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Category: Geometry

Question-64459

Question Number 64459 by Tawa1 last updated on 18/Jul/19 Commented by Tony Lin last updated on 18/Jul/19 AR=AP=2,RC=QC=5letPB=BQ=x$${cos}\mathrm{60}°=\frac{\left(\mathrm{2}+{x}\right)^{\mathrm{2}} +\left(\mathrm{5}+{x}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{2}+{x}\right)\left(\mathrm{5}+{x}\right)}…

Question-64425

Question Number 64425 by Tawa1 last updated on 17/Jul/19 Commented by Prithwish sen last updated on 17/Jul/19 Areaofshadedportion=AreaofsquareAreaofquartercircle$$=\:\mathrm{a}^{\mathrm{2}} \:−\:\frac{\pi\mathrm{a}^{\mathrm{2}} }{\mathrm{4}}\:\left[\:\mathrm{if}\:\boldsymbol{\mathrm{a}}\:\mathrm{be}\:\mathrm{the}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\right. \