Question
Complete the following activity to form a quadratic equation.
Activity :
Activity :
✓
Answer
Standard form : $ax^2 + bx + c = 0$
Sum of roots: $5 + 12 = 17$
Product of roots: $5 \times 12 = 60$
Equation: $x^2 - (Sum)x + (Product) = 0$ i.e., $x^2 - 17x + 60 = 0$
Sum of roots: $5 + 12 = 17$
Product of roots: $5 \times 12 = 60$
Equation: $x^2 - (Sum)x + (Product) = 0$ i.e., $x^2 - 17x + 60 = 0$
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Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. Complete the following activity.
Activity: As shown in figure GLMNT is a reactangle.
$\therefore$ Area of rectangle $=$ length $\times$ breadth
$\therefore$ Area of rectangle $=\square \times$ breadth
$\therefore 192=\square \times$ breadth
$\therefore$ Breadth $=12 cm$
Also,
$\angle TLM =90^{\circ}$ [Each angle of reactangle is right angle]
In $\triangle T L M$,
By Pythagoras theorem
$\therefore TM ^2= TL ^2+\square$
$\therefore TM ^2=12^2+\square$
$\therefore TM ^2=144+\square$
$\therefore TM ^2=400$
$\therefore TM =20$
→Activity: As shown in figure GLMNT is a reactangle.
$\therefore$ Area of rectangle $=$ length $\times$ breadth
$\therefore$ Area of rectangle $=\square \times$ breadth
$\therefore 192=\square \times$ breadth
$\therefore$ Breadth $=12 cm$
Also,
$\angle TLM =90^{\circ}$ [Each angle of reactangle is right angle]
In $\triangle T L M$,
By Pythagoras theorem
$\therefore TM ^2= TL ^2+\square$
$\therefore TM ^2=12^2+\square$
$\therefore TM ^2=144+\square$
$\therefore TM ^2=400$
$\therefore TM =20$
If α, β are roots of quadratic equation,
Fill in the blanks with correct number
$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
→$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
In figure 3.52 , chords $\mathrm{PQ}$ and $\mathrm{RS}$ intersect at $\mathrm{T}$.
(i) Find $m$ (arc SQ) if $m \angle \mathrm{STQ}=58^{\circ}, m \angle \mathrm{PSR}=24^{\circ}$.
(ii) Verify,
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$
(iii) Prove that :
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$ for any measure of $\angle \mathrm{STQ}$.
(iv) Write in words the property in (iii).
→(i) Find $m$ (arc SQ) if $m \angle \mathrm{STQ}=58^{\circ}, m \angle \mathrm{PSR}=24^{\circ}$.
(ii) Verify,
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$
(iii) Prove that :
$\angle \mathrm{STQ}=\frac{1}{2}[m(\operatorname{arc} \mathrm{PR})+m(\operatorname{arcSQ})]$ for any measure of $\angle \mathrm{STQ}$.
(iv) Write in words the property in (iii).
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 $
Write the correct number in the given boxes from the following A. P.
3, 6, 9, 12, . . .
$\text { Here } t _1=\square, t _2=\square, t _3=\square, t _4=\square, $
$ t _2- t _1=\square, t _3- t _2=\square$
$\therefore d =\square$
→3, 6, 9, 12, . . .
$\text { Here } t _1=\square, t _2=\square, t _3=\square, t _4=\square, $
$ t _2- t _1=\square, t _3- t _2=\square$
$\therefore d =\square$
Complete the following activity to solve the quadratic equation
Complete the following table to draw the graph of $2x – 6y = 3$
→Pushpmala has invested ₹ 24,000 and purchased share of FV ₹ 20 at a premium of ₹ 4. Complete the following activity to find the number of shares she purchased.
Activity :
FV = ₹ 20
Premium = ₹ 4
MV = FV + ⬜
= 20 + ⬜
= ₹ 24
$\text { Number of shares } = \frac{\text { Total investmen }}{ MV } $
$=\frac{24,000}{\square}$
= ⬜ shares.
→Activity :
FV = ₹ 20
Premium = ₹ 4
MV = FV + ⬜
= 20 + ⬜
= ₹ 24
$\text { Number of shares } = \frac{\text { Total investmen }}{ MV } $
$=\frac{24,000}{\square}$
= ⬜ shares.
