2019 CBSE (Class 10) Board Paper Solution Mathematics : 30-3-1

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 30 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 30 questions divided into four sections – A, B, C and D

(iii) Section A consists of 6 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 8 questions of 4 marks each.

(iv) There is no overall choice. However, an internal choice has been provided in two questions of 1 mark, two questions of 2 marks, four questions of 3 marks each and three questions of 4 marks each. You have to attempt only one of the alternative in all such questions.

(v) Use of calculator is not permitted.

SECTION – A

Question number 1 to 6 carry 1 mark each.

Question 1: Write the discriminant of the quadratic equation $(x + 5)^2 = 2 (5x - 3)$ .

Answer:

Given Equation: $(x + 5)^2 = 2 (5x - 3)$

$\Rightarrow x^2 + 25 + 10x = 10x - 6$

$\Rightarrow x^2 + 31 = 0$

Comparing it with $ax^2 + bx + c= 0$ . $a = 1, b = 0 \text{ and } c = 31$

Determinant $= b^2 - 4ac = -4(1)(31) = -124$

$\\$

Question 2: Find after how many places of decimal the decimal form of $\displaystyle\text{the number } \frac{27}{2^3 . 5^4 . 3^2} \text{will terminate.}$

Express $429$ as a product of its prime factors.

Answer:

$\displaystyle \text{Simplifying } \frac{27}{2^3 . 5^4 . 3^2} = \frac{3^3}{2^3 5^4 3^2} = \frac{3}{2^3 5^4} = \frac{3}{5000} = 0.0006$

So the decimal form will terminate are $4$ digits.

$429 = 3 \times 11 \times 13$

$\\$

Question 3: Find the sum of first $10$ multiples of $6$ .

Answer:

$AP$ is $6, 12, 18, \cdots$

$a = 6, d = 12-6 = 6, n = 10$

$\displaystyle S_{10} = \frac{n}{2} (2a + (n-1)d)$

$\displaystyle = \frac{10}{2} (2 \times 6 + (10-1) \times 6)$

$= 5 (12 + 54) = 330$

$\\$

Question 4: Find the value(s) of $x$ , if the distance between the points $A(0, 0)$ and $B(x, - 4)$ is $5$ units.

Answer:

Given: $A(0,0)$ and $B(x, -4)$

$AB = 5$ units

$\therefore 5 = \sqrt{(x-0)^2 + (-4-0)^2}$

$\Rightarrow 25 = x^2 + 16$

$\Rightarrow x^2 = 9$

$\Rightarrow x = \pm 3$

Therefore $B$ could be $(3, -4) \text{ or} (-3, -4)$

$\\$

Question 5: Two concentric circles of radii $a$ and $b (a > b)$ are given. Find the length of the chord of the larger circle which touches the smaller circle.

Answer:

$\text{Length of the chord } = 2 \Big( \sqrt{a^2 - b^2} \Big)$

$\\$

Question 6: In Figure 1, $PS = 3$ cm, $QS = 4$ cm, $\angle PRQ = \theta, \angle PSQ = 90^o, PQ \perp RQ$ and $RQ = 9$ cm. Evaluate $\tan \theta$ .

$\displaystyle \text{If } \tan \alpha = \frac{5}{12} \text{, find the value of } \sec \alpha$

Answer:

$PQ = \sqrt{4^2 + 3^2} = 5$ cm

$\displaystyle \tan \theta = \frac{5}{9}$

$\displaystyle \tan \alpha = \frac{5}{12}$

$AB = \sqrt{12^2 + 5^2} = 13$

$\displaystyle \therefore \cos \alpha = \frac{12}{13} \Rightarrow \sec \alpha = \frac{13}{12}$

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: Points $\displaystyle A(3, 1), B(5, 1), C(a, b)$ and $\displaystyle D(4, 3)$ are vertices of a parallelogram $\displaystyle ABCD$ . Find the values of $\displaystyle a$ and $\displaystyle b$ .

Points $\displaystyle P$ and $\displaystyle Q$ trisect the line segment joining the points $\displaystyle A(- 2, 0)$ and $\displaystyle B(0, 8)$ such that $\displaystyle P$ is near to $\displaystyle A$ . Find the coordinates of points $\displaystyle P$ and $\displaystyle Q$ .

