+\mathrm{4}}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{3}{k}+\mathrm{4}=\mathrm{10} \\ $$$$\Rightarrow{k}=\mathrm{2} \\…

Question-185938

Question Number 185938 by Mingma last updated on 30/Jan/23 Answered by HeferH last updated on 30/Jan/23 $${Let}\:{S}\:{be}\:{the}\:{pink}\:{area}\:{and}\:{let}\:\mathrm{5}{k}\:{be}\:{the}\:{area} \\ $$$$\:{of}\:{ABC}. \\ $$$$\:{S}\left(\frac{\mathrm{3}{k}}{\mathrm{2}}\:+\:{S}\right)\:=\:\left({k}\:−\:{S}\right)\left(\frac{\mathrm{3}{k}}{\mathrm{2}}−{S}\right) \\ $$$$\:{S}\:\:=\:\frac{\mathrm{3}{k}}{\mathrm{8}} \\ $$$$\:\frac{{Pink}}{{Area}\:{of}\:{ABC}}\:=\:\frac{\frac{\mathrm{3}{k}}{\mathrm{8}}}{\mathrm{5}{k}}\:=\:\frac{\mathrm{3}}{\mathrm{40}}\:\:…

Question-185907

Question Number 185907 by Mingma last updated on 29/Jan/23 Answered by HeferH last updated on 29/Jan/23 $$\:{If}\:{yellow}\:=\:{orange}\:{then}\: \\ $$$$\:{Area}\:\left({ACB}\right)\:=\:{Area}\:\left({PQRS}\right). \\ $$$$\:{let}\:{O}\:{be}\:{the}\:{center}\:{of}\:{the}\:{semicircle}. \\ $$$$\:{OP}\:=\:{a} \\ $$$$\:{AO}\:=\:{R}…

Question-185800

Question Number 185800 by Rupesh123 last updated on 27/Jan/23 Answered by ajfour last updated on 27/Jan/23 $${let}\:{center}\:{of}\:{small}\:{circle}\:\left(−{h},−{k}\right) \\ $$$${hence}\:{eq}.\:\:\:\left({x}+{h}\right)^{\mathrm{2}} +\left({y}+{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${side}\:{of}\:{eql}.\:\:{triangle}\:{a}={R}\sqrt{\mathrm{3}} \\ $$$${length}\:{of}\:{tangent}\:{from}\:{A}\left(\mathrm{0},{R}\right)…