2 Marks Questions

Question 12 Marks

A sector of a circle of radius $4 \ cm$ contains an angle of $30^{\circ}$. Find the area of the sector.

Answer

$\text { Radius of cirlce }=4 \ cm$
$\theta=30^{\circ}$
$\therefore \text { Area of sector }=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{30^{\circ}}{360^{\circ}} \times \pi \times 4 \times 4$
$=\frac{4 \pi}{3} \ cm^2$

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Question 22 Marks

Verify that if $\tan ^2 \theta+\sin \theta=\cos ^2 \theta$ is an identity or not.

Answer

$\tan ^2 \theta+\sin \theta=\cos ^2 \theta$
Taking $\theta=45^{\circ}$, we have
$\text { L.H.S. }=\tan ^2 45^{\circ}+\sin 45^{\circ}$
$=(1)^2+\frac{1}{\sqrt{2}}$
$=1+\frac{1}{\sqrt{2}}$
$=\frac{\sqrt{2}+1}{\sqrt{2}}$
$\text { R.H.S. }=\cos ^2 45$
$=\left(\frac{1}{\sqrt{2}}\right)^2$
$=\frac{1}{2}$
$\Rightarrow \text { LH.S. } \neq \text { R.H.S. }$
Hence given expression is not an identity.

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Question 32 Marks

Four cows are tethered at the four corners of a square field of side $50 m$ such that each can graze the maximum unshared area. What area will be left ungrazed? $[$Take $\pi=3.14].$

Answer

Shaded area $=$ area of square $-4 ($area of sector$)$
$=\left[(50 \times 50)-\frac{4 \times \pi \times(25)^2 \times 90}{360}\right] m^2$
$=[2500-3.14 \times 25 \times 25] m^2$
$=[2500-1962.5] m^2$
$=537.5 m^2$

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Question 42 Marks

If $\sin X+\sin ^2 X=1,$ prove that $\cos ^2 X+\cos ^4 X=1$.

Answer

Given $\sin X+\sin ^2 X=1......... (i)$
$\Rightarrow \sin X=1-\sin ^2 X=\cos ^2 X .........(ii)$
Now we show that $\cos ^2 X+\cos ^4 X=1$
$\text { L.H.S }=\cos ^2 X+\cos ^4 X$
$=1-\sin ^2 X+\left(1-\sin ^2 X\right)^2 \text { [Using (ii)] }$
$=\sin X+\sin ^2 X[\text { Using (ii)] }$
$=1[\text { Using (i)] }$
$=\text { R.H.S }$

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Question 52 Marks

In figure, $\triangle ABC$ is circumscribing a circle. Find the length of $BC$ .

Answer

Given,

$A R=4 \ cm$
Also $, A R=A Q $
$\Rightarrow A Q=4 \ cm$
Now $, Q C=A C-A Q$
$=11 \ cm-4 \ cm=7 \ cm \ldots \text { (i) }$
Also, $B P=B R$
$\therefore BP=3 \ cm$ and $ PC=QC$
$\therefore PC=7 \ cm\ [$ From $ (i)]$
$B C=B P+P C$
$=3 \ cm+7 \ cm$
$=10 \ cm$

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Question 62 Marks

In Figure, $\triangle ODC \sim \triangle OBA , \angle BOC =125^{\circ}$ and $\angle CDO =70^{\circ}$. Find $\angle DOC , \angle DCO$ and $\angle OAB$.

Answer

From the given figure,
$\angle D O C+125^{\circ}=180^{\circ} \ [$linear pair$]$
$\angle D O C=55^{\circ}$
Now, in $\triangle DOC,$
$\angle D C O+\angle O D C+\angle D O C=180^{\circ} \ [$angle sum property of a triangle$]$
$\angle D C O+70^{\circ}+55^{\circ}=180^{\circ}$
$\angle D C O=55^{\circ}$
Now, $\triangle O D C \cong \triangle O B A \ [$given$]$
$\therefore \angle O A B=\angle O C D=55^{\circ}$

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Question 72 Marks

The difference of the square of two numbers is $45$. The square of the smaller number is $4$ times the larger number. Find the number.

Answer

Let the larger number be $x$ and the smaller number be $y$.
Them, $x=x^2-y_2^2=45 \ldots \ldots(i)$
and $ y^2=4 x \ldots \ldots \text { (ii ) }$
substituting $ y ^2=5 x $ in $(i),$ we have
$x^2-4 x=45$
$\Rightarrow x^2-4 x=45=0$
$\Rightarrow x^2-9 x+5 x-45=0$
$\Rightarrow x(x-9)+5(x-9)=0$
$\Rightarrow(x-0)(x+5)=0$
$\Rightarrow$ either $ x-9=0 $ or $ x+5=0$
$\Rightarrow x=9$ or $x=-5$
$\therefore $ smaller number is $ \sqrt{36} $ or $ \sqrt{-20}\ [$From $(ii)]$
$\Rightarrow $ smaller number is $= \pm 6\ [\because \sqrt{-20} $ is not real$]$
$\therefore$ Larger number $=9$ and smaller number $= \pm 6$

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