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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 80

General Instructions:

(i) All questions are compulsory

(ii) The question paper consists of 31 questions divided into four sections – A, B, C and D

(iii) Section A consists of 4 questions of 1 mark each. Section B consists of 6 questions of 2 marks each. Section C consists of 10 questions of 3 marks each. Section D consists of 11 questions of 4 marks each.

(iv) Use of calculator is not permitted.

SECTION – A

Question number 1 to 4 carry 1 mark each.

Question 1: What is the common difference of an A.P. in which $\displaystyle a_{21} - a_7 = 84 ?$

Given: $\displaystyle a_{21} -a_7 = 84$

$\displaystyle a-{21} = a + (21-1) d = a + 20d$

$\displaystyle a_7 = a + (7-1) d = a + 6 d$

$\displaystyle \therefore (a + 20 d ) - (a + 6d) = 84$

$\displaystyle 14 d = 84$

$\displaystyle \Rightarrow d = 6$

$\displaystyle \\$

Question 2: If the angle between two tangents drawn from an external point $\displaystyle P$ to a circle of radius $\displaystyle a$ and center $\displaystyle O$, is $\displaystyle 60^{\circ}$, then find the length of $\displaystyle OP$.

Radius $\displaystyle = a$

$\displaystyle \therefore \frac{OA}{PA} = \tan 30^{\circ}$

$\displaystyle \Rightarrow \frac{OA}{PA} = \frac{1}{\sqrt{3}}$

$\displaystyle \Rightarrow PA = \sqrt{3} a$

$\displaystyle \text{Also, } \frac{OA}{OP} = \sin 30^{\circ} = \frac{1}{2}$

$\displaystyle \Rightarrow OP = 2a$

$\displaystyle \\$

Question 3: If a tower $\displaystyle 30 \text{ m }$ high, casts a shadow $\displaystyle 10 \sqrt{3} \text{ m }$ long on the ground, then what is the angle of elevation of the sun?

$\displaystyle \frac{30}{10\sqrt{3}} = \tan \alpha$

$\displaystyle \Rightarrow \tan \alpha = \sqrt{3} = \tan 60^{\circ}$

$\displaystyle \Rightarrow \alpha = 60^{\circ}$

$\displaystyle \\$

Question 4: The probability of selecting a rotten apple randomly from a heap of $\displaystyle 900$ apples is $\displaystyle 0.18$. What is the number of rotten apples in the heap?

Let the number of rotten apples $\displaystyle = x$

$\displaystyle \therefore \frac{x}{900} = 0.18$

$\displaystyle \Rightarrow x = 0.18 \times 900 = 162$

Hence the number of rotten apples $\displaystyle = 162$

$\displaystyle \\$

Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: Find the value of $\displaystyle p$, for which one root of the quadratic equation $\displaystyle px^2 - 14x + 8 = 0$ is $\displaystyle 6$ times the other.

Given equation: $\displaystyle px^2 - 14x + 8 = 0$

$\displaystyle \Rightarrow a = p, b = -14, c = 8$

Let $\displaystyle \alpha \text{ and } \beta$ be the roots

Given $\displaystyle \beta = 6 \alpha$

$\displaystyle \alpha . \beta = \frac{c}{a}$

$\displaystyle \Rightarrow \alpha (6 \alpha) = \frac{8}{p}$

$\displaystyle \Rightarrow 6 {\alpha}^2 = \frac{8}{p}$ … … … … … i)

$\displaystyle \alpha + \beta = - \frac{b}{a}$

$\displaystyle 6 \alpha + \alpha = \frac{14}{p}$

$\displaystyle 7 \alpha = \frac{14}{p}$

$\displaystyle \Rightarrow \alpha = \frac{2}{p}$ … … … … … ii)

Solving i) and ii)

$\displaystyle 6 \Big( \frac{2}{p} \Big)^2 = \frac{8}{p}$

$\displaystyle \frac{6 \times 4}{p^2} = \frac{8}{p}$

$\displaystyle \Rightarrow p = \frac{6 \times 4}{8^2} = 3$

$\displaystyle \\$

$\displaystyle \text{Question 6: Which term of the progression } 20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots \\ \\ \text{is the first negative term?}$

$\displaystyle \text{Given AP is } 20, 19 \frac{1}{4} , 18 \frac{1}{2} , 17 \frac{3}{4} , \ldots$

