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यूजर का नामheureka
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 #1
avatar+24883 
+1

In an isosceles triangle, the sum of its base and its height is 10,
and the radius of its circumscribed circle is 25/8.
Find the sides of the triangle.

 

\(\begin{array}{|lrcll|} \hline & c+h &=& 10 \\ \text{or} & \mathbf{h} &=& \mathbf{10-c} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(\dfrac{c}{2}\right)^2+h^2 &=& a^2 \\ \left(\dfrac{c}{2}\right)^2+(10-c)^2 &=& a^2 \\ \dfrac{c^2}{4}+(10-c)^2 &=& a^2 \\ \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline A &=& \dfrac{ch}{2} \\\\ \mathbf{A} &=& \mathbf{\dfrac{c(10-c)}{2}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{abc}{4A}} \quad | \quad b=a \\\\ r &=& \dfrac{ca^2}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4*\dfrac{c(10-c)}{2}} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{2c(10-c)} \\\\ r &=& \dfrac{ \dfrac{c^2}{4}+(10-c)^2 }{2(10-c)} \\\\ 2(10-c)r &=& \dfrac{c^2}{4}+(10-c)^2 \\\\ (10-c)^2 - 2r(10-c) + \dfrac{c^2}{4} &=& 0 \\\\ 100-20c+c^2 -20r + 2rc - \dfrac{c^2}{4} &=& 0 \\ \ldots \\ \mathbf{\dfrac{5}{4}c^2-c(20-2r) + 100-20r} &=& \mathbf{0} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2-4*\dfrac{5}{4}*(100-20r)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2- 5(100-20r)} }{ 2*\dfrac{5}{4}} \\ && \ldots \\\\ c &=& \dfrac{20-2r \pm \sqrt{4(r^2+5r-25)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm 2\sqrt{ r^2+5r-25 } }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{10-r \pm \sqrt{ r^2+5r-25 } }{ \dfrac{5}{4}}\quad | \quad r=\dfrac{25}{8} \\\\ c &=& \dfrac{6.875 \pm \sqrt{ 0.390625 } }{ 1.25} \\\\ c &=& \dfrac{6.875 \pm 0.625 }{ 1.25} \\\\ c_1 = \dfrac{7.5}{1.25} && c_2 = \dfrac{6.25}{1.25} \\\\ \mathbf{c_1=6} && \mathbf{c_2=5} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \mathbf{a_1^2} &=& \mathbf{\dfrac{6^2}{4}+(10-6)^2} \quad | \quad c_1=6 \\ a_1^2 &=& \dfrac{36}{4}+4^2 \\ a_1^2 &=& 9+16 \\ a_1^2 &=& 25 \\ \mathbf{a_1} &=& \mathbf{5} \\ \hline \mathbf{a_2^2} &=& \mathbf{\dfrac{5^2}{4}+(10-5)^2} \quad | \quad c_2=5 \\ a_2^2 &=& \dfrac{25}{4}+5^2 \\ a_2^2 &=& \dfrac{25*5}{4} \\ \mathbf{a_2} &=& \mathbf{\dfrac{5}{2}\sqrt{5}} \\ \hline \end{array} \)

 

laugh

26 Mei 2020 11.07.06
 #1
avatar+24883 
+3

Solve

\(|x - 1| - 2|x| = 3|x + 1|\).

 

\(\begin{array}{|rcll|} \hline \text{zero in }~ |x - 1|: \\ \hline x - 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ 1 } \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x|: \\ \hline \mathbf{x} &=&\mathbf{0} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x + 1|: \\ \hline \hline x + 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ -1 } \\ \hline \end{array}\)

 

We create a sign table:

\(\begin{array}{|l|c|c\|\|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-1\right) & -1 & \left(-1,0\right) & 0 & \left(0,1\right) & 1 & \left(1,\infty\right) \\ \hline \text{sign of } (x+1): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x): & - & - & - & 0 & +& +& + \\ \hline \text{sign of } (x-1): &- & - & - & - & - & 0 & + \\ \hline \end{array} \)

 

For the 4 cases  \((-\infty, -1] ,\ (-1 , 0] ,\ (0, 1] ,\ (1, \infty)\)
we can therefore always resolve all amounts and solve the corresponding equation without amounts,
taking into account the respective case condition.
The negative terms from the amounts are highlighted in green:

 

