2017 CBSE (Class 10) Board Paper Solution Mathematics : 30-1-1

Instructions:

Please check that this question paper consists of 11 pages.
Code number given on the right hand side of the question paper should be written on the title page of the answer book by the candidate.
Please check that this question paper consists of 31 questions.
Please write down the serial number of the question before attempting it.
15 minutes times has been allotted to read this question paper. The question paper will be distributed at 10:15 am. From 10:15 am to 10:30 am, the students will read the question paper only and will not write any answer on the answer book during this period.

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 $\displaystyle a_{21} - a_7 = 84 ?$

Answer:

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 $\displaystyle OP$ .

Answer:

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?

Answer:

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

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

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

$\displaystyle \\$

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?

Answer:

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.

Answer:

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)

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

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

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

$\displaystyle \\$

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

Answer:

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

$\displaystyle \therefore a = 20$

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

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

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

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.

Answer:

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 $\displaystyle AB + CD = BC + DA$

Answer:

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 $\displaystyle P \text{ and } Q$ .

Answer:

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 $\displaystyle 3x = 2y$ .

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

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

Answer:

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 $\displaystyle y$ .

Answer:

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

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.

Answer:

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

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

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

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

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

$\displaystyle \\$

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 $\displaystyle \pi = \frac{22}{7}$ ]

Answer:

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

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

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 ?

Answer:

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.

Answer:

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.

Answer:

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.

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

Answer:

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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 ?

Answer:

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.

Answer:

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

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

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

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

Answer:

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

Answer:

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 $\displaystyle \triangle ABC$ .

Answer:

$\displaystyle \\$

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 $\displaystyle \sqrt{3} = 1.732$ ]

Answer:

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 $\displaystyle k$ .

Answer:

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.

Answer:

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.

Answer:

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.