CBSE Class 10 Mathematics Solved Question Paper 2013

Let r be the radius of the circle,

From the given information, we have

$$\begin{gathered} 2\mathrm{\pi r} - \mathrm{r} =37 \mathrm{cm} \\ \Rightarrow \mathrm{r} ( 2\mathrm{\pi } - 1 ) = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} \left( 2 \mathrm{x} \frac{22}{7} - 1\right) = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} \mathrm{x} \frac{37}{7} = 37 \mathrm{cm} \\ \Rightarrow \mathrm{r} = 7 \mathrm{cm} \\ \therefore \mathrm{Circumference} \mathrm{of} \mathrm{the} \mathrm{circle} = 2\mathrm{\pi r} \\ = 2 \mathrm{x} \frac{22}{7} \mathrm{x} 7 \mathrm{cm} = 44 \mathrm{cm} \end{gathered}$$

Let AB be the tower of height 75 m and C be position of the car.

In 

$$\begin{gathered} ∆\mathrm{ABC}, \\ \cot 30°= \frac{\mathrm{AC}}{\mathrm{AB}} \\ \Rightarrow \mathrm{AC}=\mathrm{AB} \cot 30° \\ \Rightarrow \mathrm{AC}=75\mathrm{m} \mathrm{x} \sqrt{3} \\ \Rightarrow \mathrm{AC}=75\sqrt{3}\mathrm{m} \\ \mathrm{Thus}, \mathrm{the} \mathrm{distance} \mathrm{of} \mathrm{the} \mathrm{car} \mathrm{from} \mathrm{the} \mathrm{base} \mathrm{of} \mathrm{the} \mathrm{tower} \mathrm{is} 75\sqrt{3 }\mathrm{m}. \end{gathered}$$ 

 $$\begin{gathered} Comman difference = \frac{1 - 6q}{3q} - \frac{1}{3q} = \frac{1-6q-1}{3q} \\ = \frac{-6q}{3q} \\ = -2 \end{gathered}$$

From the figure, the coordinates of A, B, and C are (1,3), (-1, 0) and (4, 0)

respectively.

$$\begin{gathered} \mathrm{Area} \mathrm{of}∆\mathrm{ABC} \\ = \frac{1}{2} \left|1(0-0) + (-1)(0-3) +(3-0)\right| \\ =\frac{1}{2} \left|0+3+12\right| \\ =\frac{1}{2} \left|15\right| \\ = 7.5 \mathrm{sq}. \mathrm{units} \end{gathered}$$

S = { 1, 2, 3,........90 }

n(s) = 90

The prime number less than 23 are 2, 3, 5, 7, 11, 13, 17 and 19

Let event E be defined as 'getting a prime number less than 23'.

n(E) = 8

$$\begin{gathered} \therefore \mathrm{p}\left(\mathrm{E}\right) =\frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes}} = \frac{8}{90} \\ =\frac{4}{45} \end{gathered}$$

S = { 1, 2, 3, 4, 5, 6 }

let event E be defined as 'getting an even number'.

n(E) = { 2, 4, 6 }

 $$\begin{gathered} \therefore \mathrm{P}\hspace{0.17em}\left(\mathrm{E}\right) = \frac{\mathrm{Number} \mathrm{of} \mathrm{favourable} \mathrm{outcomes}}{\mathrm{Number} \mathrm{of} \mathrm{possible} \mathrm{outcomes}} =\frac{3}{6} \\ = \frac{1}{2} \end{gathered}$$

Given: Ab, BC,CD,and AD tangents to the circle with centre O at

Q,P,S and R respectively.

AB=29 cm, AD=23 cm, DS=5 cm and $\angle$B=90⁰

Construction: Join PQ.

We know that, the lengths of the tangents drawn from an external

point to a circle are equal.

DS=DR=5 cm

$\therefore$AR= AD - DR= 23 cm- 5 cm= 18 cm 

AQ=AR= 18 cm

$\therefore$ QB = AB - AQ = 29 cm - 18 cm= 11 cm

QB = BP = 11 cm

In $∆$$$\begin{gathered} \mathrm{In}∆\mathrm{PQB} \\ {\mathrm{PQ}}^2 = {\mathrm{QB}}^2 +{\mathrm{BP}}^2 =(11 \mathrm{cm}{)}^2 +(11 \mathrm{cm}{)}^2 = 2\mathrm{x}(11 \mathrm{cm}{)}^2 \\ \mathrm{PQ}=11\sqrt{2} \mathrm{cm} ..........(1) \\ \mathrm{IN}∆\mathrm{OPQ}, \\ {\mathrm{PQ}}^2 = {\mathrm{OQ}}^2 +{\mathrm{OP}}^2 = {\mathrm{r}}^2 +{\mathrm{r}}^2 = 2{{\mathrm{r}}^2}^{ } \\ (11\sqrt{2}{)}^2 = 2{{\mathrm{r}}^2}^{ } \\ 121 = {\mathrm{r}}^2 \\ \mathrm{r} = 11 \mathrm{cm} \\ \mathrm{Thus}, \mathrm{the} \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circleis} 11 \mathrm{cm}. \end{gathered}$$

$$\begin{gathered} \mathrm{AP}\perp \mathrm{PB} \left(\mathrm{Given}\right) \\ \mathrm{CA}\perp \hspace{0.17em}\mathrm{AP}, \mathrm{CB} \perp \mathrm{BP} (\mathrm{Since} \mathrm{radius} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{tangent}) \\ \mathrm{AC}= \mathrm{CB}\hspace{0.17em}= \mathrm{radius} \mathrm{of} \mathrm{the} \mathrm{circle} \end{gathered}$$
Therefore, APBC is a square having side equal to 4 cm.
Therefore, length of each tangent is 4 cm.