\(\begin{array}{|rcll|} \hline (-\infty, -1] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3({\color{green}-(x + 1)}) \\ -x+1 +2x &=& -3(x + 1) \\ -x+1 +2x &=& -3x -3 \\ 4x &=& -4 \\ \mathbf{x} &=& \mathbf{-1}\ \checkmark \qquad -\infty < x \le -1 \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline (-1 , 0] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3(x + 1) \\ -x+1 +2x &=& 3x+3 \\ 2x &=& -2 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad -1 < x \le 0 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (0, 1] \\ \hline {\color{green}-(x - 1)} - 2(x) &=& 3(x + 1) \\ -x+1 -2x &=& 3x+3 \\ 6x &= & -2 \\ x &=& -\dfrac{1}{3} \quad \mathbf{\text{no}} \qquad 0 < x \le 1 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline (1, \infty) \\ \hline (x - 1) - 2(x) &=& 3(x + 1) \\ x-1 -2x &=& 3x+3 \\ 4x &= & -4 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad 1 < x < \infty \\ \hline \end{array}\)

 

\(\mathbf{x = -1}\)

 

laugh

26 Mei 2020 09.16.28
 #1
avatar+24883 
+2

One of their common tangents is touching as shown in the figure. 

Find the radius of the circle with center R.

 

\(\text{Let $PR=p+r$} \\ \text{Let $PB=p-r$} \\ \text{Let $QR=q+r$} \\ \text{Let $QC=q-r$} \\ \text{Let $PQ=p+q$} \\ \text{Let $PA=p-q$} \\ \text{Let $DF=AQ$} \\ \text{Let $DF=DE+EF$}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{DE^2 +PB^2} &=& \mathbf{PR^2} \\ DE^2 + (p-r)^2 &=& (p+r)^2 \\ DE^2 &=& (p+r)^2 - (p-r)^2 \\ DE^2 &=& p^2+2pr+r^2-p^2+2pr-r^2 \\ DE^2 &=& 2*2pr \\ \mathbf{DE} &=& \mathbf{2\sqrt{pr} } \\ \\ \hline \mathbf{EF^2 +QC^2} &=& \mathbf{QR^2} \\ EF^2 + (q-r)^2 &=& (q+r)^2 \\ EF^2 &=& (q+r)^2 - (q-r)^2 \\ EF^2 &=& q^2+2qr+r^2-q^2+2qr-r^2 \\ EF^2 &=& 2*2qr \\ \mathbf{EF} &=& \mathbf{2\sqrt{qr} } \\ \\ \hline \mathbf{DF^2 + PA^2} &=& \mathbf{PQ^2} \\ DF^2 + (p-q)^2 &=& (p+q)^2 \\ DF^2 &=& (p+q)^2 - (p-q)^2 \\ DF^2 &=& p^2+2pq+q^2-p^2+2pq-q^2 \\ DF^2 &=& 2*2pq \\ \mathbf{DF} &=& \mathbf{2\sqrt{pq} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{DF} &=& \mathbf{DE+EF} \\ 2\sqrt{pq} &=& 2\sqrt{pr}+2\sqrt{qr} \quad | \quad : 2 \\ \sqrt{pq} &=& \sqrt{pr}+\sqrt{qr} \\ \sqrt{p}\sqrt{q} &=& \sqrt{p}\sqrt{r}+\sqrt{q}\sqrt{r} \quad | \quad : \sqrt{r} \\\\ \dfrac{\sqrt{p}\sqrt{q}}{\sqrt{r}} &=& \sqrt{p} +\sqrt{q} \quad | \quad : \sqrt{p}\sqrt{q} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{\sqrt{p}}{\sqrt{p}\sqrt{q}} +\dfrac{\sqrt{q}}{\sqrt{p}\sqrt{q}} \\\\ \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \quad | \quad p=16,\ q=4 \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{\sqrt{4}} +\dfrac{1}{\sqrt{16}} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{2} +\dfrac{1}{4} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{3}{4} \\\\ \sqrt{r} &=& \dfrac{4}{3} \quad | \quad \text{square both sides}\\\\ r &=& \dfrac{16}{9} \\\\ r &=& 1.\bar{7}\ \text{cm} \\ \hline \end{array} \)

 

The radius of the circle with center R is \(\mathbf{1.\bar{7}\ \text{cm}}\)

 

laugh

26 Mei 2020 06.58.41