Answer:

Let midpoint of $\displaystyle BD$ be $\displaystyle O$

$\displaystyle \therefore O (x, y) = \Big( \frac{5+4}{2} , \frac{1+3}{2} \Big) = (4.5, 2)$

$\displaystyle O$ is also the midpoint of $\displaystyle AC$

$\displaystyle O(x, y) = \Big( \frac{3+a}{2} , \frac{1+b}{2} \Big)$

$\displaystyle \text{Therefore } \frac{3+a}{2} = 4.5 \Rightarrow a = 9-3 = 6$

$\displaystyle \frac{1+b}{2} = 2 \Rightarrow b = 4-1 = 3$

$\displaystyle \frac{AP}{PB} = \frac{1}{2}$

Using section formula [ $\displaystyle P$ divides $\displaystyle AB$ in the ratio of $\displaystyle 1:2$ ]

$\displaystyle \Rightarrow P(x_1, y_1) = \Big( \frac{1 \times 0 + 2 \times (-2)}{1+2} , \frac{1 \times 8 + 2 \times 0}{1+2} \Big) = \Big( \frac{-4}{3} , \frac{8}{3} \Big)$

Similarly,

Using section formula [ $\displaystyle Q$ divides $\displaystyle AB$ in the ratio of $\displaystyle 2:1$ ]

$\displaystyle \Rightarrow P(x_2, y_2) = \Big( \frac{2 \times 0 + 1 \times (-2)}{2+1} , \frac{2 \times 8 + 1 \times 0}{2+1} \Big) = \Big( \frac{-2}{3} , \frac{16}{3} \Big)$

$\displaystyle \\$

Question 8: Solve the following pair of linear equations :

$\displaystyle 3x - 5y = 4$

$\displaystyle 2y + 7 = 9x$

Answer:

Given equations:

$\displaystyle 3x - 5y = 4$ … … … … … i)

$\displaystyle 2y + 7 = 9x$

$\displaystyle \Rightarrow 9x - 2y = 7$ … … … … … ii)

Multiplying i) by $\displaystyle 3$ and subtracting ii) from the resultant equation

$\displaystyle \hspace*{1cm} 9x - 15y = 12 \\ \underline {(-) \hspace*{0.4cm} 9x - 2y = \hspace*{0.2cm} 7} \\ \hspace*{1.3cm} -13y = \hspace*{0.2cm} 5$

$\displaystyle \Rightarrow y = \frac{-5}{13}$

Substituting into i)

$\displaystyle 3x - 5 \Big( \frac{-5}{13} \Big) = 4$

$\displaystyle \Rightarrow x = \frac{9}{13}$

$\displaystyle \text{Hence } x = \frac{9}{13} \text{ and } y = \frac{-5}{13}$

$\displaystyle \\$

Question 9: If HCF of $\displaystyle 65$ and $\displaystyle 117$ is expressible in the form $\displaystyle 65n - 117$ , then find the value of $\displaystyle n$ .

On a morning walk, three persons step out together and their steps measure $\displaystyle 30$ cm, $\displaystyle 36$ cm and $\displaystyle 40$ cm respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps ?

Answer:

$\displaystyle 117 = 3 \times 3 \times 13$

$\displaystyle 65 = 5 \times 13$

Therefore HCF of $\displaystyle 117$ and $\displaystyle 65$ is $\displaystyle 13$

$\displaystyle \therefore 65n - 117 = 13$

$\displaystyle \Rightarrow 65n = 130$

$\displaystyle \Rightarrow n = 2$

Finding LCM of $\displaystyle 30, 36$ and $\displaystyle 40$

$\displaystyle 2 | \underline{30, 36, 40} \\ 5 | \underline{15, 18, 20} \\ 3 | \underline{3, 18, 4} \\ 2 | \underline{1, 6, 4} \\ \hspace*{0.2cm}| 1, 3, 2$

Therefore LCM $\displaystyle = 2 \times 5 \times 3 \times 2 \times 3 \times 2 = 360$

Therefore the minimum distance each should walk should be $\displaystyle 360$ cm.

$\displaystyle \\$

Question 10: A die is thrown once. Find the probability of getting (i) a composite number, (ii) a prime number.

Answer:

Total possible outcomes are $\displaystyle 1, 2, 3, 4, 5,6$

Therefore total number of events $\displaystyle = 6$

i) Composite numbers are $\displaystyle 4$ and $\displaystyle 6$

$\displaystyle \text{Therefore Probability (Composite Numbers) } = \frac{2}{6} = \frac{1}{3}$

ii) Prime numbers are $\displaystyle 2, 3, 5$

$\displaystyle \text{Therefore Probability (Prime Numbers) } = \frac{3}{6} = \frac{1}{2}$

$\displaystyle \\$

Question 11: Using completing the square method, show that the equation $\displaystyle x^2 - 8x + 18 = 0$ has no solution.

Answer:

Given equation $\displaystyle x^2 - 8x + 18 = 0$

$\displaystyle \Rightarrow x^2 - 8x = -18$

$\displaystyle \Rightarrow x^2 - 8x+16 = -18 + 16$

$\displaystyle \Rightarrow (x-4)^2 = -2$

A square of a number is always positive. Hence the given equation has no solution.

$\displaystyle \\$

Question 12: Cards numbered $\displaystyle 7$ to $\displaystyle 40$ were put in a box. Poonam selects a card at random. What is the probability that Poonam selects a card which is a multiple of $\displaystyle 7$ ?

