Question Number 16874 by Tinkutara last updated on 27/Jun/17 $$\mathrm{Let}\:{ABC}\:\mathrm{be}\:\mathrm{an}\:\mathrm{acute}\:\mathrm{triangle}.\:\mathrm{The} \\ $$$$\mathrm{interior}\:\mathrm{bisectors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angles}\:\angle{B}\:\mathrm{and} \\ $$$$\angle{C}\:\mathrm{meet}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{points}\:{L}\:\mathrm{and}\:{M},\:\mathrm{respectively}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a}\:\mathrm{point}\:{K}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{interior}\:\mathrm{of}\:\mathrm{the}\:\mathrm{side}\:{BC}\:\mathrm{such}\:\mathrm{that} \\ $$$$\Delta{KLM}\:\mathrm{is}\:\mathrm{equilateral}\:\mathrm{if}\:\mathrm{and}\:\mathrm{only}\:\mathrm{if} \\ $$$$\angle{A}\:=\:\mathrm{60}°. \\…
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Question Number 82378 by ajfour last updated on 20/Feb/20 Commented by ajfour last updated on 20/Feb/20 $$\mathrm{If}\:\mathrm{all}\:\mathrm{5}\:\mathrm{regions}\:\mathrm{have}\:\mathrm{equal}\:\mathrm{areas}, \\ $$$$\mathrm{find}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{rectangle},\:\mathrm{given} \\ $$$$\mathrm{radius}=\mathrm{1}. \\ $$ Answered by…
Question Number 16803 by ajfour last updated on 26/Jun/17 Commented by ajfour last updated on 26/Jun/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangle}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distances} \\ $$$$\mathrm{from}\:\mathrm{any}\:\mathrm{point}\:\mathrm{to}\:\mathrm{the}\:\mathrm{three}\:\mathrm{sides} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$…
Question Number 16771 by Tinkutara last updated on 26/Jun/17 Commented by Tinkutara last updated on 26/Jun/17 $$\left(\mathrm{1}\right)\:{O}\:\mathrm{is}\:\mathrm{circumcenter}\:\mathrm{and}\:{H}\:\mathrm{is}\:\mathrm{orthocenter} \\ $$$$\mathrm{of}\:\Delta{ABC}.\:{AOA}'\:\mathrm{is}\:\mathrm{the}\:\mathrm{diameter}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{A}'{BHC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parallelogram}. \\ $$ Answered by…
Question Number 82308 by Power last updated on 20/Feb/20 Commented by mr W last updated on 20/Feb/20 $$\left.{D}\right)\:\mathrm{25} \\ $$ Terms of Service Privacy Policy…
Question Number 16785 by RasheedSoomro last updated on 26/Jun/17 Commented by RasheedSoomro last updated on 26/Jun/17 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{area}-\mathrm{formula}\:\mathrm{find}\:\mathrm{out} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}\:\mathrm{to}\:\mathrm{the}\:\mathrm{whole}. \\ $$ Commented by mrW1 last…
Question Number 82302 by Power last updated on 20/Feb/20 Commented by mr W last updated on 20/Feb/20 $${x}=\mathrm{20}° \\ $$$${see}\:{Q}\mathrm{61347} \\ $$ Commented by Power…
Question Number 16756 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Alternate}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{are}\:\mathrm{joined}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{What}\:\mathrm{fraction}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is} \\ $$$$\mathrm{shaded}?\:\left(\mathrm{Justify}\:\mathrm{your}\:\mathrm{answer}.\right) \\ $$ Commented by Tinkutara last updated on 26/Jun/17…
Question Number 16754 by ajfour last updated on 26/Jun/17 Commented by ajfour last updated on 26/Jun/17 $$\mathrm{Q}.\mathrm{16748}\:\left(\mathrm{solution}\right) \\ $$$$\:\mathrm{by}\:\mathrm{fault}\:\mathrm{it}\:\mathrm{gets}\:\mathrm{uploaded}\:\mathrm{as}\: \\ $$$$\mathrm{question}. \\ $$ Answered by…