$\displaystyle \therefore a = 20$

$\displaystyle d = 19 \frac{1}{4} - 20 = - \frac{3}{4}$

$\displaystyle \text{Confirming it once again: } d = 18 \frac{1}{2} - 19 \frac{1}{4} = - \frac{3}{4}$

$\displaystyle \therefore a = 20 \text{ and } d= - \frac{3}{4}$

Let $\displaystyle n^{th}$ term be zero

$\displaystyle \therefore a _n = a + (n-1)d$

$\displaystyle \Rightarrow 0 = 20 + (n-1) (- \frac{3}{4} )$

$\displaystyle \Rightarrow 3(n-1) = 80$

$\displaystyle 3n = 83$

$\displaystyle n = \frac{83}{3} = 27.67$

$\displaystyle \therefore n^{th}$ term where the value is very close is zero

$\displaystyle \therefore (n+1)^{th}$ term or $\displaystyle 28^{th}$ term will be negative

Also $\displaystyle 28^{th}$ term

$\displaystyle a_{28} = 20 + (28 - 1)(- \frac{3}{4} ) = - \frac{1}{4}$

$\displaystyle \\$

Question 7: Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

Given: A circle with center $\displaystyle O, PA \text{ and } PB$ are tangents drawn at the end $\displaystyle A \text{ and } B$ on chord $\displaystyle AB$

To prove: $\displaystyle \angle PAB = \angle PBA$

Construction: Join $\displaystyle OA \text{ and } OB$

Proof: In $\displaystyle \triangle OAB$, we have

$\displaystyle OA = OB$ (radius of the same circle)

$\displaystyle \Rightarrow \angle 2 = \angle 1$ (angles opposite to equal sides of a triangle)

$\displaystyle \angle OBP = \angle OAP = 90^{\circ}$ (radius is $\displaystyle \perp$ tangents)

$\displaystyle \Rightarrow \angle 2 + \angle 3 = \angle 1 + \angle 4$

$\displaystyle \Rightarrow \angle 3 = \angle 4$

$\displaystyle \Rightarrow \angle PAB = \angle PBA$

Hence proved.

$\displaystyle \\$

Question 8: A circle touches all the four sides of a quadrilateral $\displaystyle ABCD$. Prove that $\displaystyle AB + CD = BC + DA$

Given: A circle touches quadrilateral $\displaystyle ABCD$ on all four sides.

To prove: $\displaystyle AB + CD = AD + BC$

Proof: Since tangents to a circle from external points are equal.

$\displaystyle \therefore AP = AS, DQ = DR, CS = CR, BS = BP$

$\displaystyle \therefore AP + DQ + CS + BS = AP + DR + RC + PB$

$\displaystyle \Rightarrow AD + BC = AB + CD$

Hence proved.

$\displaystyle \\$

Question 9: A line intersects the $\displaystyle y-axis \text{ and } x-axis$ at the points $\displaystyle P \text{ and } Q$ respectively. If $\displaystyle (2, - 5)$ is the mid-point of $\displaystyle PQ$, then find the coordinates of $\displaystyle P \text{ and } Q$.

Let $\displaystyle P$ be $\displaystyle (0, y) \text{ and } Q$ be $\displaystyle (x, 0)$

$\displaystyle \therefore \frac{x+0}{2} = 2 \Rightarrow x = 4$

$\displaystyle \text{Also } \therefore \frac{0+y}{2} = -5 \Rightarrow y = -10$

$\displaystyle \therefore P(0, -10) \text{ and } Q(4, 0)$

$\displaystyle \\$

Question 10: If the distances of $\displaystyle P(x, y)$ from $\displaystyle A(5, 1) \text{ and } B(- 1, 5)$ are equal, then prove that $\displaystyle 3x = 2y$.

Given: $\displaystyle AP = BP \Rightarrow AP^2 = BP^2$

$\displaystyle \therefore (5-x)^2 + (1-y)^2 = (-1-x)^2 + (5-y)^2$

$\displaystyle \Rightarrow 25 + x^2 - 10x + 1 + y^2 -2y = 1 + x^2 + 2x + 25 + y^2 - 10y$

$\displaystyle \Rightarrow 25 - 10x + 1 - 2y = 1 + 2x + 25 - 10y$

$\displaystyle \Rightarrow -10x - 2y = 2x - 10y$

$\displaystyle \Rightarrow 8y = 12x$

$\displaystyle \Rightarrow 2y = 3x \ or \ 3x = 2y$

Hence proved.

