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$x^2+7 x+5=0$
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To draw the graph of $4 x+5 y=19$, complete the following activity to find $y$, when $x=1$.
→A ladder $10\ m$ long reaches a window $8\ m$ above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.
Activity: As shown in figure suppose
PR is the length of ladder $=10 m$
At P - window, At Q - base of wall, At R - foot of ladder
$\therefore PQ =8 m$
$\therefore Q R=?$
In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$\therefore PQ ^2+\square= PR ^2$
Here, $P R=10, P Q=\square$
From equation (l)
$ 8^2+Q^2=10^2$
$Q R^2=10^2-8^2$
$Q R^2=100-64$
$Q R^2=\square$
$Q R=6 $
$\therefore$ The distance of foot of the ladder from the base of wall is $6 m$.
→Activity: As shown in figure suppose
PR is the length of ladder $=10 m$
At P - window, At Q - base of wall, At R - foot of ladder
$\therefore PQ =8 m$
$\therefore Q R=?$
In $\triangle P Q R, m \angle P Q R=90^{\circ}$
By Pythagoras Theorem,
$\therefore PQ ^2+\square= PR ^2$
Here, $P R=10, P Q=\square$
From equation (l)
$ 8^2+Q^2=10^2$
$Q R^2=10^2-8^2$
$Q R^2=100-64$
$Q R^2=\square$
$Q R=6 $
$\therefore$ The distance of foot of the ladder from the base of wall is $6 m$.
In the adjoining figure, AB || CD || EF. If AC = 5.4, CE = 9, BD = 7.5, then find DF.
$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
→$AB \| CD \| EF$
$\therefore \frac{ AC }{ ⬜ }=\frac{ ⬜ }{ DF }$
$\therefore \quad \frac{5.4}{9}=\frac{⬜}{ DF }$
$\therefore \quad DF =\frac{7.5 \times 9}{5.4}$
∴ DF = ⬜
[Given]
⬜
Complete the following activity to find the 27 th term or the A.P. 9. 4. -1,-6,-11, ........
To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
4, 16, 64,…
4, 16, 64,…
Write the values of the following trigonometric ratios.
$\sin 30^{\circ}=\frac{1}{⬜}$
→$\sin 30^{\circ}=\frac{1}{⬜}$
Smt. Malhotra purchased solar panels for the taxable value of Rs. 85,000. She sold them for Rs. 90,000. The rate of GST is 5%. Find the ITC of Smt. Malhotra. What is the amount of GST payable by her?
....(2)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.
From the given figure, in $\triangle A B C$, if $A D \perp B C, \angle C=45^{\circ}, A C=8 \sqrt{2}, B D=5$, then for finding value of $A D$ and $BC$, complete the following activity.
Activity: In $\triangle ADC$, if $\angle ADC =90^{\circ}, \angle C =45^{\circ}$ [Given]
$\therefore \angle DAC =\square \quad$..... [Remaining angle of $\triangle ADC ]$
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$ \therefore \square=\frac{1}{\sqrt{2}} AC \text { and } \square=\frac{1}{\sqrt{2}} AC$
$\therefore AD =\frac{1}{\sqrt{2}} \times \square \text { and } DC =\frac{1}{\sqrt{2}} \times 8 \sqrt{2}$
$\therefore AD =8 \text { and } DC =8$
$\therefore BC = BD + DC$
$=5+8$
$=13 $
→Activity: In $\triangle ADC$, if $\angle ADC =90^{\circ}, \angle C =45^{\circ}$ [Given]
$\therefore \angle DAC =\square \quad$..... [Remaining angle of $\triangle ADC ]$
By theorem of $45^{\circ}-45^{\circ}-90^{\circ}$ triangle,
$ \therefore \square=\frac{1}{\sqrt{2}} AC \text { and } \square=\frac{1}{\sqrt{2}} AC$
$\therefore AD =\frac{1}{\sqrt{2}} \times \square \text { and } DC =\frac{1}{\sqrt{2}} \times 8 \sqrt{2}$
$\therefore AD =8 \text { and } DC =8$
$\therefore BC = BD + DC$
$=5+8$
$=13 $