Answer:

Total number of cards $\displaystyle = (40-7)+1 = 33 + 1 = 34$

Favorable outcomes $\displaystyle = 7, 14, 21, 28, 35 \Rightarrow 5$

Therefore Probability $\displaystyle = \frac{5}{34}$

$\displaystyle \\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: The perpendicular from $\displaystyle A$ on side $\displaystyle BC$ of a $\displaystyle \triangle ABC$ meets $\displaystyle BC$ at $\displaystyle D$ such that $\displaystyle DB = 3CD$ . Prove that $\displaystyle 2AB^2 = 2AC^2 + BC^2$ .

$\displaystyle AD$ and $\displaystyle PM$ are medians of triangles $\displaystyle ABC$ and $\displaystyle PQR$ respectively $\displaystyle \text{where } \triangle ABC \sim \triangle PQR \text{. Prove that } \frac{AB}{PQ} = \frac{AD}{PM}$

Answer:

Given: $\displaystyle \triangle ABC$ with $\displaystyle AD \perp BC$ Also $\displaystyle DB = 2 CD$

To Prove: $\displaystyle 2AB^2 = 2AC^2 + BC^2$ .

Proof: Let $\displaystyle BC = x$

$\displaystyle \Rightarrow CD + DB = x \Rightarrow CD + 3 CD = x \Rightarrow 4 CD = x \Rightarrow CD = \frac{x}{4}$

$\displaystyle \text{Also } DB = 3CD \Rightarrow DB = \frac{3x}{4}$

Using Pythagoras theorem, in right angled triangle $\displaystyle \triangle ADC$

$\displaystyle AC^2 = AD^2 + DC^2$

$\displaystyle \Rightarrow AC^2 = AD^2 + \Big( \frac{x}{4} \Big)^2$

$\displaystyle \Rightarrow AC^2 = AD^2 + \Big( \frac{x^2}{16} \Big)$ … … … … … i)

Using Pythagoras theorem, in right angled triangle $\displaystyle \triangle ADB$

$\displaystyle AB^2 = AD^2 + DB^2$

$\displaystyle \Rightarrow AB^2 = AD^2 + \Big( \frac{3x}{4} \Big)^2$

$\displaystyle \Rightarrow AB^2 = AD^2 + \Big( \frac{9x^2}{16} \Big)$ … … … … … ii)

Subtracting ii) from i)

$\displaystyle AC^2 - AB^2 = AD^2 + \Big( \frac{x^2}{16} \Big) - AD^2 - \Big( \frac{9x^2}{16} \Big)$

$\displaystyle \Rightarrow AC^2 - AB^2 = \Big( \frac{x^2}{16} \Big) - \Big( \frac{9x^2}{16} \Big)$

$\displaystyle \Rightarrow AC^2 - AB^2 = - \frac{8}{16} x^2$

$\displaystyle \Rightarrow AC^2 - AB^2 = - \frac{1}{2} x^2$

$\displaystyle \Rightarrow 2AC^2 - 2AB^2 = -x^2$ (Putting $\displaystyle x = BC$ back in the equation)

$\displaystyle \Rightarrow 2AC^2 - 2AB^2 = -BC^2$

$\displaystyle \Rightarrow 2AB^2 = 2AC^2 + BC^2$ .

Hence Proved

Given: $\displaystyle \triangle ABC$ and $\displaystyle \triangle PQR$

$\displaystyle AD$ is the median of $\displaystyle \triangle ABC$ and $\displaystyle PM$ is the median of $\displaystyle \triangle PQR$

Also $\displaystyle \triangle ABC \sim \triangle PQR$

$\displaystyle \text{To Prove: } \frac{AB}{PQ} = \frac{AD}{PM}$

$\displaystyle \text{Proof: Since } AD \text{ is a median } \Rightarrow BD = DC = \frac{1}{2} BC$

$\displaystyle \text{Similarly, } PM \text{ is the median } \Rightarrow QM = MR = \frac{1}{2} QR$

Now, $\displaystyle \triangle ABC \sim \triangle PQR$

$\displaystyle \Rightarrow \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \text{ (corresponding sides of similar triangles)}$

So, $\displaystyle \frac{AB}{PQ} = \frac{BC}{QR}$

$\displaystyle \Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM} \text{ (Since AD and PM are medians)}$

$\displaystyle \Rightarrow \frac{AB}{PQ} = \frac{BD}{QM}$ … … … … … i)

Also since $\displaystyle \triangle ABC \sim \triangle PQR$

$\displaystyle \angle B = \angle Q$ (corresponding angles of similar triangles)

Now in $\displaystyle \triangle ABD$ and $\displaystyle \triangle PQM$

$\displaystyle \angle B = \angle Q$

$\displaystyle \frac{AB}{PQ} = \frac{BD}{QM}$ from i)

Hence $\displaystyle \triangle ABD \sim \triangle PQM$ (By SAS criterion)

Since corresponding sides of similar triangles are proportional

$\displaystyle \frac{AB}{PQ} = \frac{AD}{PM} \text{ . Hence proved.}$

$\displaystyle \\$

Question 14: Check whether $\displaystyle g(x)$ is a factor of $\displaystyle p(x)$ by dividing polynomial $\displaystyle p(x)$ by polynomial $\displaystyle g(x)$ , where $\displaystyle p(x) = x^5 - 4x^3 + x^2 + 3x + 1$ , $\displaystyle g(x) = x^3 - 3x + 1$