$\displaystyle \\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 11: If $\displaystyle ad \neq bc$, then prove that the equation $\displaystyle (a^2 + b^2) x^2 + 2 (ac + bd) x + (c^2 + d^2) = 0$ has no real roots.

The equation is of the form $\displaystyle ax^2 + bx+ c =0$

$\displaystyle \therefore D = b^2 - 4ac$

$\displaystyle \Rightarrow D = [ 2(ac+bd) ]^2 - 4 [ (a^2+b^2)(c^2+d^2) ]$

$\displaystyle \Rightarrow D = 4 [ a^2c^2+b^2d^2+2abcd - a^2c^2-a^2d^2-b^2c^2-b^2d^2 ]$

$\displaystyle \Rightarrow D = 4 [ 2abcd - a^2d^2 - b^2c^2 ]$

$\displaystyle \Rightarrow D = - (ad-bc)^2$

Since $\displaystyle ad \neq bc$, the equation has no real roots.

$\displaystyle \\$

Question 12: The first term of an A.P. is $\displaystyle 5$, the last term is $\displaystyle 45$ and the sum of all its terms is $\displaystyle 400$. Find the number of terms and the common difference of the A.P.

We know $\displaystyle a = 5$

Let $\displaystyle n^{th}$ term be the last

$\displaystyle \therefore 45 = 5 + (n-1) d$

$\displaystyle \Rightarrow (n-1)d = 40$ … … … … … i)

Also the sum of the $\displaystyle n$ terms is $\displaystyle 400$

$\displaystyle \therefore 400 = \frac{n}{2} [ 2 \times 5 + (n-1)d ]$ … … … … … ii)

Substituting i) in ii) we get

$\displaystyle 400 = \frac{n}{2} [10 + 40]$

$\displaystyle \Rightarrow n = \frac{400}{25} = 16$

$\displaystyle \therefore d = \frac{40}{n-1} = \frac{40}{16-1} = \frac{8}{3}$

$\displaystyle \\$

Question 13: On a straight line passing through the foot of a tower, two points $\displaystyle C \text{ and } D$ are at distances of $\displaystyle 4 \text{ m }$ and $\displaystyle 16 \text{ m }$ from the foot respectively. If the angles of elevation from $\displaystyle C \text{ and } D$ of the top of the tower are complementary, then find the height of the tower.

$\displaystyle {\alpha}_1 + {\alpha}_2 = 90^{\circ}$

$\displaystyle \text{From } \triangle ABC: \frac{h}{4} = \tan {\alpha}_1$ … … … … … i)

From $\displaystyle \triangle ABD: \frac{h}{16} = \tan {\alpha}_2 = \tan (90^{\circ} - {\alpha}_1) = \cot {\alpha}_1$

$\displaystyle \therefore \frac{h}{16} = \frac{1}{\tan {\alpha}_1}$

$\displaystyle \Rightarrow \tan {\alpha}_1 = \frac{16}{h}$ … … … … … ii)

Substituting ii) in i) we get

$\displaystyle \frac{h}{4} = \frac{16}{h}$

$\displaystyle \Rightarrow h^2 = 4 \times 16$

$\displaystyle \Rightarrow h = 2 \times 4 = 8 \text{ m }$

$\displaystyle \\$

Question 14: A bag contains $\displaystyle 15$ white and some black balls. If the probability of drawing a black ball from the bag is thrice that of drawing a white ball, find the number of black balls in the bag.

Let the number of black balls $\displaystyle = x$

$\displaystyle \therefore P(Black \ Balls) = \frac{x}{x+15}$

$\displaystyle \text{and } P(White \ Balls) = \frac{15}{x+15}$

$\displaystyle \text{Given } \frac{x}{x+15} = 3 \times \frac{15}{x+15}$

$\displaystyle \Rightarrow x = 45$

Therefore number of Black balls $\displaystyle = 45$

$\displaystyle \\$

Question 15: In what ratio does the point $\displaystyle \Big( \frac{24}{11} , y \Big)$ divide the line segment joining the points divide the line segment joining the points $\displaystyle P(2, - 2) \text{ and } Q(3, 7)$ ? Also find the value of $\displaystyle y$.

