2 Marks Questions

Question 12 Marks

If $\sqrt{2} \sin \theta=1$, find the value of $\sec ^2 \theta-\operatorname{cosec}^2 \theta$.

Answer

$\text { Given, } \sqrt{2} \sin \theta=1$
$\sin \theta=\frac{1}{\sqrt{2}}$
$\sin 45^{\circ}=\frac{1}{\sqrt{2}}$
$\therefore \sin \theta=\sin 45^{\circ}$
$\theta=45^{\circ}$
$\text { put } \theta=45^{\circ} i n$
$\sec ^2 \theta-\operatorname{cosec}^2 \theta$
$=\sec ^2 45^{\circ}-\operatorname{cosec}^2 45^{\circ}$
$=(\sqrt{2})^2-(\sqrt{2})^2$
$=2-2$
$=0$

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Question 22 Marks

In Figure, a chord $AB$ of a circle of radius $10 \ cm$ subtends a right angle at the centre

Find
$i$. Area of sector $\text{OAPB}$
$ii$. Area of minor segment $\text{APB}. \ ($Use $\pi=3.14)$

Answer

$i$. Area of sector $\text{OAPB} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 100$
$=78.5 \ cm^2$
$ii$. Area minor segment $\text{APB} =$ Area sector $\text{OAPB} \ - $ Area $\triangle OAB$
$=78.5-\frac{1}{2} \times 100$
$=28.5 \ cm^2$

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Question 32 Marks

$\Lambda B C D$ is a flower bed. If $O \Lambda=21 m$ and $OC =14 m$, find the area of the bed.

Answer

We have, $OA = R =21 m$ and $OC = r =14 m$
$\therefore$ Area of the flower bed $=$ Area of a quadrant of a circle of radius $R -$ Area of a quadrant of a circle of radius r
$=\frac{1}{4} \pi R^2-\frac{1}{4} \pi r^2$
$=\frac{\pi}{4}\left(R^2-r^2\right)$
$=\frac{1}{4} \times \frac{22}{7}\left(21^2-14^2\right) \ cm^2$
$=\left\{\frac{1}{4} \times \frac{22}{7} \times(21+14)(21-14)\right\} m^2$
$=\left\{\frac{1}{4} \times \frac{22}{7} \times 35 \times 7\right\} m^2$
$=192.5 m^2$

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Question 42 Marks

If $x=a \cos ^3 \theta, y=b \sin ^3 \theta$, prove that $\left(\frac{x}{a}\right)^{2 / 3}+\left(\frac{y}{b}\right)^{2 / 3}=1$

Answer

We have,
$x=a \cos ^3 \theta, y=b \sin ^3 \theta$
$\frac{x}{a}=\cos ^3 \theta \text { and } \frac{y}{b}=\sin ^3 \theta$
$\text { L.H.S }=\left[\frac{x}{a}\right]^{2 / 3}+\left[\frac{y}{b}\right]^{2 / 3}$
$=\left[\cos ^3 \theta\right]^{2 / 3}+\left[\sin ^3 \theta\right]^{2 / 3}\left[\because\left(a^m\right)^n=a^{m \times n}\right]$
$=\cos ^2 \theta+\sin ^2 \theta$
$=1\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]$
$=\text { R.H.S }$
Hence proved.

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Question 52 Marks

A quadrilateral $ABCD$ is drawn to the circumference of a circle. Prove that: $AB + CD = AD + BC$

Answer

Let the sides of the quadrilateral $A B C D$ touch the circle at $P, Q, R$ and $S$.
Since, the lengths of the tangents from an external point to a given circle are equal.
$\therefore AP=AS$
$\Rightarrow BP=BQ$
$CR=CQ$
$\Rightarrow DR=DS$
Adding, $( AP + BP )+( CR + DR )$
$=( BQ + CQ )+( AS + DS )$
$\Rightarrow AB+CD=BC+AD.$
Hence proved

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Question 62 Marks

In the given figure $,XZ$ is parallel to $BC . AZ =3 \ cm, ZC =2 \ cm, BM =3 \ cm$ and $MC =5 \ cm$. Find the length of $XY$.

Answer

Given that,
In the figure the triangle $\text{ABC}$
$XZ \| BC$ and the length of $AZ =3 \ cm, ZC =2 \ cm, BM =3 \ cm$ and $MC =5 \ cm$.
From $\triangle ABC$ and $\triangle AXZ$
$\angle AXZ =\angle ABC\ [$by corresponding angles$]$
$\angle AZX =\angle ACB \ [$by corresponding angles$]$
By basic proportionality theorem $\triangle ABC$ and $\triangle AXZ$ are similar.
So,
$\frac{Y Z}{M C}=\frac{A Z}{Z C}$
$\frac{Y Z}{5}=\frac{3}{2}$
$Y Z=\frac{15}{2}$
Then,
$\frac{X Z}{B C}=\frac{A Z}{Z C}$
$\frac{X Y+Y Z}{B M+M C}=\frac{A Z}{Z C}$
$\frac{X Y+\frac{15}{2}}{3+5}=\frac{3}{2}$
$X Y+\frac{15}{2}=\frac{24}{2}$
$X Y=\frac{9}{2}=4.5 \ cm$

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Question 72 Marks

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

Answer

Let us assume that $2+3 \sqrt{3}$ is a rational number $2+3 \sqrt{3}=\frac{p}{q} ; p , q$ are integers and $q \neq 0$
$\Rightarrow \sqrt{3}=\frac{p-2 q}{3 q}$
RHS is rational but LHS is irrational
$\therefore$ Our assumption is wrong. Hence $2+3 \sqrt{3}$ is an irrational number.

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Maths 2 Marks Questions

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