2 Marks Questions

Question 12 Marks

Find the area of the minor and the major sectors of a circle with radius $6 \ cm ,$ if the angle subtended by the minor arc at the centre is $60^{\circ}. ($Use $\pi=3.14 )$

Answer

$\text { Area of minor sector }=\frac{3.14 \times(6)^2 \times 60^{\circ}}{360^{\circ}}$
$=18.84$
Hence, area of minor sector is $18.84 \ cm^2$
Area of major sector $=$ Area of circle $-$ Area of minor sector
$=3.14 \times(6)^2-18.84$
$=94.2$
Hence, area of major sector is $94.2 \ cm^2$

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Question 22 Marks

A chord of a circle of radius $10 \ cm$ subtends a right angle at the centre. Find the area of the minor segment. $[$Use $\pi=3.14].$

Answer

Let $O$ be the centre of the circle and $AB$ be the chord.
Now, Area of the minor segment $=$ Area of the sector $\ce{OACBO}-$ Area of $\triangle AOB$
$=3.14 \times 10 \times 10 \times \frac{90}{360}-\frac{1}{2} \times 10 \times 10$
$=78.5-50$
$=28.5 \ cm^2$

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Question 32 Marks

Prove that $(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)=\sec \theta+\operatorname{cosec} \theta$.

Answer

$\text { L.H.S. }=(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)$
$=(\sin \theta+\cos \theta) \times \frac{1}{\sin \theta \cos \theta}\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}$
$=\frac{\sin \theta}{\cos \theta \sin \theta}+\frac{\cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\operatorname{cosec} \theta$
$=\text { R.H.S. }$
Hence proved.

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Question 42 Marks

In $\triangle \text{ABC}$, right$-$angled at $\text{B , AB} =5 \ cm$ and $\angle \text{ACB} =30^{\circ}$. Determine the lengths of sides $\text{BC}$ and $\text{AC}.$

Answer

Given $AB =5 \ cm$
$ \angle \text{ACB}=30^{\circ} $

According to diagram,
$\tan C=\frac{\text { side opposite to angle } C}{\text { side adjacent to angle } C}$
$\tan 30^{\circ}=\frac{A B}{B C}$
$\frac{1}{\sqrt{3}}=\frac{5}{B C}$
$BC=5 \sqrt{3} \ cm$
$\sin C=\frac{\text { side of angle } C}{\text { hypotenuse }}$
$\sin 30^{\circ}=\frac{A B}{A C}$
$\frac{1}{2}=\frac{5}{A C}$
$AC=10 \ cm$

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Question 52 Marks

From a point $P, 10 \ cm$ away from the centre of a circle, a tangent $PT$ of length $8 \ cm$ is drawn. Find the radius of the circle.

Answer

Let $O$ be the centre of the given circle.
Then, $OP =10 \ cm$.
Also, $PT =8 \ cm$.
Join $OT$.
Now $, PT$ is a tangent at $T$ and $OT$ is the radius through the point of contact $T$ .
$\therefore O T \perp P T$
In the right $\triangle O T P$ we have
$OP^2=OT^2+PT^2 \ [$by Pythagoras' theorem$]$
$\Rightarrow OT=\sqrt{O P^2-P T^2}$
$=\sqrt{(10)^2-(8)^2} \ cm$
$=\sqrt{36} \ cm=6 \ cm$
Hence, the radius of the circle is $6 \ cm$ .

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Question 62 Marks

State which pairs of triangles in the given figure are similar? Also, state the similarity criterion used.

Answer

Here, $\frac{A B}{D F}=\frac{4}{6}=\frac{2}{3}, \frac{B C}{E F}=\frac{5}{7.5}=\frac{2}{3}, \frac{A C}{D E}=\frac{3}{4.5}=\frac{2}{3}$
As, $\frac{A B}{D F}=\frac{B C}{E F}=\frac{A C}{D E}$
So, $\triangle A B C \sim \triangle D F E$ [by SSS similarity criterion]
Hence ABC and DFE are similar triangles, but no other pairs of triangles in the given figure are similar.

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Question 72 Marks

Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons.

Answer

$
\frac{175}{15}=11.667
$
Hence 175 is not divisible by 15
But LCM of two numbers should be divisible by their HCF.
$\therefore$ Two numbers cannot have their HCF as 15 and LCM as 175.

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Maths 2 Marks Questions

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