$\displaystyle \text{Given } P(2, -2), Q(3, 7) \text{ and } R( \frac{24}{11} , y)$

Let $\displaystyle R$ divides $\displaystyle P \text{ and } Q$ in the ratio $\displaystyle k:1$. Using sections formula,

$\displaystyle \frac{24}{11} = \frac{3 \times k + 1 \times 2}{k+1}$

$\displaystyle \Rightarrow 24(k+1) = 33k + 22$

$\displaystyle \Rightarrow 24k + 24 = 33k + 22$

$\displaystyle \Rightarrow k = \frac{2}{9}$

Hence $\displaystyle R$ divides $\displaystyle P \text{ and } Q$ in the ratio of $\displaystyle 2:9$

$\displaystyle \text{Similarly, } y = \frac{k \times 7 + 1 \times (-2)}{k+1}$

$\displaystyle \Rightarrow y = \frac{\frac{2}{9} \times 7 + 1 \times (-2)}{\frac{2}{9}+1}$

$\displaystyle \Rightarrow y = \frac{14-18}{2+9} = - \frac{4}{11}$

$\displaystyle \\$

Question 16: Three semicircles each of diameter 3 cm, a circle of diameter 4·5 cm and a semicircle of radius 4·5 cm are drawn in the given figure. Find the area of the shaded region.

$\displaystyle \text{Area of large semi circle } = \frac{1}{2} \times \pi (4.5)^2 = 10.125 \pi \ cm^2$

$\displaystyle \text{Area of large circle } = \pi ( \frac{4.5}{2} )^2 = 5.0625 \pi \ cm^2$

$\displaystyle \text{Area of two small semi circles } = 2 \times [ \frac{1}{2} \pi \times (1.5)^2 ] = 2.25 \pi \ cm^2$

$\displaystyle \text{Area of shaded small semi circle } = \frac{1}{2} \pi (1.5)^2 = 1.125 \pi \ cm^2$

$\displaystyle \text{Therefore shaded area } = 10.125 \pi - 5.0625 \pi - 2.25 \pi + 1.125 \pi = 3.9375 \pi = 12.375 \ cm^2$

$\displaystyle \\$

Question 17: In the given figure, two concentric circles with centre $\displaystyle O$ have radii $\displaystyle 21 \text{ cm }$ and $\displaystyle 42 \text{ cm }$. If $\displaystyle \angle AOB = 60^{\circ}$, find the area of the shaded region. [ Use $\displaystyle \pi = \frac{22}{7}$ ]

$\displaystyle \text{Area of } OAB = \frac{60}{360} \times \pi (42)^2 = 294 \pi \ cm^2$

$\displaystyle \text{Area of } OCD = \frac{60}{360} \times \pi (21)^2 = 73.5 \pi \ cm^2$

Therefore area of $\displaystyle CDBA = 294 \pi - 73.3 \pi = 220.5 \pi \ cm^2$

$\displaystyle \text{Shaded Area } = \frac{300}{360} \times \pi (42)^2 - \frac{300}{360} \times \pi (21)^2$

$\displaystyle = 1470 \pi - 367.5 \pi = 1102.5 \times \frac{22}{7} = 3465 \ cm^2$

$\displaystyle \\$

Question 18: Water in a canal, $\displaystyle 5.4 \text{ m }$ wide and $\displaystyle 1.8 \text{ m }$ deep, is flowing with a speed of $\displaystyle 25 \text{ km }$/hour. How much area can it irrigate in $\displaystyle 40 \text{ m }$inutes, if $\displaystyle 10 \text{ cm }$ of standing water is required for irrigation ?

Cross Section of canal $\displaystyle = 5.4 \times 1.8 = 9.72 \ m^2$

$\displaystyle \text{Speed of water } = 25 \ \frac{km}{hr} = 25 \times \frac{1000 \ m}{60 \ min} = \frac{2500}{6}\ \frac{m}{min}$

$\displaystyle \text{Therefore volume of water flow in 1 minute } = 9.72 \times \frac{2500}{6} = 4050 \frac{m^3}{min}$

Let the area irrigated in $\displaystyle 40 \ min = x \ m^2$

$\displaystyle \therefore x \times \frac{10}{100} = 40 \times 4050$

$\displaystyle \Rightarrow x = 1620000 \ m^2$

$\displaystyle \\$

Question 19: The slant height of a frustum of a cone is $\displaystyle 4 \text{ cm }$ and the perimeters of its circular ends are $\displaystyle 18 \text{ cm }$ and $\displaystyle 6 \text{ cm }$. Find the curved surface area of the frustum.

