Question
Complete the following activity to solve the quadratic equation
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The first term and the common difference of an A.P. are 6 and 3 respectively. Find $S_{27}$.
→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)]
Write the correct number in the given boxes from the following $A. P.$
$70, 60, 50, 40, . . .$
Here $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$
→$70, 60, 50, 40, . . .$
Here $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$
To draw tangents to the circle from the endpoints of the diameter of the circle.
| Construct a circle with center O. Draw any diameter AB of it |
| ↓ |
| Draw ray OA and ray OB |
| ↓ |
| Construct perpendicular to ray OA from point A |
| ↓ |
| Construct perpendicular to Ray OB from point B |
Find distance between point $Q(3,-7)$ and point $R(3,3)$
Solution: Suppose $Q\left(x_1, y_1\right)$ and point $R\left(x_2, y_2\right)$
$x_1=3, y_1=-7 \text { and } x_2=3, y_2=3$
Using distance formula,
$ d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\sqrt{\square-100}$
$\therefore d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\square $
→Solution: Suppose $Q\left(x_1, y_1\right)$ and point $R\left(x_2, y_2\right)$
$x_1=3, y_1=-7 \text { and } x_2=3, y_2=3$
Using distance formula,
$ d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\sqrt{\square-100}$
$\therefore d(Q, R)=\sqrt{\square}$
$\therefore d(Q, R)=\square $
Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are $9 cm$ and $12 cm$.
Activity: In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$P Q^2+\square=P R^2$
$\therefore P R^2=9^2+12^2$
$\therefore P R^2=\square+144$
$\therefore P R^2=\square$
$\therefore P R=15$
$\therefore$ Length of hypotenuse of triangle PQR is $\square cm$.
→Activity: In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$P Q^2+\square=P R^2$
$\therefore P R^2=9^2+12^2$
$\therefore P R^2=\square+144$
$\therefore P R^2=\square$
$\therefore P R=15$
$\therefore$ Length of hypotenuse of triangle PQR is $\square cm$.
Write the values of the following trigonometric ratios.
$\sin 60^{\circ}=\frac{⬜}{⬜}$
→$\sin 60^{\circ}=\frac{⬜}{⬜}$
In fig., $PM = 10 cm, A(\triangle PQS) = 100$ sq.cm, $A(\triangle QRS) = 110$ sq.cm, then $NR?$
$\triangle P Q S$ and $\triangle Q R S$ having seg $Q S$ common base.
Areas of two triangles whose base is common are in proportion of their corresponding
$ \frac{ A ( PQS )}{ A ( QRS )}=\frac{[}{ NR }$
$\frac{100}{110}=\frac{[}{ NR },$
$NR =[\ldots] cm $
→$\triangle P Q S$ and $\triangle Q R S$ having seg $Q S$ common base.
Areas of two triangles whose base is common are in proportion of their corresponding
$ \frac{ A ( PQS )}{ A ( QRS )}=\frac{[}{ NR }$
$\frac{100}{110}=\frac{[}{ NR },$
$NR =[\ldots] cm $
In $\Delta LMN , I =5, m =13, n =12$ then complete the activity to show that whether the given triangle is right angled triangle or not.
*(l, $m , n$ are opposite sides of $\angle L , \angle M , \angle N$ respectively)
Activity: In $\triangle LMN , I =5, m =13, n =\square$
$
\therefore 1^2=\square, m^2=169, n^2=144 \text {. }
$
$
\therefore 1^2+n^2=25+144=\square
$
$
\therefore \square+1^2=m^2
$
$\therefore$ By Converse of Pythagoras theorem, $\triangle LMN$ is right angled triangle.
→→*(l, $m , n$ are opposite sides of $\angle L , \angle M , \angle N$ respectively)
Activity: In $\triangle LMN , I =5, m =13, n =\square$
$
\therefore 1^2=\square, m^2=169, n^2=144 \text {. }
$
$
\therefore 1^2+n^2=25+144=\square
$
$
\therefore \square+1^2=m^2
$
$\therefore$ By Converse of Pythagoras theorem, $\triangle LMN$ is right angled triangle.
