2 Marks Questions

Question 12 Marks

If a chord of a circle of radius $10 \ cm$ subtends an angle of $60^{\circ}$ at the centre of the circle, find the area of the corresponding minor segment of the circle. $($Use $\pi=3.14$ and $\sqrt{3}=1.73 )$

Answer

Area of minor segment $=\frac{3.14 \times(10)^2 \times 60^{\circ}}{360^{\circ}}-\frac{1}{2} \times(10)^2 \times \frac{\sqrt{3}}{2}$
$=\frac{314}{6}-\frac{173}{4}$
$=9 \frac{1}{12} \text { or } 9.08$
Hence, area of minor segment is $9.08 \ cm ^2$

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Question 22 Marks

In a circle of radius $21 \ cm,$ an arc subtends an angle of $60^{\circ}$ at the centre. Find the area of the sector formed by the arc. Also, find the length of the arc.

Answer

Radius $(r)$ of circle $= 21 \ cm$
Angle subtended by the given arc = $60^{\circ}$
Length of an arc of a sector of angle $\theta=\frac{\theta}{360^{\circ}} \times 2 \pi r$

Area of sector $\text{OACB}=\frac{60^{\circ}}{360^{\circ}} \times \pi r^2$
$=\frac{1}{6} \times \frac{22}{7} \times 21 \times 2$
$=231 \ cm^2$
Length of arc $\text{ACB}=\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21$
$=\frac{1}{6} \times 2 \times 22 \times 3$
$=22 \ cm$

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Question 32 Marks

If $m \sin A+n \cos A=p$ and $m \cos A-n \sin A=q$, prove that $m^2+n^2=p^2+q^2$

Answer

Given
$m \sin A + n \cos A = p......(1)$
$m \cos A - n \sin A = q........(2)$
Squaring $(1)$ and $(2)$ we get,
$m^2 \sin ^2 A+n^2 \cos ^2 A+2\ m n \sin A \cos A=p^2 \ldots \ldots(3)$
$m^2 \cos ^2 A+n^2 \sin ^2 A-2\ m n \sin A \cos A=q^2 \ldots \ldots . \text { (4) }$
Adding $(3)$ and $(4)$ we get,
$ m ^2\left(\sin ^2 A+\cos ^2 A\right)+ n ^2\left(\sin ^2 A+\cos ^2 A\right)= p ^2+ q ^2$
$\Rightarrow m^2+ n ^2= p ^2+ q ^2\left[\because \sin ^2 A+\cos ^2 A=1\right]$

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Question 42 Marks

Prove that: $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$

Answer

$=\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
$=\sqrt{\frac{(1+\sin A)^2}{1-\sin ^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos ^2 A}}$
$=\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$
$=\sec A +\tan A$
$=\text { RHS }$

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Question 52 Marks

In figure $, PQ$ is a tangent from an external point $P$ to a circle with centre $O$ and $OP$ cuts the circle at $T$ and $\text{QOR}$ is a diameter. If $\angle POR =130^{\circ}$ and $S$ is a point on the circle, find $\angle 1+\angle 2$.

Answer

$\angle 2=\frac{1}{2} \angle R O T \ ($Angle subtended at the center by same arc$)$
$\angle 2=\frac{1}{2} \times 130^{\circ}=65^{\circ}$
$\angle R O T=\angle 1+\angle P Q O$
$\angle 1=130^{\circ}-90^{\circ}=40^{\circ}$
$\therefore \angle 1+\angle 2=65^{\circ}+40^{\circ}=105^{\circ}$

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Question 62 Marks

If $D$ and $E$ are points on sides $\text{AB}$ and $\text{AC}$ respectively of a $\triangle \text{ABC}$ such that $\ce{DE \| BC}$ and $\text{BD = CE}$. Prove that $\Delta \text{ABC}$ is isosceles.

Answer

We have, $\ce{DE \| BC}$

Therefore, by $\ce{BPT, \frac{AD}{BD}=\frac{AE}{EC}}$
$\Rightarrow \text{AD=AE}$
Adding $\text{DB}$ on both sides
$\Rightarrow \text{AD + DB = AE + DB}$
$\Rightarrow \text{AD + DB = AE + EC} [\therefore \text{BD = CE} ]$
$\Rightarrow \text{AB = AC}$
$\therefore \triangle \text{ABC}$ is isosceles triangle.

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Question 72 Marks

Prove that $5+2 \sqrt{3}$ is an irrational number, given that $\sqrt{3}$ is an irrational number.

Answer

Given
$\sqrt{3}$ is an irrational number
Let $5+2 \sqrt{3}$ is a rational number
$\therefore$ we can write $5+2 \sqrt{3}=\frac{p}{q}$,
where $p$ and $q$ are integers
$\Rightarrow 2 \sqrt{3}=\frac{p}{q}-5=\frac{p-5 q}{q}$
$\sqrt{3}=\frac{p-5 q}{2 q}$
Here, $\frac{p-5 q}{2 q}$ is a rational number
So, $\sqrt{3}$ is also a rational number.
But it is given that $\sqrt{3}$ is irrational number.
$\Rightarrow$ our assumption was wrong
$\Rightarrow 5+2 \sqrt{3}$ is an irrational number.

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Maths 2 Marks Questions

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