We know that the curved surface area of frustum $\displaystyle = \pi (r_1+r_2) l$

Here $\displaystyle l = 4 \ cm$

$\displaystyle \text{For } r_1: 2 \pi r_1 = 18 \Rightarrow r_1 = \frac{9}{\pi}$

$\displaystyle \text{For } r_2: 2 \pi r_2 = 6 \Rightarrow r_1 = \frac{3}{\pi}$

$\displaystyle \text{Now curved surface area } = \pi (r_1+r_2) l = \pi \Big( \frac{9}{\pi} + \frac{3}{\pi} \Big) l = 12 \times 4 = 48 \ cm^2$

Hence the curved surface area $\displaystyle = 48 \ cm^2$

$\displaystyle \\$

Question 20: The dimensions of a solid iron cuboid are $\displaystyle 4.4 \text{ m }$ $\displaystyle \times 2.6 \text{ m }$ $\displaystyle \times 1.0 \text{ m }$. It is melted and recast into a hollow cylindrical pipe of $\displaystyle 30 \text{ cm }$ inner radius and thickness $\displaystyle 5 \text{ cm }$. Find the length of the pipe.

Dimensions of cuboid: $\displaystyle l = 4.4 \ m, b = 2.6 \ m, h = 1 \ m$

Therefore volume of cuboid $\displaystyle = lbh = 4.4 \times 2.6 \times 1 = 11.44 \ m^3$

Dimension of the pipe: Inner radius $\displaystyle (r) = 30 \ cm = 0.3 \ m$

Thickness $\displaystyle = 5 \ cm = 0.05 \ m$

Therefore outer radius $\displaystyle (R) = 0.30 + 0.05 = 0.35 \ m$

Let l be the length of the pipe

$\displaystyle \therefore \pi (R^2 - r^2) \times l = 11.44$

$\displaystyle \Rightarrow \frac{22}{7} \Big( (0.35)^2 - (0.30)^2 \Big) \times l = 11.44$

$\displaystyle \Rightarrow \frac{22}{7} \times 0.0325 \times l = 11.44$

$\displaystyle \Rightarrow l = \frac{7 \times 11.44}{22\times 0.0325} = 112 \ m$

$\displaystyle \\$

Section – D

Question number 21 to 30 carry 4 mark each.

$\displaystyle \text{Question 21: Solve for } x: \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4} , x \neq -1, - \frac{1}{5} , -4$

$\displaystyle \frac{1}{x+1} + \frac{3}{5x+1} = \frac{5}{x+4}$

$\displaystyle \Rightarrow \frac{5x+1 + 3(x+1)}{(x+1)(5x+1)} = \frac{5}{x+4}$

$\displaystyle \Rightarrow (8x+4)(x+4) = 5(x+1)(5x+1)$

$\displaystyle \Rightarrow 8x^2 + 4x + 32x + 16 = 25x^2 + 25x + 5x + 5$

$\displaystyle \Rightarrow 6x + 11 = 17x^2$

$\displaystyle \Rightarrow 17x^2 - 6x - 11 = 0$

$\displaystyle \Rightarrow 17x^2 - 17x + 11x - 11 = 0$

$\displaystyle \Rightarrow 17x(x-1) + 11 (x-1) = 0$

$\displaystyle \Rightarrow (x-1)(17x + 11) = 0$

$\displaystyle \Rightarrow x = 1 \ or \ x = - \frac{11}{17}$

$\displaystyle \\$

Question 22: Two taps running together can fill a tank in $\displaystyle 3 \frac{1}{3}$ hours. If one tap takes $\displaystyle 3$ hours more than the other to fill the tank, then how much time will each tap take to fill the tank ?

