Question
Complete the following activity to find the value of x.
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Complete the following activity to find the number of natural numbers from 1 to 171 which are divisible by 5 .
Find coordinates of the midpoint of a segment joining point $A(-1,1)$ and point $B(5,-7)$
Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$
$x_1=-1, y_1=1 \text { and } x_2=5, y_2=-7$
Using midpoint formula,
$\therefore$ Coordinates of midpoint of segment $A B$
$ =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
$=\left(\frac{\square}{2}, \frac{\square}{2}\right) $
$\therefore$ Coordinates of the midpoint $=\left(\frac{4}{2}, \frac{\square}{2}\right)$
$\therefore$ Coordinates of the midpoint $=(2, \square)$
→Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$
$x_1=-1, y_1=1 \text { and } x_2=5, y_2=-7$
Using midpoint formula,
$\therefore$ Coordinates of midpoint of segment $A B$
$ =\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
$=\left(\frac{\square}{2}, \frac{\square}{2}\right) $
$\therefore$ Coordinates of the midpoint $=\left(\frac{4}{2}, \frac{\square}{2}\right)$
$\therefore$ Coordinates of the midpoint $=(2, \square)$
Complete the following activity to form a quadratic equation.
Activity :
Activity :
First term and common difference of an A.P. are 6 and 3 respectively ; find S27.
$a=6, d=3, S_{27}=?$
$S_n=\frac{n}{2}$[⬜ + (n - 1)d]
$\left.\therefore \quad S _{27}=\frac{27}{2}[12+(27-1) ⬜\right]$
$=\frac{27}{2} \times ⬜$
$=27 \times 45$ = ⬜
→$a=6, d=3, S_{27}=?$
$S_n=\frac{n}{2}$[⬜ + (n - 1)d]
$\left.\therefore \quad S _{27}=\frac{27}{2}[12+(27-1) ⬜\right]$
$=\frac{27}{2} \times ⬜$
$=27 \times 45$ = ⬜
The coordinates of the vertices of a triangle $ABC$ are $A (-7,6), B (2,-2)$ and $C (8,5)$. Find coordinates of its centroid.
Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$
$x_1=-7, y_1=6 \text { and } x_2=2, y_2=-2 \text { and } x_3=8, y_3=5$
Using Centroid formula
$\therefore$ Coordinates of the centroid of a traingle
$ ABC =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
$=\left(\frac{\square}{3}, \frac{\square}{3}\right) $
$\therefore$ Coordinates of the centroid of a triangle $A B C=\left(\frac{3}{3}, \square\right)$
$\therefore$ Coordinates of the centroid of a triangle $A B C=(1, \square)$
→Solution: Suppose $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$
$x_1=-7, y_1=6 \text { and } x_2=2, y_2=-2 \text { and } x_3=8, y_3=5$
Using Centroid formula
$\therefore$ Coordinates of the centroid of a traingle
$ ABC =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$
$=\left(\frac{\square}{3}, \frac{\square}{3}\right) $
$\therefore$ Coordinates of the centroid of a triangle $A B C=\left(\frac{3}{3}, \square\right)$
$\therefore$ Coordinates of the centroid of a triangle $A B C=(1, \square)$
In $\triangle A B C$, ray $B D$ bisects $\angle A B C . A-D-C$, side $D E \|$ side $B C, A-E-B$, then prove that $\frac{A B}{B C}=\frac{A E}{E B}$
In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
⬜
[From (i) and (ii)]
→In $\triangle A B C$, ray $B D$ bisects $\angle B$.
$\therefore \quad \frac{ AB }{ BC }=\frac{ AD }{ DC }$
In $\triangle ABC , DE \| BC$
$\therefore \quad \frac{ AE }{ EB }=\frac{ AD }{ DC }$
$\therefore \frac{ AB }{ ⬜ }=\frac{ ⬜ }{ EB }$
[Given]
[Angle biscetor theorem]
[Given]
⬜
[From (i) and (ii)]
From given figure, in $\triangle P Q R$, if $\angle Q P R=90^{\circ}, P M \perp Q R, P M=10, Q M=8$, then for finding the value of $Q R$, complete the following activity.
Activity: In $\triangle P Q R$, if $\angle Q P R=90^{\circ}, P M \perp Q R$,
[Given]
In $\triangle P M Q^1$ by Pythagoras Theorem,
$\therefore PM ^2+\square= PQ ^2$
$\therefore P Q^2=10^2+8^2$
$\therefore P Q^2=\square+64$
$\therefore P Q^2=\square$
$\therefore P Q=\sqrt{164}$
Here, $\triangle QPR \sim \triangle QMP \sim \triangle PMR$
$\therefore \triangle QMP \sim \triangle PMR$
$\therefore \frac{ PM }{ RM }=\frac{ QM }{ PM }$
$\therefore PM ^2= RM \times QM$
$\therefore 10^2= RM \times 8$
$RM =\frac{100}{8}=\square$
And,
$Q R=Q M+M R$
$QR =\square+\frac{25}{2}=\frac{41}{2}$
→→→Activity: In $\triangle P Q R$, if $\angle Q P R=90^{\circ}, P M \perp Q R$,
[Given]
In $\triangle P M Q^1$ by Pythagoras Theorem,
$\therefore PM ^2+\square= PQ ^2$
$\therefore P Q^2=10^2+8^2$
$\therefore P Q^2=\square+64$
$\therefore P Q^2=\square$
$\therefore P Q=\sqrt{164}$
Here, $\triangle QPR \sim \triangle QMP \sim \triangle PMR$
$\therefore \triangle QMP \sim \triangle PMR$
$\therefore \frac{ PM }{ RM }=\frac{ QM }{ PM }$
$\therefore PM ^2= RM \times QM$
$\therefore 10^2= RM \times 8$
$RM =\frac{100}{8}=\square$
And,
$Q R=Q M+M R$
$QR =\square+\frac{25}{2}=\frac{41}{2}$
