2 Marks Questions

Question 12 Marks

The perimeter of a sector of a circle of radius $5.2 \ cm$ is $16.4 \ cm$. Find the area of the sector.

Answer

Let $\text{OAB}$ be the given sector.
It is given that Perimeter of sector $\text{OAB} = 16.4 \ cm$
$\Rightarrow OA + OB +\operatorname{arc~AB}=16.4 \ cm$
$\Rightarrow 5.2+5.2+\operatorname{arc~AB}=16.4$
$\Rightarrow \operatorname{arc} AB =6 \ cm$
$\Rightarrow l=6 \ cm$
$\therefore$ Area of sector $OAB =\frac{1}{2} l r$
$=\frac{1}{2} \times 6 \times 5.2 \ cm^2$
$=15.6 \ cm^2$

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Question 22 Marks

Find the area of a quadrant of a circle , whose circumference is $22 \ cm $.

Answer

Given , Circumference $= 22 \ cm$
$\Rightarrow 2 \pi r=22$
$\Rightarrow r=\frac{7}{2}=3.5 \ cm$
$\text { Area of Circle }=\pi r^2=\frac{22}{7} \times(3.5)^2=38.5 \ cm^2$
$\text { Area of quadrant of circle }=\frac{\text { Area of circle }}{4}$
$=\frac{38.5}{4}=9.625 \ cm^2$
$\therefore$ Area of the quadrant of cicle $=9.625 \ cm^2$

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Question 32 Marks

If $\cot \theta=\frac{15}{8}$, then evaluate: $\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$

Answer

Given $\cot \theta=\frac{15}{8}$
To evaluate: $\frac{(2+2 \sin \theta)(1-\sin \theta)}{(1+\cos \theta)(2-2 \cos \theta)}$
$=\frac{2(1+\sin \theta)(1-\sin \theta)}{2(1+\cos \theta)(1-\cos \theta)}$
$=\frac{\left(1-\sin ^2 \theta\right)}{\left(1-\cos ^2 \theta\right)}=\frac{\cos ^2 \theta}{\sin ^2 \theta}=\cot ^2 \theta$
$=(\cot \theta)^2=\left(\frac{15}{8}\right)^2=\frac{225}{64}$
Hence, the value of the given expression is $\frac{225}{64}$.

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Question 42 Marks

Prove that:
$\frac{1+\tan A}{2 \sin A}+\frac{1+\cot A}{2 \cos A}=\operatorname{cosec} A+\sec A$

Answer

LHS $=\frac{1+\tan A}{2 \sin A}+\frac{1+\cot A}{2 \cos A}$
$=\frac{\cos A+\sin A}{2 \sin A \cos A}+\frac{\sin A+\cos A}{2 \cos A \sin A}$
$=\frac{2(\cos A+\sin A)}{2 \sin A \cos A}$
= cosec A + sec A
= RHS

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Question 52 Marks

In the given figure, $O$ is the centre of the circle. $\text{PA}$ and $\text{PB}$ are tangents. Show that $\text{AOBP}$ is a cyclic quadrilateral.

Answer

We know that the radius and tangent are perpendicular at their point of contact
$\because \angle \text{OBP}=\angle \text{OAP}=90^{\circ}$
Now, In quadrilateral $\text{AOBP}$
$\angle \text{APB}+\angle \text{AOB} +\angle \text{OBP} +\angle \text{OAP}=360^{\circ}$
$\Rightarrow \angle \text{APB}+\angle \text{AOB} +90^{\circ}+90^{\circ}=360^{\circ}$
$\Rightarrow \angle \text{APB} +\angle \text{AOB}=180^{\circ}$
Since, the sum of the opposite angles of the quadrilateral is $180^\circ$
Hence, $\text{AOBP}$ is a cyclic quadrilateral.

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Question 62 Marks

In figure $DE \| AC$ and $DF \| AE$. Prove that $\frac{B F}{F E}=\frac{B E}{E C}$

Answer

In $\triangle A B E$, we have $D F \| A E$, then
$\frac{B D}{A D}=\frac{B F}{F E}[$ By BPT $] \ldots \ldots$ (i)
In $\triangle A B C$, we have $D E \| A C$, then
$\frac{B D}{A D}=\frac{B E}{E C}[$ By BPT $] \ldots \ldots$ (2)
From (i) and (2), We get
$\frac{B F}{F E}=\frac{B E}{E C}$

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Question 72 Marks

Find the $\text{HCF}$ and $\text{tLCM}$ of $90$ and $144$ by the method of prime factorization.

Answer

The prime factorization of $90$ and $140$ are as follows
$90=2 \times 3 \times 3 \times 5=2 \times 3^2 \times 5$
$144=2 \times 2 \times 2 \times 2 \times 3 \times 3=2^4 \times 3^2$
Hence $\text{HCF }(90,144)=2 \times 3^2=18$
and $\text{LCM }(90,144)=2^4 \times 3^2 \times 5=720$

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