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Question Number 17963 by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

$$\mathrm{AB}=2\mathrm{a},\mathrm{CD}=2\mathrm{b}$$$$\mathrm{Find}\mathbf{r}\mathrm{in}\mathrm{terms}\mathrm{of}\mathrm{a},\mathrm{b},\varphi ,\mathrm{and}\mathrm{R}.$$$$\mathrm{Hence}\mathrm{find}{\mathrm{AP}}^2+{\mathrm{BP}}^2+{\mathrm{CP}}^2+{\mathrm{DP}}^2.$$

Commented by ajfour last updated on 13/Jul/17

Commented by ajfour last updated on 13/Jul/17

$$\mathrm{x}=\sqrt{{\mathrm{R}}^2-{\mathrm{a}}^2},\mathrm{y}=\sqrt{{\mathrm{R}}^2-{\mathrm{b}}^2}$$$$\sin {\mathit{\varphi }}_1=\frac{\mathrm{x}}{\mathrm{r}},\sin {\mathit{\varphi }}_2=\frac{\mathrm{y}}{\mathrm{r}}$$$${\mathit{\varphi }}_1+{\mathit{\varphi }}_2=\mathit{\varphi },\mathrm{so}$$$${\sin }^{-1}\left(\frac{\sqrt{{\mathrm{R}}^2-{\mathrm{a}}^2}}{\mathrm{r}}\right)+{\sin }^{-1}\left(\frac{\sqrt{{\mathrm{R}}^2-{\mathrm{b}}^2}}{\mathrm{r}}\right)=\mathit{\varphi }...\left(\mathrm{i}\right)$$$$\mathrm{cannot}\mathrm{r}\mathrm{be}\mathrm{obtained}\mathrm{explicitly}?$$$$\mathrm{AP}=\mathrm{a}-\mathrm{PM},\mathrm{BP}=\mathrm{a}+\mathrm{PM}$$$$\mathrm{CP}=\mathrm{b}-\mathrm{NP},\mathrm{DP}=\mathrm{b}+\mathrm{NP}$$$${\mathrm{AP}}^2+{\mathrm{BP}}^2+{\mathrm{CP}}^2+{\mathrm{DP}}^2=$$$${2\mathrm{a}}^2+{2\mathrm{PM}}^2+{2\mathrm{b}}^2+{2\mathrm{NP}}^2...\left(\mathrm{ii}\right)$$$$\mathrm{Now},{\mathrm{PM}}^2+{\mathrm{NP}}^2={\mathrm{r}}^2-{\mathrm{x}}^2+{\mathrm{r}}^2-{\mathrm{y}}^2$$$$\mathrm{and}\mathrm{with}{\mathrm{x}}^2={\mathrm{R}}^2-{\mathrm{a}}^2,{\mathrm{y}}^2={\mathrm{R}}^2-{\mathrm{b}}^2$$$$\mathrm{we}\mathrm{have}$$$${\mathrm{PM}}^2+{\mathrm{NP}}^2={2\mathrm{r}}^2-({\mathrm{x}}^2+{\mathrm{y}}^2)$$$$={2\mathrm{r}}^2-({2\mathrm{R}}^2-{\mathrm{a}}^2-{\mathrm{b}}^2)$$$$={2\mathrm{r}}^2+{\mathrm{a}}^2+{\mathrm{b}}^2-{2\mathrm{R}}^2$$$$\mathrm{substituting}\mathrm{in}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}$$$${\mathrm{AP}}^2+{\mathrm{BP}}^2+{\mathrm{CP}}^2+{\mathrm{DP}}^2$$$$={\left(2\mathrm{a}\right)}^2+{\left(2\mathrm{b}\right)}^2-4({\mathrm{R}}^2-{\mathrm{r}}^2);$$$$\mathrm{r}\mathrm{is}\mathrm{obtained}\mathrm{from}\left(\mathrm{i}\right).$$

Commented by b.e.h.i.8.3.417@gmail.com last updated on 14/Jul/17

$$niceandbeautifulmrAjfour.$$

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