Answer:

$\displaystyle \text{If } p(x) \text{ divisible by } g(x) \Rightarrow \text{ the remainder of } \frac{p(x)}{g(x)} = 0$

$\displaystyle x^3-3x+1 ) \overline{x^5 - 4x^3 + x^2 + 3x + 1} ( x^2-1 \\ \hspace*{1.5cm}(-) { x^5-3x^3+x^2} \\ \hspace*{2cm} \overline{-x^3+3x+1} \\ \hspace*{1.5cm}(-) \underline{ -x^3+3x-1} \\ \hspace*{4.0cm} {2}$

Therefore remainder is $\displaystyle 2$ .

Therefore $\displaystyle g(x)$ is not a factor of $\displaystyle p(x)$ .

$\displaystyle \\$

Question 15: Find the area of the triangle formed by joining the mid-points of the sides of the triangle $\displaystyle ABC$ , whose vertices are $\displaystyle A(0, - 1), B(2, 1)$ and $\displaystyle C(0, 3)$ .

Answer:

$\displaystyle \text{Mid-point of } AB (x_1, y_1) = \Big( \frac{0+2}{2} , \frac{-1+1}{2} \Big) = (1, 0)$

$\displaystyle \text{Mid-point of } BC (x_2, y_2) = \Big( \frac{2+0}{2} , \frac{1+3}{2} \Big) = (1, 2)$

$\displaystyle \text{Mid-point of } AC (x_3, y_3) = \Big( \frac{0+0}{2} , \frac{-1+3}{2} \Big) = (0, 1)$

$\displaystyle \text{Area of } \triangle PQR = \frac{1}{2} \Big[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \Big]$

$\displaystyle = \frac{1}{2} \Big[ 1(2-1) + 1(1-0) + 0(0-2) \Big]$

$\displaystyle = \frac{1}{2} [ 1+1] = 1$ sq. units.

$\displaystyle \\$

Question 16: Draw the graph of the equations $x - y + 1 = 0$ and $3x + 2y - 12 = 0$ . Using this graph, find the values of $x$ and $y$ which satisfy both the equations.

Answer:

For $x - y + 1 = 0$ : When $x =0, y = 1$ and when $y = 0, x = -1$

For $3x + 2y - 12 = 0$ : When $x = 0, y = 6$ and when $y = 0, x = 4$

From the graph the two lines intersect at $(2, 3)$ . This is the point that satisfy both the equations.

$\\$

Question 17: Prove that $\displaystyle \sqrt{3}$ is an irrational number.

Find the largest number which on dividing $\displaystyle 1251, 9377$ and $\displaystyle 15628$ leaves remainders $\displaystyle 1, 2$ and $\displaystyle 3$ respectively.

Answer:

Assume $\displaystyle \sqrt{3}$ is a rational number

i.e. it can be expressed as a rational fraction of the form $\displaystyle \frac{a}{b}$ where $\displaystyle a, b$ are relatively prime numbers.

$\displaystyle \text{Since } \sqrt{3} = \frac{a}{b}$

$\displaystyle \text{We have } 3 = \frac{a^2}{b^2} \text{ or } a^2 = 3 b^2$

If $\displaystyle b$ is even, then $\displaystyle a$ is also even in which case $\displaystyle a/b$ is not in simplest form. If $\displaystyle b$ is odd then $\displaystyle a$ is also odd. Therefore:

$\displaystyle a = 2n+1$

$\displaystyle b = 2m+1$

$\displaystyle (2n+1)^2 = 3 (2m+1)^2$

$\displaystyle 4n^2 + 1 + 4n = 12m^2 + 3 + 12m$

$\displaystyle 4n^2 + 4n = 12m^2 + 2 + 12m$

$\displaystyle 2n^2 + 2n = 6m^2 + 1 + 6m$

$\displaystyle 2(n^2+n) = 2(3m^2+3) + 1$

Since $\displaystyle (n^2+n)$ is an integer, the left hand side is even. Since $\displaystyle (3m^2+3m)$ is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

Therefore $\displaystyle \sqrt{3}$ is an irrational number

Since, $\displaystyle 1, 2$ and $\displaystyle 3$ are the remainders of $\displaystyle 1251, 9377$ and $\displaystyle 15628$ respectively. Thus, after subtracting these remainders from the numbers.

We have the numbers, $\displaystyle 1251 - 1 = 1250, 9377 - 2 = 9375$ and $\displaystyle 15628 - 3 = 15625$

Which is divisible by the required number.

