Question
Two coins are tossed simultaneously. Complete the following activity of writing the sample space $(S)$ and expected outcomes of the events :
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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,…
From given figure, In ∆ABC, If AC = 12 cm. then AB =?
Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
→Activity: From given figure, In $\triangle ABC , \angle ABC =90^{\circ}, \angle ACB =30^{\circ}$
$\therefore \angle BAC =\square$
$\therefore \triangle ABC$ is $30^{\circ}-60^{\circ}-90^{\circ}$ triangle
$\therefore$ In $\triangle ABC$ by property of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.
$ \therefore A B=\frac{1}{2} A C \text { and } \square=\frac{\sqrt{3}}{2} A C$
$\therefore \square=\frac{1}{2} \times 12 \text { and } B C=\frac{\sqrt{3}}{2} \times 12$
$\therefore \square=6 \text { and } B C=6 \sqrt{3} $
Smita invested ₹ 12,000 to purchase shares of FV 10 at a premium of 2. Find the number of shares she purchased. Complete the given activity to get the answer:
FV = ₹ 10, premium = ₹ 2
FV = ₹ 10, premium = ₹ 2
Write the values of the following trigonometric ratios.
$\sin 30^{\circ}=\frac{1}{⬜}$
→$\sin 30^{\circ}=\frac{1}{⬜}$
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 $=(2, \square)$
To draw the graph of $4 x+5 y=19$, complete the following activity to find $y$, when $x=1$.
→Complete the following table to draw the graph of $2x – 6y = 3$
→In the adjoining figure BP ⊥ AC, CQ ⊥ AB, A – P- C, A- Q- B , then prove that △ APB and △ AQC are similar.
Some sequences are given below. Show the positions of the terms by $t_1, t_2, t_3, \ldots$
i. $7,7,7,7, \ldots$ Here $t_1=7, t_2=$⬜, $t_3=$ ⬜, ...
ii. $-2,-6,-10,-14, \ldots$ Here $t _1=-2, t _2=$⬜, $t_3=$⬜, ....
→i. $7,7,7,7, \ldots$ Here $t_1=7, t_2=$⬜, $t_3=$ ⬜, ...
ii. $-2,-6,-10,-14, \ldots$ Here $t _1=-2, t _2=$⬜, $t_3=$⬜, ....