2 Marks Questions

Question 12 Marks

Prove that $7-3 \sqrt{5}$ is an irrational number, given that $\sqrt{5}$ is an irrational number.

Answer

Assuming $7-3 \sqrt{5}$ to be a rational number.
Let $7-3 \sqrt{5}=\frac{a}{b}$ where $a$ and $b$ are integers $\& b \neq 0$
$
\Rightarrow \sqrt{5}=\frac{7 b-a}{3 b}
$
Here RHS is rational but LHS is irrational.
Therefore our assumption is wrong.
Hence, $7-3 \sqrt{5}$ is an irrational number.

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Question 22 Marks

PQRS is a trapezium with $P Q \| S R$. If $M$ and $N$ are two points on the non-parallel sides PS and QR respectively, such that MN is parallel to PQ , then show that $\frac{ PM }{ MS }=\frac{ QN }{ NR }$.

Answer

Join PR
$PQ \| SR$ and $MN \| PQ \Rightarrow MN \| SR$
In $\Delta$ PSR,
$\frac{P M}{M S}=\frac{P O}{O R}......(i)$
In $\triangle PQR$,
$\frac{P O}{O R}=\frac{Q N}{N R}......(ii)$
From (i) and (ii), $\frac{P M}{M S}=\frac{Q N}{N R})$

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Question 32 Marks

Find the value of $x$ such that,
$3 \tan ^2 60^{\circ}-x \sin ^2 45^{\circ}+\frac{3}{4} \sec ^2 30^{\circ}=2 \operatorname{cosec}^2 30^{\circ}$

Answer

$3 \tan ^2 60^{\circ}-x \sin ^2 45^{\circ}+\frac{3}{4} \sec ^2 30^{\circ}=2 \operatorname{cosec}^2 30^{\circ}$
$\Rightarrow 3(\sqrt{3})^2-x\left(\frac{1}{\sqrt{2}}\right)^2+\frac{3}{4}\left(\frac{2}{\sqrt{3}}\right)^2=2(2)^2$
$\Rightarrow 9-\frac{x}{2}+1=8$
$\Rightarrow x=4$

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Question 42 Marks

If $\cos (A+B)=\frac{1}{2}$ and $\tan (A-B)=\frac{1}{\sqrt{3}}$, where $0 \leq A+B \leq 90^{\circ}$, then find the value of $\sec (2 A-3 B)$.

Answer

$\cos (A+B)=\frac{1}{2}$
$\Rightarrow A+B=60^{\circ} \ldots \text { (i) }$
$\tan (A-B)=\frac{1}{\sqrt{3}}$
$\Rightarrow A-B=30^{\circ} \ldots \text { (ii) }$
$\text { Solving (i) and (ii), we get } A=45^{\circ} \text { and } B=15^{\circ}$
$\Rightarrow \sec (2 A-3 B)=\sec \left(90^{\circ}-45^{\circ}\right)$
$=\sec 45^{\circ}=\sqrt{2}$

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Question 52 Marks

In the given figure, PAQ and PBR are tangents to the circle with centre ' $O$ ' at the points $A$ and $B$ respectively. If T is a point on the circle such that $\angle QAT =45^{\circ}$ and $\angle TBR =65^{\circ}$, then find $\angle ATB$.

Answer

Join OA, OB and OT

Now $\angle A T O=\angle T A O=90^{\circ}-45=45^{\circ}$
and $\angle B T O=\angle T B O=90^{\circ}-65^{\circ}=25^{\circ}$
$
\Rightarrow \angle A T B=\angle A T O+\angle B T O
$
$
=45^{\circ}+25^{\circ}=70^{\circ}
$

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Question 62 Marks

Find the area of the shaded region if length of radius of each circle is $7 \ cm$ .Each circle touches the other two externally.

Answer

$\text { Side of square }=14 \ cm$
$\text { Area of shaded region } =\text { area of square }- \text { area of } 4 \text { quadrants }$
$ =14^2-4 \times \frac{22}{7} \times 7^2 \times \frac{90}{360}$
$ =(196-154)=42$
$\text{ Hence, area of shaded region }=42 \ cm^2$

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Question 72 Marks

A chord is subtending an angle of $90^{\circ}$ at the centre of a circle of radius 14 cm . Find the area of the corresponding minor segment of the circle.

Answer

$
\begin{aligned}
\text { Area of minor segment } & =\pi \times 14^2 \times \frac{1}{4}-\frac{1}{2} \times 14^2 \\
& =(154-98)=56
\end{aligned}
$
Hence, area of minor segment $=56 cm^2$

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Maths 2 Marks Questions

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