Now, required number $\displaystyle =$ HCF of $\displaystyle 1250, 9375$ and $\displaystyle 15625$

By Euclid’s division algorithm $\displaystyle a = bq + r , 0 \leq r < b$ … … … … … (i)

$\displaystyle [ \because$ dividend $\displaystyle =$ divisor $\displaystyle \times$ quotient $\displaystyle +$ remainder $\displaystyle ]$

For largest number, put $\displaystyle a = 15625$ and $\displaystyle b = 9375$

$\displaystyle 15625 = 9375 \times 1 + 6250 \ \ \ [ \because r \neq 0 ]$

$\displaystyle \Rightarrow 9375 = 6250 \times 1 + 3125 \ \ \ [ \because r \neq 0 ]$

$\displaystyle \Rightarrow 6250 = 3125 \times 2 + 0$

Now $\displaystyle r = 0$

$\displaystyle \therefore$ HCF $\displaystyle (15625$ and $\displaystyle 9375) = 3125$

Now, we take $\displaystyle c = 1250$ and $\displaystyle d = 3125$ , then again using Euclid’s division algorithm, $\displaystyle d = cq + r, 0 \leq r < c$

$\displaystyle \Rightarrow 3125 = 1250 \times 2 + 625$ $\displaystyle [ \because r \neq 0 ]$

$\displaystyle \Rightarrow 1250 = 625 \times 2 + 0$ [ Now $\displaystyle r = 0$ ]

$\displaystyle \therefore$ HCF $\displaystyle (1250, 9375$ and $\displaystyle 15625) = 625$

Hence, $\displaystyle 625$ is the largest number which divides $\displaystyle 1251, 9377$ and $\displaystyle 15628$ leaving remainder $\displaystyle 1, 2$ and $\displaystyle 3$ respectively

$\displaystyle \\$

Question 18: $\displaystyle A, B$ and $\displaystyle C$ are interior angles of a triangle $\displaystyle ABC$ . Show that

$\displaystyle \text{(i) } \sin \Big( \frac{B+C}{2} \Big) = \cos \frac{A}{2}$

$\displaystyle \text{(ii) If } \angle A = 90^o , \text{ then find the value of} \tan \Big( \frac{B+C}{2} \Big)$

$\displaystyle \text{(iii) If } \tan (A + B) = 1 \text{ and } \tan (A - B) = \frac{1}{\sqrt{3}} , 0^o < A + B < 90^o , A > B ,$

then find the values of $\displaystyle A$ and $\displaystyle B$ .

Answer:

$\displaystyle \text{i) To prove: } \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)$

In $\displaystyle \triangle ABC$ , sum of angles $\displaystyle = 180^o$

$\displaystyle A + B + C = 180^o$

$\displaystyle \Rightarrow B+ C = 180^o - A$

$\displaystyle \Rightarrow \frac{B+C}{2} = 90^o - \frac{A}{2}$

$\displaystyle \Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \sin \Big( 90 - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \sin \Big( \frac{B+C}{2} \Big) = \cos \Big( \frac{A}{2} \Big)$ … … … … … i)

Hence Proved.

$\displaystyle \text{ii) Similarly, } \cos \Big( \frac{B+C}{2} \Big) = \cos \Big( 90 - \frac{A}{2} \Big)$

$\displaystyle \Rightarrow \cos \Big( \frac{B+C}{2} \Big) = \sin \Big( \frac{A}{2} \Big)$ … … … … … ii)

From i) and ii) and $\displaystyle \angle A = 90^o$

$\displaystyle \tan \Big( \frac{B+C}{2} \Big) = \frac{\cos 45^o}{\sin 45^o} = \frac{1}{\tan 45^o} = 1$

$\displaystyle \therefore \tan \Big( \frac{B+C}{2} \Big) = 1$

Given $\displaystyle \tan (A+B) = 1 \Rightarrow \tan (A+B) = \tan 45^o$

$\displaystyle \therefore A+B = 45^o$ … … … … … i)

$\displaystyle \text{Similarly, } \tan (A-B) = \frac{1}{\sqrt{3}} \Rightarrow \tan (A-B) = \tan 30^o$

$\displaystyle \therefore A-B = 30^o$ … … … … … ii)

Solving i) and ii)

$\displaystyle 2A = 75^o \Rightarrow A = 37.5^o$

From i) $\displaystyle B = 45^o - A = 45^o - 37.5^o = 7.5^o$

Hence $\displaystyle A = 37.5^o$ and $\displaystyle B = 7.5^o$

$\displaystyle \\$

Question 19: In Figure 2, $PQ$ is a chord of length $8$ cm of a circle of radius $5$ cm. The tangents at $P$ and $Q$ intersect at a point $T$ . Find the length $TP$ .

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.