Time taken by larger tap to fill the tank $\displaystyle = x$ hours

Therefore Time taken by smaller tap to fill the tank $\displaystyle = (x+3)$ hours

$\displaystyle \text{Portion of the tank filled by larger tap in 1 hour } = \frac{1}{x}$

$\displaystyle \text{Portion of the tank filled by Smaller tap in 1 hour } = \frac{1}{x+3}$

$\displaystyle \text{Given: Time taken when both taps are running to fill the tank } = 3 \frac{1}{13} = \frac{40}{13}$ hours

$\displaystyle \text{Portion of tank filled in 1 hour when both taps are running } = \frac{1}{\frac{40}{13}} = \frac{13}{40}$

$\displaystyle \therefore \frac{1}{x} + \frac{1}{x+3} = \frac{13}{40}$

$\displaystyle \Rightarrow 40(x + 3 + x) = 13x(x + 3)$

$\displaystyle \Rightarrow 40 (2x + 3) = 13x(x+3)$

$\displaystyle \Rightarrow 80x + 120 = 13x^2 + 39x$

$\displaystyle \Rightarrow 13x^2 - 41x- 120 = 0$

$\displaystyle \Rightarrow 13x^2 - 65x + 24x - 120 = 0$

$\displaystyle \Rightarrow 13x( x - 5) + 24( x - 5) = 0$

$\displaystyle \Rightarrow (x-5) ( 13x + 24) = 0$

$\displaystyle \Rightarrow x = 5 \ or \ x = - \frac{24}{13}$ (not possible as time cannot be negative)

Therefore the time taken by the larger tap to fill the tank $\displaystyle = 5$ hours

and the time taken by the smaller tap to fill the tank $\displaystyle = 8$ hours

$\displaystyle \\$

Question 23: If the ratio of the sum of the first $\displaystyle n$ terms of two A.Ps is $\displaystyle (7n + 1) : (4n + 27)$, then find the ratio of their $\displaystyle 9^{th}$ terms.

$\displaystyle \text{Given: } \frac{S_n}{ {S'}_n} = \frac{7n+1}{4n+27}$

$\displaystyle \Rightarrow \frac{ \frac{n}{2} [ 2a + (n-1)d ] }{\frac{n}{2} [ 2a' + (n-1)d' ]} = \frac{7n+1}{4n+27}$

$\displaystyle \frac{2a + (n-1)d}{2a' + (n-1)d'} = \frac{7n+1}{4n+27}$

$\displaystyle \frac{a + \frac{(n-1)}{2} d}{a' + \frac{(n-1)}{2} d'} = \frac{7n+1}{4n+27}$ … … … … … i)

We need to find:

$\displaystyle \frac{T_9}{ {T'}_9} = \frac{a + (9-1)d}{a' + (9-1)d'} = \frac{a + 8d}{a' + 8d'}$ … … … … … ii)

$\displaystyle \text{If we put } \frac{n-1}{2} = 8 \Rightarrow n = 17$ … … … … … iii)

Substituting $\displaystyle n = 17$ in i) we get

$\displaystyle \frac{a + 8d}{a' + 8 d'} = \frac{7 \times 17 + 1}{4 \times 17 + 27} = \frac{120}{146} = \frac{24}{19}$

$\displaystyle \text{Hence } \frac{T_9}{ {T'}_9} = \frac{24}{19}$

$\displaystyle \\$

Question 24: Prove that the lengths of two tangents drawn from an external point to a circle are equal.

Given: Let the circle with center $\displaystyle O$

Let $\displaystyle P$ be an external point from which tangents $\displaystyle PQ \text{ and } PR$ are drawn as shown in the diagram

To prove: $\displaystyle PQ = PR$

Construction: Join $\displaystyle OQ, OR \text{ and } OP$

Proof: As $\displaystyle PQ$ is tangent $\displaystyle OQ \perp PQ$. Therefore $\displaystyle \angle OQP = 90^{\circ}$

Similarly, As $\displaystyle PR$ is tangent $\displaystyle OR \perp PR$. Therefore $\displaystyle \angle ORP = 90^{\circ}$

(Note: Tangents at any point on a circle is perpendicular to the radius through the point of contact)

In $\displaystyle \triangle OQP \text{ and } \triangle ORP$

$\displaystyle \angle OQP = \angle ORP = 90^{\circ}$

$\displaystyle OP$ is common

$\displaystyle OQ = OR$ (radius of the same circle)

$\displaystyle \therefore \triangle OQP \cong \triangle ORP$

$\displaystyle \Rightarrow PQ = PR$

Therefore both tangents are equal in length.

