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x2 + 7x + 5 = 0
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Write the values of the following trigonometric ratios.
$\cos 30^{\circ}=\frac{⬜}{⬜}$
→$\cos 30^{\circ}=\frac{⬜}{⬜}$
In figure 1.66, seg PQ || seg DE, A(Δ PQF) = 20 units, PF = 2 DP, then find A(DPQE) by completing the following activity.
A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x
$DF = DP +\square=\square+\square=3 x$
In Δ FDE and Δ FPQ,
∠FDE ≅ ∠ .......... corresponding angles
∠FED ≅ ∠ ......... corresponding angles
∴ Δ FDE ~ Δ FPQ .......... AA test
$\therefore \frac{ A (\Delta FDE )}{ A (\Delta FPQ )}=\frac{\square}{\square}=\frac{(3 x )^2}{(2 x )^2}=\frac{9}{4}$
$A (\Delta FDE )=\frac{9}{4} A(\Delta FPQ )=\frac{9}{4} \times \square=\square$
$A (\square DPQE )= A (\Delta FDE )- A (\Delta FPQ )$
$\quad=\square-\square$
$\quad=\square$
→A(Δ PQF) = 20 units, PF = 2 DP, Let us assume DP = x. ∴ PF = 2x
$DF = DP +\square=\square+\square=3 x$
In Δ FDE and Δ FPQ,
∠FDE ≅ ∠ .......... corresponding angles
∠FED ≅ ∠ ......... corresponding angles
∴ Δ FDE ~ Δ FPQ .......... AA test
$\therefore \frac{ A (\Delta FDE )}{ A (\Delta FPQ )}=\frac{\square}{\square}=\frac{(3 x )^2}{(2 x )^2}=\frac{9}{4}$
$A (\Delta FDE )=\frac{9}{4} A(\Delta FPQ )=\frac{9}{4} \times \square=\square$
$A (\square DPQE )= A (\Delta FDE )- A (\Delta FPQ )$
$\quad=\square-\square$
$\quad=\square$
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 $t _1=\square, t _2=\square, t _3=\square, \ldots$
$\therefore a =\square, d =\square$
Find $\frac{A(\triangle A B C)}{A(\triangle A P Q)}$
Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$
→Solution :
In ∆ABC, BC is the base and AR is the height.
In ∆APQ, PQ is the base and AR is the height.
[The ratio of areas of two triangles is equal to the ratio of the product of their bases and corresponding heights]
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ \times ⬜ }{ ⬜ \times ⬜ }$
$\therefore \frac{ A (\triangle ABC )}{ A (\triangle APQ )}=\frac{ ⬜ }{⬜}$
If ΔABC ~ ΔPQR, A (ΔABC) = 80, A(ΔPQR) = 125, then fill in the blanks.
$\frac{ A (\triangle ABC )}{ A (\Delta \ldots)}=\frac{80}{125}$
$\therefore \frac{ AB }{ PQ }$ = $[-]$
→$\frac{ A (\triangle ABC )}{ A (\Delta \ldots)}=\frac{80}{125}$
$\therefore \frac{ AB }{ PQ }$ = $[-]$
Complete the following table:
| Sr. No. | Face Value | Type | Market Value |
| (i) | ₹ 100 | Premium ₹ 25 | |
| (ii) | At par | ₹ 175 | |
| (iii) | ₹ 100 | Discount ₹ 40 |
Fifty cards bearing numbers 1 to 50 are placed in a box. One card is drawn at random. Complete the following activity to find the probability of the event A that the number on the card is divisible by 11.
Fill in the blanks with correct number
$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
→$\left|\begin{array}{ll}
3 & 2 \\
4 & 5
\end{array}\right|=3 \times \square-\square \times 4=\square-8=\square$
Complete the following activity to find the number of natural numbers from 1 to 171 which are divisible by 5 .
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).
