2 Marks Questions

Question 12 Marks

From a circular piece of carboard of radius $3 \ cm$ two sectors of $90^\circ$ have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters. $($Take $\pi=22 / 7 ).$

Answer

Radius of the circular piece of cardboard$(r) =3 \ cm$
$\therefore$ Two sectors of $90^{\circ}$ each have been cut off
$\therefore$ We get a semicircular cardboard piece
$\therefore$ Perimeter of arc $ACB$
$=\frac{1}{2}(2 \pi r)=\pi r$
$=\frac{22}{7} \times 3$
$=\frac{66}{7}$
$=9.428 \ cm$

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Question 22 Marks

Prove that: $\tan ^2 \theta-\sin ^2 \theta=\tan ^2 \theta \sin ^2 \theta$

Answer

$\tan ^2 \theta-\sin ^2 \theta=\tan ^2 \theta \sin ^2 \theta$
$\text { R.H.S. }=\tan ^2 \theta \cdot \sin ^2 \theta$
$=\tan ^2 \theta\left(1-\cos ^2 \theta\right)\left[\because \sin ^2 \theta=1-\cos ^2 \theta\right]$
$=\tan ^2 \theta-\tan ^2 \theta \cos ^2 \theta$
$=\tan ^2 \theta-\frac{\sin ^2 \theta}{\cos ^2 \theta} \cdot \cos ^2 \theta\left[\because \tan ^2 \theta=\frac{\sin ^2 \theta}{\cos ^2 \theta}\right]$
$=\tan ^2 \theta-\sin ^2 \theta$
$=\text { L.H.S. }$
Hence proved.

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Question 32 Marks

Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii $4 \ cm$ and $3 \ cm.$

Answer

Let the radius of the large circle be $R .$
Then, we have
Area of large circle of radius $R =$ Area of a circle of radius $4 \ cm+$ Area of circle of radius $3 \ cm$
$\Rightarrow \pi R^2=\left(\pi \times 4^2+\pi \times 3^2\right)$
$\Rightarrow \pi R^2=(16 \pi+9 \pi)$
$\Rightarrow \pi R^2=25 \pi$
$\Rightarrow R^2=25$
$\Rightarrow R=5 \ cm$
$\Rightarrow \text { Diameter }=2 R=10 \ cm$

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Question 42 Marks

Prove the trigonometric identity: $\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta$

Answer

$\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}=2 \sec ^2 \theta$
$\text { L.H.S. }=\frac{1}{1-\sin \theta}+\frac{1}{1+\sin \theta}$
$=\frac{1+\sin \theta+1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}=\frac{2}{1-\sin ^2 \theta}$
$=\frac{2}{\cos ^2 \theta}\left[\because 1-\sin ^2 \theta=\cos ^2 \theta\right]$
$=2 \sec ^2 \theta\left[\because \sec (x)=\frac{1}{\cos (x)}\right]$
$=\text { R.H.S. Proved }$

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Question 52 Marks

The tangent at a point $A$ of a circle with centre $O$ intersects the diameter $PQ$ of the circle$($when extended$)$ at the point $B.$ If $\angle \text{BAQ}=105^{\circ}$, find $\text{APQ}$.

Answer

According to the question,

$\text{PAQ}=90^{\circ}[$Angle in semicircle$]$
$\therefore \angle \text{APQ} +\angle 1=90^{\circ} [$ Sum of acute angels of right angle $\triangle ]$
$\left.\Rightarrow \angle \text{APQ}+\angle 2=90^{\circ} \text { [OA = OQ} \therefore \angle 1=\angle 2\right]$
$\Rightarrow \angle \text{APQ} +(\angle \text{BAQ} -\angle \text{BAO} )=90^{\circ}$
$\Rightarrow \angle \text{APQ}+\left(105^{\circ}-90^{\circ}\right)=90^{\circ}[\because OA \perp AB ]$
$\Rightarrow \angle \text{APQ} =90^{\circ}-15^{\circ}=75^{\circ}$

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Question 62 Marks

$\triangle ABD$ is a right triangle right $-$ angled at $A$ and $AC \perp BD$. Show that $\frac{A B^2}{A C^2}=\frac{B D}{D C}$

Answer

Given: $\triangle ABD$ is a right triangle right angled at $A$ and $AC \perp BD$.
To Prove: $\frac{A B^2}{A C^2}=\frac{B D}{D C}$

Proof: We know that if a perpendicular is drawn from the vertex of the right angle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
So, $\triangle B A D \sim \triangle B C A............(i)$
and $\triangle A C B \sim \triangle D C A.............(ii)$
If two triangles are similar, then the ratio of their corresponding sides are equal.
$\frac{B A}{B C}=\frac{B D}{B A}\ [$from $(i)]$
$BA^2=BC \times BD........(iii)$
Also, $\frac{A C}{D C}=\frac{B C}{A C} \ [$from $(ii)]$
$AC ^2= DC \times BC ........... (iv)$
Hence $\frac{A B^2}{A C^2}=\frac{C B \times B D}{D C \times B C}$
$\frac{A B^2}{A C^2}=\frac{B D}{D C}$
Hence proved.

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Question 72 Marks

Find by prime factorisation the $\text{LCM}$ of the numbers $18180$ and $7575.$ Also, find the $\text{HCF}$ of the two numbers.

Answer

$18180=2^2 \times 3^2 \times 5 \times 101$
$7575=3 \times 5^2 \times 101$
$\text{LCM}=2^2 \times 3^2 \times 5^2 \times 101=90900$
$\text{HCF}=3 \times 5 \times 101=1515$

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Maths 2 Marks Questions

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