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Question Number 176195 by adhigenz last updated on 14/Sep/22

Answered by behi834171 last updated on 15/Sep/22

$$BC=BP=PC=\frac{\sqrt{3}}{3},AB=AR=BR=\frac{2\sqrt{3}}{3}$$$${PQ}^2=\frac{1}{3}+1-2\times \frac{\sqrt{3}}{3}\times 1\times cos\left(150\right)=$$$$=\frac{1}{3}+1+2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}\Rightarrow PQ=\sqrt{\frac{7}{3}}$$$$cos\measuredangle CPQ=\frac{{\left(\frac{\sqrt{3}}{3}\right)}^2+{\left(\sqrt{\frac{7}{3}}\right)}^2-1^2}{2\frac{\sqrt{3}}{3}.\sqrt{\frac{7}{3}}}=\frac{5}{2\sqrt{7}}$$$$\Rightarrow \measuredangle CPQ=19.{12}^{\bullet }$$$${RQ}^2={PR}^2+{PQ}^2-2PR.PQ.cos(60+19.12)=$$$$={\left(\sqrt{3}\right)}^2+{\left(\sqrt{\frac{7}{3}}\right)}^2-2\times \sqrt{3}\times \sqrt{\frac{7}{3}}.[\frac{1}{2}.\frac{5}{2\sqrt{7}}-\frac{\sqrt{3}}{2}.\frac{\sqrt{3}}{2\sqrt{7}}]=$$$$=3+\frac{7}{3}-1=\frac{13}{3}\Rightarrow \mathit{R}\mathit{Q}=\sqrt{\frac{13}{3}}$$$$\mathit{c}\mathit{o}\mathit{s}\mathit{A}\mathit{Q}\mathit{T}=\frac{{\left(\sqrt{\frac{13}{3}}\right)}^2+1^2-{\left(2\frac{\sqrt{3}}{3}\right)}^2}{2\times \frac{\sqrt{13}}{\sqrt{3}}\times 1}=\frac{\frac{13}{3}+1-\frac{4}{3}}{\frac{2\sqrt{13}}{\sqrt{3}}}=$$$$=\frac{\frac{12}{3}}{\frac{2\sqrt{13}}{\sqrt{3}}}=\frac{6}{\sqrt{39}}$$$$TQ=\frac{1}{\mathit{c}\mathit{o}\mathit{s}AQT}=\frac{\sqrt{39}}{6}$$$$TR=QR-QT=\sqrt{\frac{13}{3}}-\frac{\sqrt{39}}{6}=\frac{\sqrt{39}}{3}$$$$QR({PT}^2+QT.TR)={PR}^2.QT+{PQ}^2.TR\Rightarrow$$$$\sqrt{\frac{13}{3}}({PT}^2+\frac{\sqrt{39}}{6}.\frac{\sqrt{39}}{3})={\left(\sqrt{3}\right)}^2.\frac{\sqrt{39}}{6}+{\left(\sqrt{\frac{7}{3}}\right)}^2.\frac{\sqrt{39}}{3}\Rightarrow$$$$\Rightarrow \sqrt{\frac{13}{3}}({PT}^2+\frac{39}{18})=\frac{\sqrt{39}}{2}+\frac{7\sqrt{39}}{9}=\frac{23\sqrt{39}}{18}$$$$\Rightarrow {PT}^2=\frac{23\sqrt{39}}{18}\times \frac{\sqrt{3}}{\sqrt{13}}-\frac{39}{18}=\frac{30}{18}=\frac{5}{3}$$$$\Rightarrow \mathit{P}\mathit{T}=\sqrt{\frac{5}{3}}$$$$\mathit{n}\mathit{o}\mathit{w}:$$$$\mathit{P}\mathit{T}=\sqrt{\frac{5}{3}}=\frac{\sqrt{15}}{3},\mathit{T}\mathit{Q}=\frac{\sqrt{39}}{6},\mathit{P}\mathit{Q}=\sqrt{\frac{7}{3}}=\frac{\sqrt{21}}{3}$$$$\mathit{P}\mathit{T}=1.29,\mathit{T}\mathit{Q}=1.04,\mathit{P}\mathit{Q}=1.53$$$$\mathit{p}=\frac{1.29+1.04+1.53}{2}=1.93$$$$\mathit{A}rea=\sqrt{1.93\times 0.64\times 0.89\times 0.40}=0.66.◼$$$$[\mathit{a}\mathit{r}\mathit{e}\mathit{a}=\frac{\sqrt{1015}}{48}]$$

Commented by adhigenz last updated on 15/Sep/22

$$\mathrm{Appreciate}\mathrm{it}\mathrm{sir},\mathrm{thanks}\mathrm{so}\mathrm{much}..$$$$\mathrm{But}\mathrm{why}\measuredangle AQT=30°?$$

Commented by adhigenz last updated on 15/Sep/22

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\measuredangle BTR=60°,\mathrm{since}\measuredangle BAR=60°$$

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}.$$

Commented by behi834171 last updated on 15/Sep/22

$$youarerightsir.thisisatypo,but,\mathit{f}\mathit{i}\mathit{x}\mathit{e}\mathit{d}!$$

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