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A line is parallel to one side of triangle which intersects remaining two sides in two distinct points then that line divides sides in same proportion.
Given: In $\triangle A B C$ line I || side $B C$ and line I intersect side $A B$ in $P$ and side $A C$ in $Q$.
To prove: $\frac{ AP }{ PB }=\frac{ AQ }{ QC }$
Construction: Draw $CP$ and $BQ$
Proof: $\triangle APQ$ and $\triangle PQB$ have equal height.
$\left.\frac{ A (\Delta APQ )}{ A (\Delta PQB )}=\frac{[}{ PB } \quad \ldots . . \text { (i) [areas in proportion of base }\right]$
$\frac{ A (\Delta APQ )}{ A (\Delta PQC )}=\frac{[}{ QC } \ldots . . . \text { (ii) [areas in proportion of base] }$
$\triangle P Q C$ and $\triangle P Q B$ have [____ ] is common base.
Seg PQ \| Seg BC, hence height of $\triangle A P Q$ and $\triangle P Q B$.
$A (\triangle PQC )= A (\Delta \ldots, \ldots \text {.....(iii) }$
$\frac{ A (\Delta APQ )}{ A (\Delta PQB )}=\frac{ A (\Delta \ldots)}{ A (\Delta \ldots \ldots[( i ) \text {, (ii), and (iii)] }}$
$\frac{ AP }{ PB }=\frac{ AQ }{ QC } \quad \ldots . . .[( i ) \text { and (iii)] }$
→Given: In $\triangle A B C$ line I || side $B C$ and line I intersect side $A B$ in $P$ and side $A C$ in $Q$.
To prove: $\frac{ AP }{ PB }=\frac{ AQ }{ QC }$
Construction: Draw $CP$ and $BQ$
Proof: $\triangle APQ$ and $\triangle PQB$ have equal height.
$\left.\frac{ A (\Delta APQ )}{ A (\Delta PQB )}=\frac{[}{ PB } \quad \ldots . . \text { (i) [areas in proportion of base }\right]$
$\frac{ A (\Delta APQ )}{ A (\Delta PQC )}=\frac{[}{ QC } \ldots . . . \text { (ii) [areas in proportion of base] }$
$\triangle P Q C$ and $\triangle P Q B$ have [____ ] is common base.
Seg PQ \| Seg BC, hence height of $\triangle A P Q$ and $\triangle P Q B$.
$A (\triangle PQC )= A (\Delta \ldots, \ldots \text {.....(iii) }$
$\frac{ A (\Delta APQ )}{ A (\Delta PQB )}=\frac{ A (\Delta \ldots)}{ A (\Delta \ldots \ldots[( i ) \text {, (ii), and (iii)] }}$
$\frac{ AP }{ PB }=\frac{ AQ }{ QC } \quad \ldots . . .[( i ) \text { and (iii)] }$
Solve the following simultaneous equations by graphical method.
The distances covered by 250 public transport buses in a day is shown in the following frequency distribution table. Find the median of the distances.
Complete the following table to draw the graph of 3 𝑥 − 2 𝑦 = 18
X | 0 | 4 | 2 | -1 |
Y | -9 | ------ | ------- | ------ |
x, y | (0,-9) | (--,--) | (--,--) | ------ |
If the point $P (6,7)$ divides the segment joining $A (8,9)$ and $B (1,2)$ in some ratio, find that ratio
Solution:
Point $P$ divides segment $A B$ in the ratio $m$ : $n$.
$A(8,9)=\left(x_1, y_1\right), B(1,2)=\left(x_2, y_2\right) \text { and } P(6,7)=(x, y)$
Using Section formula of internal division,
$ \therefore 7=\frac{m(\square)-n(9)}{m+n}$
$\therefore 7 m+7 n=\square+9 n$
$\therefore 7 m-\square=9 n-\square$
$\therefore \square=2 n$
$\therefore \frac{m}{n}=\square $
→Solution:
Point $P$ divides segment $A B$ in the ratio $m$ : $n$.
$A(8,9)=\left(x_1, y_1\right), B(1,2)=\left(x_2, y_2\right) \text { and } P(6,7)=(x, y)$
Using Section formula of internal division,
$ \therefore 7=\frac{m(\square)-n(9)}{m+n}$
$\therefore 7 m+7 n=\square+9 n$
$\therefore 7 m-\square=9 n-\square$
$\therefore \square=2 n$
$\therefore \frac{m}{n}=\square $
The following frequency distribution table gives the ages of 200 patients treated in a hospital in a week. Find the mode of ages of the patients.
The following table gives the information of frequency distribution of weekly wages of 150 workers of a company. Find the mean of the weekly wages by 'step deviation' method.
In an A.P., the first term is -5 and the last term is $4 5$. If the sum of $n$ terms in the A.P. is $1 2 0$, then complete the activity to find $n$.
→One of the roots of equation 5m2 + 2m + k = 0 is $\frac{-7}{5}$ Complete the following activity to find the value of 'k'.
$[-]$ is a root of quadratic equation $5 m^2+2 m+k=0$
∴ Put m $[-]$ in the equation.
∴ 5 $ \times$ $[-]$ + 2 $\times$$[-]$ + k = 0
∴ $[-]$ + $[-]$ + k = 0
∴ ⬜ + k = 0
∴ k = ⬜
→$[-]$ is a root of quadratic equation $5 m^2+2 m+k=0$
∴ Put m $[-]$ in the equation.
∴ 5 $ \times$ $[-]$ + 2 $\times$$[-]$ + k = 0
∴ $[-]$ + $[-]$ + k = 0
∴ ⬜ + k = 0
∴ k = ⬜
The weekly wages of 120 workers in a factory are shown in the following frequency distribution table. Find the mean of the weekly wages.