Answer:

$TP = TQ$ (Length of the tangents from an external point to a circle are equal)

In $\triangle TPQ TP = TQ$

$OT$ is the bisector of $\angle PTQ$

$\therefore OT \perp PQ$

Since $OT \perp PQ$

$PR = RQ$ ( perpendicular from center to a chord bisects the chord)

$\therefore PR = QR = \frac{1}{2} PQ = 4 \text{ cm }$

In Right triangle $ORP$

$OP^2 = OR^2+ PR^2$

$\Rightarrow 5^2 = OR^2 + 4^2$

$\Rightarrow OR^2 = 25 - 16 = 9$

$\Rightarrow OR = 3$ cm

Let $TP = x$

In Right triangle $PRT$

$TP^2 = PR^2 + RT^2$

$\Rightarrow x^2 = 4^2 + RT^2$ … … … … … i)

Since $TP$ is a tangent, $OP \perp TP$

$\angle OPT = 90^o$

In Right triangle $OPT$

$OT^2 = OP^2 + TP^2$

$\Rightarrow OT^2 = 5^2 + x^2$

$\Rightarrow (OR+RT)^2 = 5^2 + x^2$

$\Rightarrow (3+RT)^2 = 5^2 + x^2$

$\Rightarrow 9 + RT^2 + 6 RT = 25 + x^2$

From i) $x^2 = 16+RT^2$

$9 + RT^2 + 6 RT = 25 + 16+RT^2$

$\Rightarrow 6RT = 32$

$\displaystyle \Rightarrow RT = \frac{32}{6} = \frac{16}{3}$

$\displaystyle \therefore x^2 = 16 + \Big( \frac{16}{3} \Big)^2 = \frac{400}{9}$

$\displaystyle \Rightarrow x = \frac{20}{3}$

$\displaystyle \text{Hence TP is } \frac{20}{3} \text{ cm }$

Given: A circle with center $O$ touches the side $AB, BC, CD$ and $DA$ of a quadrilateral $ABCD$ at points $P, Q, R$ and $S$ .

To Prove: $\angle AOB + \angle COD = 180^o$ and $\angle AOD + \angle BOC = 180^o$

Construction: Join $OP, OQ, OR$ and $OS$

Proof: Since the two tangents from an external point to a circle subtend equal angles at the center;

$\angle 1 = \angle 2, \angle 3 = \angle 4, \angle 5 = \angle 6$ and $\angle 7 = \angle 8$

Since the sum of all the angles subtended at the point is $360^o$

$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^o$

$\Rightarrow 2 (\angle 2 + \angle 4 + \angle 6 + \angle 7) = 360^o$

$\Rightarrow \angle 2 + \angle 3 + \angle 6 + \angle 7 = 180^o$

$\Rightarrow (\angle 2+\angle 3) + ( \angle 6+\angle 7) = 180^o$

$\Rightarrow \angle COD + \angle AOB = 180^o$

Similarly

$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^o$

$\Rightarrow 2 (\angle 1 + \angle 4 + \angle 5 + \angle 8) = 360^o$

$\Rightarrow \angle 1 + \angle 4 + \angle 5 + \angle 8 = 180^o$

$\Rightarrow (\angle 1+\angle 8) + ( \angle 4+\angle 5) = 180^o$

$\Rightarrow \angle AOD + \angle BOC = 180^o$

$\\$

Question 20: Water in a canal, $6$ m wide and $1.5$ m deep, is flowing with a speed of $10$ km/h. How much area will it irrigate in $30$ minutes if $8$ cm of standing water is needed ?

Answer:

$\displaystyle \text{Speed of water } = \frac{10 \ km}{hr} = \frac{10 \times 1000 \ m}{60 \ min} = \frac{1000 \ m}{6 \ min}$

$\displaystyle \text{Rate of flow } = 6 \times 1.5 \ m^2 \times \frac{1000 \ m}{6 \ min} = \frac{1500 \ m^3}{min}$

$\displaystyle \text{Therefore volume of water in 30 mins } = 1500 \frac{m^3}{min} \times 30 \ min = 45000 \ \ m^3$

Let Area of irrigation $= x$

$\therefore x \times \frac{8}{100} \ m = 45000 \ m^3$

$\displaystyle x = \frac{45000 \times 100}{8} \ m^2 = 562500 \ m^2$

$\\$

Question 21: A class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of days a student was absent.

Number of Days:	0-6	6-12	12-18	18-24	24-30	30-36	35-42
Number of Students:	10	11	7	4	4	3	1

Answer:

Class Interval	$f_i$	$x_i$	$f_ix_i$
0-6	10	3	30
6-12	11	9	99
12-18	7	15	105
18-24	4	21	84
24-30	4	27	108
30-36	3	33	99
36-42	1	39	39
	$\Sigma f_i = 40$		$\Sigma f_ix_i = 564$

$\displaystyle \text{Mean value } = \frac{\Sigma f_ix_i}{\Sigma f_i} = \frac{564}{40} = 14.1$

$\\$

Question 22: A car has two wipers which do not overlap. Each wiper has a blade of
length $21$ cm sweeping through an angle $120^o$ . Find the total area cleaned at each sweep of the blades. (Take $\pi = \frac{22}{7}$ )

Answer:

$\displaystyle \text{Total area } = 2 \Big[ \frac{120}{360} \times \pi (21)^2 \Big]$

$\displaystyle = 2 \Big[ \frac{1}{3} \times \frac{22}{7} \times 21 \times 21\Big]$

$= 2 \times 22 \times 21$

$= 924 \ cm^2$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: A pole has to be erected at a point on the boundary of a circular park of diameter $13$ m in such a way that the difference of its distances from two diametrically opposite fixed gates $A$ and $B$ on the boundary is $7$ m. Is it possible to do so ? If yes, at what distances from the two gates should the pole be erected ?

