Question
Complete the following activity to solve the simultaneous equations
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Similar questions
Write the values of the following trigonometric ratios.
$\cos 45^{\circ}=\frac{⬜}{⬜}$
→$\cos 45^{\circ}=\frac{⬜}{⬜}$
Write the correct number in the given boxes from the following A. P.
1, 8, 15, 22, . . .
Here
a = ⬜, $t _1$ = ⬜, $t _2$ = ⬜, $t _3$ = ⬜
$t _2- t _1$ = ⬜ - ⬜ = ⬜
$t _3- t _2$ = ⬜ - ⬜ = ⬜
∴ d = ⬜
→1, 8, 15, 22, . . .
Here
a = ⬜, $t _1$ = ⬜, $t _2$ = ⬜, $t _3$ = ⬜
$t _2- t _1$ = ⬜ - ⬜ = ⬜
$t _3- t _2$ = ⬜ - ⬜ = ⬜
∴ d = ⬜
From given figure, In ∆ABC, If ∠ABC = 90° ∠CAB=30°, AC = 14 then for finding value of AB and BC, complete the following activity.
Activity: In $\triangle ABC$, If $\angle ABC =90^{\circ}, \angle CAB =30^{\circ}$
$
\therefore \angle B C A=\square
$
By theorem of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,
$\therefore \square=\frac{1}{2} AC$ and $\square=\frac{\sqrt{3}}{2} AC$
$\therefore BC =\frac{1}{2} \times \square$ and $AB =\frac{\sqrt{3}}{2} \times 14$
$\therefore BC =7$ and $AB =7 \sqrt{3}$
→Activity: In $\triangle ABC$, If $\angle ABC =90^{\circ}, \angle CAB =30^{\circ}$
$
\therefore \angle B C A=\square
$
By theorem of $30^{\circ}-60^{\circ}-90^{\circ}$ triangle,
$\therefore \square=\frac{1}{2} AC$ and $\square=\frac{\sqrt{3}}{2} AC$
$\therefore BC =\frac{1}{2} \times \square$ and $AB =\frac{\sqrt{3}}{2} \times 14$
$\therefore BC =7$ and $AB =7 \sqrt{3}$
Find distance between 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 distance formula,
$ d(A, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$\therefore d(A, B)=\sqrt{\square+[(-7)+\square]^2}$
$\therefore d(A, B)=\sqrt{\square}$
$\therefore d(A, B)=\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 distance formula,
$ d(A, B)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
$\therefore d(A, B)=\sqrt{\square+[(-7)+\square]^2}$
$\therefore d(A, B)=\sqrt{\square}$
$\therefore d(A, B)=\square $
Complete the following activity, to find the two-digit numbers which are divisible by 6.
The first term and the common difference of an A.P. are 6 and 3 respectively. Find $S_{27}$.
→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
$\therefore TM ^2= TL ^2+\square$
$\therefore TM ^2=12^2+\square$
$\therefore TM ^2=144+\square$
$\therefore TM ^2=400$
$\therefore TM =20$
Complete the following table to draw the graph of $2x – 6y = 3$
→To check the rule for the terms of the sequence look at the arrangements and fill the empty boxes suitably.
-1, -1.5, -2, -2.5,…
-1, -1.5, -2, -2.5,…