2 Marks Questions

Question 12 Marks

In the given figure, $\text{AB}$ and $\text{CD}$ are the diameters of a circle with centre $\text{O}$, perpendicular to each other. If $\text{OA}=7 \ cm$ find the area of the shaded region.

Answer

Radius of circle $( r )=\text{OA}=7 \ cm$.
Area of the semicircle $=\frac{1}{2} \times \pi r^2$
$=\frac{1}{2} \times \frac{22}{7} \times 7 \times 7$
$=11 \times 7$
$=77 \ cm^2$
Area of $\triangle \text{ABC} =\frac{1}{2} \times$ base $\times$ height
$=\frac{1}{2} \times 14 \times 7$
$=49 \ cm^2$
$\therefore$ Area of the shaded portion $=$ Area of semicircle $-$ Area of the $\triangle \text{ABC}$
$=77-49$
$=28 \ cm^2$

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Question 22 Marks

If $(3 \sin \theta+5 \cos \theta)=5$, prove that $(5 \sin \theta-3 \cos \theta)= \pm 3$

Answer

We have
$(3 \sin \theta+5 \cos \theta)^2+(5 \sin \theta-3 \cos \theta)^2$
$=9\left(\sin ^2 \theta+\cos ^2 \theta\right)+25\left(\sin ^2 \theta+\cos ^2 \theta\right)=(9+25)=34$
Therefore, $(3 \sin \theta+5 \cos \theta)^2+(5 \sin \theta-3 \cos \theta)^2=34$
$\Rightarrow 5^2+(5 \sin \theta-3 \cos \theta)^2=34[\because 3 \sin \theta+5 \cos \theta=5]$
$\Rightarrow(5 \sin \theta-3 \cos \theta)= \pm 3 [$taking square root on each side$]$
Hence, $(5 \sin \theta-3 \cos \theta)= \pm 3$.

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Question 32 Marks

Calculate the area of the shaded region common between two quadrants of circles of radius $7 \ cm$ each $($as shown in Figure$).$

Answer

Area of Shaded Region
$=2(\text {Area of one sector ABPD) }- \text {Area of square ABCD}$
$=2\left(\frac{90^{\circ} \times \pi \times 7^2}{360^{\circ}}\right)-7 \times 7$
$=28 \ cm^2$

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Question 42 Marks

Prove that: $\frac{\cos A}{1-\tan A}+\frac{\sin ^2 A}{\sin A-\cos A}=\sin A +\cos A$.

Answer

$\text { L.H.S }=\frac{\cos A}{1-\tan A}+\frac{\sin ^2 A}{\sin A-\cos A}$
$=\frac{\cos ^2 A}{\cos A-\sin A}+\frac{\sin ^2 A}{\cos A-\sin A}\left[\text { by putting } \tan A=\frac{\sin A}{\cos A}\right]$
$=\frac{\cos ^2 A-\sin ^2 A}{\cos A-\sin A}$
$=\frac{(\cos A-\sin A)(\cos A+\sin A)}{\cos A-\sin A}$
$=\cos A +\sin A$
$=\text { R.H.S }$

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Question 52 Marks

Prove that the tangents at the end of a chord of a circle make equal angles with the chord.

Answer

$\text { In } \triangle ADB $ and $ \triangle ADC ,$
$BD = DC$ and $\angle ADB =\angle ADC =90^{\circ}$
$AD = AD [$ Common $]$
$\therefore \triangle ADB \cong \triangle ADC [SAS]$
$\therefore \angle ABD =\angle ACD [$ By $\text { CPCT }]$

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Question 62 Marks

In Fig. AD and CE are two altitudes of $\triangle ABC$ intersect each other at point F . Prove that $\Delta F D C \sim \Delta B E C$

Answer

Given Altitude $A D$ and $C E$ of $\triangle A B C$ intersects each other at the point $F$.
To Prove: $\triangle F D C \sim \triangle B E C$

Proof: In $\triangle$'s FDC and BEC, we have
$\angle FDC =\angle BEC =90^{\circ}[\because A D \perp B C$ and $C E \perp A B]$
$\angle FCD =\angle ECB [$ Common angle $]$
Thus, by AA-criterion of similarity, we obtain $\triangle F D C \sim \triangle B E C$.

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Question 72 Marks

2002 cartons of Lassi bottles and 2618 cartons of Frooti are to be stacked in a storeroom. If each stack is of the same height and is to contain cartons of the same type of bottles, what would be the greatest number of cartons each stack would have?

Answer

In order to get the result we have to find the HCF of 2002 and 2618 . There prime factors are,
$2002=2 \times 7 \times 11 \times 13$ and
$2618=2 \times 7 \times 11 \times 17$
Hence HCF $=2 \times 7 \times 11=154$

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Maths 2 Marks Questions

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