Answer:

Let $AC = a$ and $BC = b$

$\therefore a - b = 7$

$\Rightarrow a = b+ 7$

We know that diameter will subtend a right angle on circumference.

$\therefore AB^2 = a^2 + b^2$

$\Rightarrow 13^2 = (b+7)^2 + b^2$

$\Rightarrow 169 = 2b^2 + 14 b + 49$

$\Rightarrow 2b^2 + 14b - 120 =0$

$\Rightarrow b^2 + 7b-60 = 0$

$\Rightarrow (b+12)(b-5) = 0$

$\Rightarrow b = 5 \text{ or} -12$ (this is not possible as b cannot be negative)

$\therefore b = 5$ m

$\Rightarrow 5+7 = 12$ m

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Question 24: If $m$ times the $m^{th}$ term of an Arithmetic Progression is equal to $n$ times its $n^{th}$ term and $m \neq n$ , show that the $(m + n)^{th}$ term of the A.P. is zero.

The sum of the first three numbers in an Arithmetic Progression is $18$ . If the product of the first and the third term is $5$ times the common difference, find the three numbers.

Answer:

Let the first term of AP $= a$

common difference $= d$

We have to show that $(m+n)^{th}$ term is zero or $a + (m+n-1)d = 0$

$m^{th}$ term $= a + (m-1)d$

$n^{th}$ term $= a + (n-1) d$

Given that $m{a +(m-1)d} = n{a + (n -1)d}$

$\Rightarrow am+m^2d-md=an+n^2d-nd$

$\Rightarrow am-an+m^2d-n^2d-md+nd=0$

$\Rightarrow a(m-n)+(m^2-n^2)d-(m-n)d=0$

$\Rightarrow a(m-n)+{(m-n)(m+n)}d-(m-n)d=0$

$\Rightarrow a(m-n) + {(m-n)(m+n) - (m-n)} d = 0$

$\Rightarrow a(m-n)+(m-n)(m+n-1)d=0$

$\Rightarrow (m-n){a + (m+n-1)d} = 0$

$\displaystyle \Rightarrow a+(m+n-1)d= \frac{0}{(m-n)}$

$\Rightarrow a+(m+n-1)d=0$

Hence Proved

Let first three numbers of A.P are $(a-d)$ , $a$ and $(a+d)$

Sum of the first three numbers $= (a-d) + a +(a+d) = 3a = 18 \text{ or} a = 6$

Product of first and third term $= (a-d)(a+d) = a^2 -d^2 = 36 -d^2 = 5d \text{ or} d^2 +5d -36 = 0 \text{ or} (d+9)(d-4) = 0$

Hence $d = 4$ and three numbers are $2, 6, 10$

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Question 25: Construct a triangle $ABC$ with side $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60^o$ . Then construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$ .

Answer:

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Question 26: In Figure 3, a decorative block is shown which is made of two solids, a cube and a hemisphere. The base of the block is a cube with edge $6$ cm and the hemisphere fixed on the top has a diameter of $4.2$ cm. Find
(a) the total surface area of the block.
(b) the volume of the block formed. (Take $\pi = \frac{22}{7}$ )

A bucket open at the top is in the form of a frustum of a cone with a capacity of $12308.8 \ cm^3$ . The radii of the top and bottom circular ends are $20$ cm and $12$ cm respectively. Find the height of the bucket and the area of metal sheet used in making the bucket. (Use $\pi = 3.14$ )

Answer:

a) Total surface area

$\displaystyle = 6 \times (6 \times 6) - \pi (2.1)^2 + \frac{1}{2} \Big( 4 \pi (2.1)^2 \Big)$

$\displaystyle = 216 - \frac{22}{7} \times 2.1 \times 2.1 + 2 \times \frac{22}{7} \times 2.1 \times 2.1$

$= 216 - 13.86 + 27.72$

$= 229.86 \ cm^2$

b) Volume of the block

$\displaystyle = 6^3 + \frac{1}{2} \Big( \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 \Big)$

$= 216 + 19.404$

$= 235.404 \ cm^3$

Volume $= 12308.80 \ cm^3$

$r_1 = 20$ cm $r_2 = 12$ cm $h = ?$

Volume of the bucket $\displaystyle = \frac{\pi}{3} h ( {r_1}^2 + {r_2}^2 +r_1. r_2 )$

$\displaystyle \Rightarrow 12308.80 = 3.14 \times \frac{h}{3} ( 20^2 + 12^2 + 20 \times 12)$

$\displaystyle \Rightarrow h = 12308.80 \times \frac{3}{3.14 \times 784} = 15$ cm

Therefore height of bucket $= 15$ cm

$l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{15^2 + 8^2} = 17$ cm

Therefore surface are of the metal sheet used

$=$ curved surface area $+$ base area

$= \pi ( r_1 + r_2) l + \pi r^2$

$= 3.14 \times 32 \times 17 + 3.14 \times 12^2 = 2160.32 \ cm^2$

Hence the height of the bucket is $15$ cm and surface are of the metal sheet is $2160.32 \ cm^2$

$\\$

Question 27: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.

