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GeometryQuestion and Answers: Page 48

Question Number 150033 Answers: 1 Comments: 0

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if M is a point on the line y=x and points P(0,1),Q(2,0) are such that PM+PQ is minimum then find P

$${if}\:{M}\:{is}\:{a}\:{point}\:{on}\:{the}\:{line}\:{y}={x}\:{and} \\ $$$${points}\:{P}\left(\mathrm{0},\mathrm{1}\right),{Q}\left(\mathrm{2},\mathrm{0}\right)\:{are}\:{such}\:{that} \\ $$$${PM}+{PQ}\:{is}\:{minimum}\:{then}\:{find}\:{P} \\ $$

Question Number 148991 Answers: 1 Comments: 0

The largest value of k for which the circle x^2 +y^2 =k^2 lies completely in the interior of the parabola y^2 =4x+16 ?

$${The}\:{largest}\:{value}\:{of}\:{k}\:{for}\:{which}\: \\ $$$${the}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={k}^{\mathrm{2}} \:{lies}\:{completely} \\ $$$${in}\:{the}\:{interior}\:{of}\:{the}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{x}+\mathrm{16}\:? \\ $$

Question Number 148932 Answers: 1 Comments: 0

(((b+c)^2 )/(bc))l_a ^2 +(((a+b)^2 )/(ab))l_c ^2 +(((a+c)^2 )/(ac))l_b ^2 =(a+b+c)^2 l_b ,l_a ,l_c −bissekterissa prove

$$\frac{\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{bc}}}\boldsymbol{{l}}_{\boldsymbol{{a}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)^{\mathrm{2}} }{\boldsymbol{{ab}}}\boldsymbol{{l}}_{\boldsymbol{{c}}} ^{\mathrm{2}} +\frac{\left(\boldsymbol{{a}}+\boldsymbol{{c}}\right)^{\mathrm{2}} }{\boldsymbol{{ac}}}\boldsymbol{{l}}_{\boldsymbol{{b}}} ^{\mathrm{2}} =\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{l}}_{\boldsymbol{{b}}} ,\boldsymbol{{l}}_{\boldsymbol{{a}}} ,\boldsymbol{{l}}_{\boldsymbol{{c}}} −\boldsymbol{{bissekterissa}} \\ $$$$\boldsymbol{{prove}} \\ $$

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S=(4/3)(√(m(m−m_a )(m−m_b )(m−m_c ))) m=((m_a +m_b +m_c )/2) m_a ;m_b ;m_c −mediani prove

$$\boldsymbol{{S}}=\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\boldsymbol{{m}}\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{a}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{b}}} \right)\left(\boldsymbol{{m}}−\boldsymbol{{m}}_{\boldsymbol{{c}}} \right)} \\ $$$$\boldsymbol{{m}}=\frac{\boldsymbol{{m}}_{\boldsymbol{{a}}} +\boldsymbol{{m}}_{\boldsymbol{{b}}} +\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\mathrm{2}} \\ $$$$\boldsymbol{{m}}_{\boldsymbol{{a}}} ;\boldsymbol{{m}}_{\boldsymbol{{b}}} ;\boldsymbol{{m}}_{\boldsymbol{{c}}} −\boldsymbol{{mediani}} \\ $$$$\boldsymbol{{prove}} \\ $$

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