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• 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 $(x + 5)^2 = 2 (5x - 3)$.

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

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

$\Rightarrow x^2 + 31 = 0$

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

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

$\\$

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

Or

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

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.

Or

$429 = 3 \times 11 \times 13$

$\\$

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

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

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

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

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

$= 5 (12 + 54) = 330$

$\\$

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

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 $(3, -4)$ 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.

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 $\tan \theta$ .

Or

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

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

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

Or

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

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

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

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

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

Or

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

Let midpoint of $BD$ be $O$

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

$O$ is also the midpoint of $AC$

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

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

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

Or

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

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

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

Similarly,

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

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

$\\$

Question 8: Solve the following pair of linear equations :
$3x - 5y = 4$
$2y + 7 = 9x$

Given equations:

$3x - 5y = 4$   … … … … … i)
$2y + 7 = 9x$

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

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

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

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

Substituting into i)

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

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

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

$\\$

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

Or

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

$117 = 3 \times 3 \times 13$

$65 = 5 \times 13$

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

$\therefore 65n - 117 = 13$

$\Rightarrow 65n = 130$

$\Rightarrow n = 2$

Or

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

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

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

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

$\\$

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

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

Therefore total number of events $= 6$

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

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

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

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

$\\$

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

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

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

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

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

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

$\\$

Question 12: Cards numbered $7$ to $40$ 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 $7$ ?

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

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

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

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

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

Or

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

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

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

Proof: Let $BC = x$

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

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

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

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

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

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

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

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

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

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

Subtracting ii) from i)

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

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

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

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

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

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

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

Hence Proved

Or

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

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

Also $\triangle ABC \sim \triangle PQR$

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

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

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

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

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

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

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

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

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

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

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

$\angle B = \angle Q$

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

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

Since corresponding sides of similar triangles are proportional

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

$\\$

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

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

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

Therefore remainder is $2$.

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

$\\$

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

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

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

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

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

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

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

$\\$

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.

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 $\sqrt{3}$ is an irrational number.

Or

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

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

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

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

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

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

$a = 2n+1$

$b = 2m+1$

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

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

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

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

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

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

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

Or

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

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

Which is divisible by the required number.

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

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

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

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

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

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

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

Now $r = 0$

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

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

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

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

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

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

$\\$

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

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

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

Or

If $\tan (A + B) = 1$ and $\tan (A - B) =$ $\frac{1}{\sqrt{3}}$ $, 0^o < A + B < 90^o$, $A > B$, then find the values of $A$ and $B$.

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

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

$A + B + C = 180^o$

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

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

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

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

Hence Proved.

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

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

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

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

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

Or

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

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

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

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

Solving i) and ii)

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

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

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

$\\$

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 $TP$.

Or

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

$TP = TQ$ (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)

$\therefore PR = QR =$ $\frac{1}{2}$ $PQ = 4$ 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) $x^2 = 16+RT^2$

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

$\Rightarrow 6RT = 32$

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

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

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

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

Or

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 ?

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

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

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

Let Area of irrigation $= x$

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

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

$\\$

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

 Class Interval $f_i$$f_i$ $x_i$$x_i$ $f_ix_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_i = 40$ $\Sigma f_ix_i = 564$$\Sigma f_ix_i = 564$

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 $\pi = \frac{22}{7}$ )

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

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

$= 2 \times 22 \times 21$

$= 924 \ cm^2$

$\\$

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 ?

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.

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

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

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

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

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

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

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

$\therefore b = 5$ m

$\Rightarrow 5+7 = 12$ m

$\\$

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.

Or

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.

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 $m{a +(m-1)d} = n{a + (n -1)d}$

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

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

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

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

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

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

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

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

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

Hence Proved

Or

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

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

Product of first and third term $= (a-d)(a+d) = a^2 -d^2 = 36 -d^2 = 5d$  or  $d^2 +5d -36 = 0$  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 $ABC$.

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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 $\pi =$ $\frac{22}{7}$)

Or

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 $\pi = 3.14$)

a) Total surface area

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

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

$= 216 - 13.86 + 27.72$

$= 229.86 \ cm^2$

b) Volume of the block

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

$= 216 + 19.404$

$= 235.404 \ cm^3$

Or

Volume $= 12308.80 \ cm^3$

$r_1 = 20$ cm  $r_2 = 12$ cm  $h = ?$

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

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

$\Rightarrow h = 12308.80 \times$ $\frac{3}{3.14 \times 784}$ $= 15$ 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 $2160.32 \ cm^2$

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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.

Or

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

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

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

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

Proof:

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

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

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

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

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

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

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

Or

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)

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

$AB^2 = AD \times AC$   … … … … … i)

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

and $BC^2 = CD \times AC$   … … … … … 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.

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

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

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

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

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

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

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

Let $\tan \theta = y$

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

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

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

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

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

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

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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

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 $\sqrt{3} = 1.732$)

Let $h$ be the height of the tower.

$\frac{h}{CB}$ $= \tan 60^o = \sqrt{3}$

$\Rightarrow CB =$ $\frac{h}{\sqrt{3}}$   … … … … … i)

Also $\frac{h}{40 + CB}$ $= \tan 30^o =$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \sqrt{3} h = 40 + CB$

$\Rightarrow CB = \sqrt{3}h - 40$  … … … … … ii)

From i) and ii)

$\frac{h}{\sqrt{3}} = \sqrt{3}h - 40$

$\Rightarrow h = 3h - 40\sqrt{3}$

$\Rightarrow 2h = 40 \sqrt{3}$

$\Rightarrow h = 20 \sqrt{3} = 20 \times 1.732 = 34.64$ m

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