Prove that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer:

Given: In $\triangle ABC, DE$ intersects $AB$ and $AC$ at $D$ and $E$ respectively.

$\displaystyle \text{To Prove: } \frac{AD}{DB} = \frac{AE}{EC}$

Construct: $DM \perp AC$ and $EN \perp AB$

Proof:

$\displaystyle \text{Area of } \triangle ADE = \frac{1}{2} \times AD \times EN$

$\triangle \text{Area of } BDE = \frac{1}{2} \times BD \times EN$

$\displaystyle \therefore \frac{ar(\triangle ADE)}{ar(\triangle BDE)} = \frac{AD}{BD}$ … … … … … i)

$\text{Now } ar(\triangle BDE) = ar(\triangle CDE)$ … … … … … ii)

(Because they have the same base $DE$ and are between the same parallel)

$\displaystyle \text{Similarly, } \frac{ar(\triangle ADE)}{ar(\triangle CDE)} = \frac{AE}{EC}$ … … … … … iii)

From i) , ii) and iii) we get $\displaystyle \frac{AD}{DB} = \frac{AE}{EC}$ . Hence proved.

Given: A right angled $\triangle ABC$ , right angled at $B$

To prove: $AC^2 = AB^2 + BC^2$

Draw: $BD \perp AC$

Proof: In $\triangle ADB \& \angle ABC$

$\angle ADB = \angle ABC = 90^o$

$\angle BAD = \angle CAB$ (common angle)

$\therefore \triangle ADB \sim \triangle ABC$ ( by AA criterion)

$\displaystyle \text{Therefore } \frac{AD}{AB} = \frac{AB}{AC} \text{ (corresponding sides are proportional) }$

$AB^2 = AD \times AC$ … … … … … i)

Similarly, $\triangle BDC \sim \triangle ABC$

and $BC^2 = CD \times AC$ … … … … … ii)

Adding i) and ii)

$AB^2 + BC^2 = AD \times AC + CD \times AC$

$\Rightarrow AB^2 + BC^2 = AC ( AD + CD)$

$\Rightarrow AB^2 + BC^2 = AC^2$ . Hence proved.

$\\$

Question 28: If $1 + \sin^2 \theta = 3 \sin \theta \cos \theta$ , then prove that $\displaystyle \tan \theta = 1 \text{ or} \tan \theta = \frac{1}{2}$

Answer:

We have $1 + \sin^2 \theta = 3 \sin \theta \cos \theta$

Dividing both sides with $\cos^2 \theta$

$\displaystyle \Rightarrow \frac{1}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{3 \sin \theta \cos \theta}{\cos^2 \theta}$

$\Rightarrow \sec^2 \theta + \tan^2 \theta = 3 \tan \theta$

$\Rightarrow 1 + \tan^2 \theta + \tan^2 \theta = 3 \tan \theta (\because \sec^2 \theta = 1 + \tan^2 \theta)$

$\Rightarrow 2 \tan^2 \theta - 3 \tan \theta + 1 = 0$

Let $\tan \theta = y$

$\therefore 2y^2 - 3y + 1= 0$

$\Rightarrow 2y^2 - 2y - y + 1 = 0$

$\Rightarrow 2y(y -1) -1(y -1) = 0$

$\Rightarrow (2y -1)(y-1) = 0$

$\displaystyle \Rightarrow y = 1 \text{ or} y = \frac{1}{2}$

$\displaystyle \text{Hence } \tan \theta = 1 \text{ or} \tan \theta = \frac{1}{2}$

$\\$

Question 29: Change the following distribution to a ‘more than type’ distribution. Hence draw the ‘more than type’ ogive for this distribution.

Class Interval:	20-30	30-40	40-50	50-60	60-70	70-80	80-90
Frequency	10	8	12	24	6	25	15

Answer:

Less Than table

Less Than	Cumulative Frequency
Less than 30	10
Less than 40	18
Less than 50	30
Less than 60	54
Less than 70	60
Less than 80	85
Less than 90	100

More than table

More Than	Cumulative Frequency
More than 30	10
More than 40	18
More than 50	30
More than 60	54
More than 70	60
More than 80	85
More than 90	100

Draw the chart. The point of intersection is :

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Question 30: The shadow of a tower standing on a level ground is found to be $40$ m longer when the Sun’s altitude is $30^o$ than when it was $60^o$ . Find the height of the tower. (Given $\sqrt{3} = 1.732$ )