$\displaystyle \\$

Question 25: In Figure 3, $\displaystyle PQ \text{ and } RS$ are two parallel tangents to a circle with center $\displaystyle O$ and another tangent $\displaystyle AB$ with point of contact $\displaystyle C$ intersecting $\displaystyle PQ$at $\displaystyle A \text{ and } RS$ at $\displaystyle B$. Prove that $\displaystyle \angle AOB = 90^{\circ}$.

Figure 3

Given: $\displaystyle PQ \parallel RS$, $\displaystyle AB$ is a tangent

To prove: $\displaystyle \angle AOB = 90^{\circ}$

Join $\displaystyle DE$ through $\displaystyle O$. $\displaystyle DOE$ is diameter of the circle

$\displaystyle \therefore \angle ADO = \angle BEO = 90^{\circ}$

$\displaystyle \Rightarrow \angle ADO + \angle BEO = 180^{\circ}$

$\displaystyle \Rightarrow DA \parallel EB$

We know that tangents to a circle from an external point are equally inclined to the line segment joining this point to the center

$\displaystyle \therefore \angle 1 = \angle 2 \text{ and } \angle 3 = \angle 4$

Now $\displaystyle DA \parallel RB \text{ and } AB$ is transversal

$\displaystyle \therefore (\angle 1 + \angle 2) + (\angle 3 + \angle 4) = 180^{\circ}$

$\displaystyle \Rightarrow 2 \angle 1 + 2 \angle 3 = 180^{\circ}$

$\displaystyle \Rightarrow \angle 1 + \angle 3 = 90^{\circ}$

From $\displaystyle \triangle AOB$

$\displaystyle \angle AOB + \angle 1 + \angle 3 = 180^{\circ}$

$\displaystyle \Rightarrow \angle AOB = 180^{\circ} - 90^0 = 90^{\circ}$

Hence $\displaystyle \angle AOB = 90^{\circ}$

$\displaystyle \\$

Question 26: Construct a triangle $\displaystyle ABC$ with side $\displaystyle BC = 7 \text{ cm }$, $\displaystyle \angle B = 45^{\circ} , \angle A = 105^{\circ}$. Then construct another triangle whose sides are $\displaystyle \frac{3}{4}$ times the corresponding sides of the $\displaystyle \triangle ABC$.

$\displaystyle \\$

Question 27: An aeroplane is flying at a height of $\displaystyle 300 \text{ m }$ above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are $\displaystyle 45^{\circ} \text{ and } 60^{\circ}$ respectively. Find the width of the river. [Use $\displaystyle \sqrt{3} = 1.732$]

Let $\displaystyle A$ be the aeroplane & $\displaystyle BD$ is the width of river

$\displaystyle \text{In } \triangle ABC, \frac{300}{BC} = \tan 45^{\circ} = 1$

$\displaystyle \Rightarrow BC = 300 \text{ m }$

In $\displaystyle \triangle ABD, \frac{300}{CD} = \tan 60^{\circ} = \sqrt{3}$

$\displaystyle \Rightarrow CD = \frac{300}{\sqrt{3}} = 100\sqrt{3} = 173.2 \text{ m }$

Therefore the width of the river $\displaystyle = BC + CD = 300 + 173.2 = 473.2 \text{ m }$

$\displaystyle \\$

Question 28: If the points $\displaystyle A(k + 1, 2k), B(3k, 2k + 3) \text{ and } C(5k - 1, 5k)$ are collinear, then find the value of $\displaystyle k$.

Given: $\displaystyle A (k+1, 2k), B( 3k, 2k+3), C(5k-1, 5k)$ are collinear

Therefore the area of $\displaystyle \triangle ABC = 0$

$\displaystyle \frac{1}{2} [ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ] = 0$

$\displaystyle \Rightarrow \frac{1}{2} [ (k+1)(2k+3-5k) + 3k(5k-2k) + (5k-1)(2k - 2k-3) ] = 0$

$\displaystyle \Rightarrow \frac{1}{2} [ (k+1)(-3k+3) + 3k(3k) + (5k-1)(-3) ] = 0$

$\displaystyle \Rightarrow \frac{1}{2} [ -3k^2+3k-3k+3+9k^2-15k+3 ] = 0$

$\displaystyle \Rightarrow \frac{1}{2} [ 6k^2 - 15k+6 ] = 0$

$\displaystyle \Rightarrow 6k^2 - 15k+6 = 0$

$\displaystyle \Rightarrow 2k^2 - 5k + 2 = 0$

$\displaystyle \Rightarrow 2k^2 - 4k - k + 2 = 0$

$\displaystyle \Rightarrow 2k(k-2)-1(k-2)=0$

$\displaystyle \Rightarrow (k-2)(2k-1) = 0$

$\displaystyle \Rightarrow k = 2 \ or \ k = \frac{1}{2}$

$\displaystyle \\$

Question 29: Two different dice are thrown together. Find the probability that the numbers obtained have

(i) even sum, and

(ii) even product.

