2 Marks Questions

Question 12 Marks

The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 9 A.M. and 9.35 A.M.

Answer

We know that:
Angle described by the minute hand in 60 minutes$=360^{\circ}$
Therefore,Angle described by the minute hand in one minute =$\frac{360}{60}=6^{\circ}$
Angle described by the minute hand in 35 minutes $=(6 \times 35)^{\circ}=210^{\circ}$
Area swept by the minute hand in 35 minutes = Area of a sector of angle 210° in a circle of radius 10 cm
$=\left\{\frac{210}{360} \times \frac{22}{7} \times(10)^2\right\} cm ^2=183.3 cm^2$

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Question 22 Marks

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer

Here, r = 14 cm and $\theta=\frac{90^{\circ}}{3}=30^{\circ}$
$\therefore$ Area swept $=\frac{\theta}{360^{\circ}} \times \pi r^2$
$=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14$
$=\frac{154}{3} cm^2$

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Question 32 Marks

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

Answer

Dividing $N ^{ r } \& D ^{ r }$ by $\sin A$ in LHS
$=\frac{\cot A-1+\operatorname{cosec} A}{\cot A+1-\operatorname{cosec} A}$
$=\frac{\cot A +\operatorname{cosec} A -\left(\operatorname{cosec}^2 A-\cot ^2 A\right)}{\cot A +1-\operatorname{cosec} A }$
= cosec A + cot A

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Question 42 Marks

The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Answer

We know that the tangent at any point of a circle is $\perp$ to the radius through the point of contact.
$\therefore \angle OPA =90^{\circ}$
$\therefore OA ^2= OP ^2+ AP ^2[$ By Pythagoras theorem $]$
$\Rightarrow(5)^2=( OP )^2+(4)^2$
$\Rightarrow 25=( OP )^2+16$
$\Rightarrow OP ^2=9$
$\Rightarrow OP =3 cm$

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Question 52 Marks

ABC is an isosceles triangle with AB = AC, circumscribed about a circle. Prove that BC is bisected at E.

Answer

ABC is an isosceles triangle.

According to the question, AB = AC ...(i)
AD = AF (Tangents from A) ...(ii)
AB - AD = AC - AF
$\Rightarrow B D=C F \ldots$ (iii)
Now, BD = BE (Tangents from B)
Also, CF = CE (Tangents from C)
$\Rightarrow BE = CE$
So, BC is bisected at the point of contact E.

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Question 62 Marks

In the given figure, $\frac{A O}{O C}=\frac{B O}{O D}=\frac{1}{2}$ and AB = 4 cm. Find the value of DC.

Answer

Given: $\frac{ AO }{ OC }=\frac{ BO }{ OD }=\frac{1}{2}$ and $AB =4 cm$
To find: DC
Proof: In AOB and COD
$\frac{ AO }{ OC }=\frac{ BO }{ OD }($ given $)$
and $\angle A O B=\angle C O D$ ( vertically opposite angles)
$\therefore \quad \triangle A O B \sim \triangle C O D$ (SAS similarity)
$\Rightarrow \quad \frac{A O}{O C}=\frac{B O}{O D}=\frac{A B}{C D}$ (corresponding sides of similar triangles are proportional)
$\Rightarrow \frac{1}{2}=\frac{4}{ CD }$
$\Rightarrow CD =8 cm$

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Question 72 Marks

Aftab tells his daughter, Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be. (Isn’t this interesting?) Represent this situation algebraically and graphically by the method of substitution.

Answer

Let the present age of Aftab and his daughter be x and y years respectively. Then, the pair of linear equations that represent the
situation is
x - 7 = 7(y - 7), i.e., x - 7y + 42 = 0 ...(1)
and x + 3 = 3(y + 3), i.e., x - 3y = 6 ...(2)
from equation (2), we get x = 3y + 6
By putting this value of x in equation (1), we get
(3y + 6) -7y + 42 = 0,
i.e., -4y = -48, which gives y = 12
Again by putting this value of y in equation (2), we get
$x=3 \times 12+6=42$
So, the present age of Aftab and his daughter are 42 and 12 years respectively.

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Maths 2 Marks Questions

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