Number of possible outcomes:

$\displaystyle (1, 1), (1, 2), (1, 3), (1,4 ), (1, 5), (1, 6)$

$\displaystyle (2, 1), (2, 2), (2, 3), (2,4 ), (2, 5), (2, 6)$

$\displaystyle (3, 1), (3, 2), (3, 3), (3,4 ), (3, 5), (3, 6)$

$\displaystyle (4, 1), (4, 2), (4, 3), (4,4 ), (4, 5), (4, 6)$

$\displaystyle (5, 1), (5, 2), (5, 3), (5,4 ), (5, 5), (5, 6)$

$\displaystyle (6, 1), (6, 2), (6, 3), (6,4 ), (6, 5), (6, 6)$

Therefore total number of possible outcomes $\displaystyle = 36$

i) Even sum outcome

$\displaystyle (1, 1), (1, 3), (1, 5)$

$\displaystyle (2, 2), (2,4 ), (2, 6)$

$\displaystyle (3, 1), (3, 3), (3, 5)$

$\displaystyle (4, 2), (4,4 ), (4, 6)$

$\displaystyle (5, 1), (5, 3), (5, 5)$

$\displaystyle (6, 2), (6,4 ), (6, 6)$

Possible outcomes $\displaystyle = 18$

$\displaystyle \text{Therefore Probability (Even sum outcome) } = \frac{18}{36} = \frac{1}{2}$

ii) Even products

$\displaystyle (1, 2), (1,4 ), (1, 6)$

$\displaystyle (2, 1), (2, 2), (2, 3), (2,4 ), (2, 5), (2, 6)$

$\displaystyle (3, 2), (3,4 ), (3, 6)$

$\displaystyle (4, 1), (4, 2), (4, 3), (4,4 ), (4, 5), (4, 6)$

$\displaystyle (5, 2), (5,4 ), (5, 6)$

$\displaystyle (6, 1), (6, 2), (6, 3), (6,4 ), (6, 5), (6, 6)$

Possible outcomes $\displaystyle = 27$

$\displaystyle \text{Therefore Probability (Even product) } = \frac{27}{36} = \frac{3}{4}$

$\displaystyle \\$

Question 30: In the given figure, $ABCD$ is a rectangle of dimensions $21$ cm $\times 14$ cm. A semicircle is drawn with $BC$ as diameter. Find the area and the perimeter of the shaded region in the figure.

Area of $\displaystyle ABCD = 21 \times 14 = 294 \ cm^2$

$\displaystyle \text{Area of semi circle } = \frac{1}{2} [ \pi (7)^2 ] = 77 \ cm^2$

Therefore area of shaded region $\displaystyle = 294 - 77 = 217 \ cm^2$

$\displaystyle \text{Perimeter of shaded region } = 21 + 14 + 21 + \frac{1}{2} (2 \pi \times 7) = 56 + 22 = 78 \ cm$

$\displaystyle \\$

Question 31: In a rain-water harvesting system, the rain-water from a roof of $\displaystyle 22 \text{ m }$ $\displaystyle \times 20 \text{ m }$ drains into a cylindrical tank having diameter of base $\displaystyle 2 \text{ m }$ and height $\displaystyle 3.5 \text{ m }$. If the tank is full, find the rainfall in cm. Write your views on water conservation.

Let the rainfall be $\displaystyle x \text{ m }$
Therefore volume of water from roof $\displaystyle =$ volume of tank
$\displaystyle \Rightarrow 22 \times 20 \times x = \pi (1)^2 \times 3.5$
$\displaystyle \Rightarrow x = \frac{22}{7} \times \frac{3.5}{22 \times 20}$
$\displaystyle \Rightarrow x = 0.025 \ m = 25 \ mm \ or \ 2.5 \ cm$