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: Given $\triangle ABC \sim \triangle PQR$, if $\frac{AB}{PQ}$ $=$ $\frac{1}{3}$, then find $\frac{ar \triangle ABC}{ar \triangle PQR}$

Given $\triangle ABC \sim \triangle PQR$

$\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle PQR)}$ $= \Big($ $\frac{AB}{PQ}$ $\Big)^2= \Big($ $\frac{1}{3}$ $\Big)^2 =$ $\frac{1}{9}$

$\\$

Question 2: What is the value of $(\cos^2 67^o - \sin^2 23^o)$ ?

$(\cos^2 67^o - \sin^2 23^o)$

$= (\cos^2 67^o - \sin^2 (90^o-67^o) )$

$= (\cos^2 67^o - \cos^2 67^o)$

$= 0$

$\\$

Question 3: If $x = 3$ is one root of the quadratic equation $x^2 - 2kx - 6 = 0$, then find the value of $k$.

If $x = 3$ is one root of the quadratic equation $x^2 - 2kx - 6 = 0$ than it should satisfy the equation.

$\Rightarrow (3)^2 - 2k(3) - 6 = 0$

$9 - 6k - 6 = 0$

$k =$ $\frac{1}{2}$

$\\$

Question 4: Find the distance of a point $P(x, y)$ from the origin.

Distance of a point $P(x, y)$ from the origin $O(0, 0) = \sqrt{(x-o)^2 + (y-0)^2} = \sqrt{x^2 + y^2}$

$\\$

Question 5: What is the HCF of smallest prime number and the smallest composite number ?

Smallest composite number is $= 4 = 2 \times 2$

Smallest prime number is $= 2 = 2 \times 1$

Therefore HCF of $4$ and $2$ is $2$

$\\$

Question 6: In an AP, if the common difference $(d) = -4$, and the seventh term $(a_7)$ is $4$, then find the first term.

Given $d = -4, a_7 = 4$

$a_n = a + (n-1) d$

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

$\Rightarrow 4 = a + (7-1)(-4)$

$\Rightarrow 4 = a - 24$

$\Rightarrow a = 28$

$\\$

Section – B

Question number 7 to 12 carry 2 mark each.

Question 7: An integer is chosen at random between $1$ and $100$. Find the probability that it is :
(i) divisible by $8$.
(ii) not divisible by $8$.

Number of integers between $1$ and $100: 2, 3 , 4, 5 , \ldots , 99$

Total number of outcomes $n(S) = 98$

(i) Numbers which are divisible by $8$ are: $8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96$

Favorable Outcomes $n(A) = 12$

Probability that integer is divisible by $8, P(E) =$ $\frac{n(A)}{n(S)}$ $=$ $\frac{12}{98}$ $=$ $\frac{6}{49}$

(ii) Probability that integer is not divisible by $8$

$P(E') = 1 - P(E)$

$= 1 -$ $\frac{6}{49}$

$=$ $\frac{43}{49}$

$\\$

Question 8:Two different dice are tossed together. Find the probability :
(i) of getting a doublet
(ii) of getting a sum $10$, of the numbers on the two dice.

Given that two different dice are tossed.

Total number of outcomes $n(S) = 6^2 = 36$

(i) Let $A$ be the event of getting a doublet.

$n(A) = \{ 1,1 \},{2,2}, \{3,3 \}, \{4,4 \}, \{5,5 \}, \{6,6 \} = 6$

Required probability $P(A) =$ $\frac{n(A)}{n(S)}$ $=$ $\frac{6}{36}$ $=$ $\frac{1}{6}$

(ii) Let $B$ be the event of getting a sum of $10$ of the numbers of two dice.

$n(B) = \{6,4 \}, \{4,6 \}, \{5,5 \} = 3$

Required probability $P(B) =$ $\frac{n(B)}{n(S)}$ $=$ $\frac{3}{36}$ $=$ $\frac{1}{12}$

$\\$

Question 9: Find the ratio in which $P(4, m)$ divides the line segment joining the points $A(2, 3)$ and $B(6, -3)$. Hence find $m$.

Let $P(4, m)$ divides $A (2, 3)$ and $B (6, -3)$ in the ratio of $k:1$

By section formula,

$4 =$ $\frac{k \times 6 + 1 \times 2}{k+1}$

$\Rightarrow 4k + 4 = 6k + 2$

$\Rightarrow k = 1$

Also $m =$ $\frac{1 \times (-3) + 1 \times 3}{1+1}$

$\Rightarrow m = 0$

$\\$

Question 10: Given that $\sqrt{2}$ is irrational, prove that $(5 + 3 \sqrt{2} )$ is an irrational number.

Given that, $\sqrt{2}$ is irrational

Let us assume $5 + 3\sqrt{2}$ is rational.

As $5 + 3\sqrt{2}$ is rational. (Assumed) They must be in the form of $\frac{p}{q}$ where $q \neq 0$, and  $p$ and  $q$ are co prime.

$\therefore 5 + \sqrt{2} =$ $\frac{p}{q}$

$\Rightarrow 3\sqrt{2} =$ $\frac{p-5q}{q}$

$\Rightarrow \sqrt{2} =$ $\frac{p-5q}{3q}$

We know $\sqrt{2}$ is irrational but $\frac{p-5q}{3q}$ is a rational number

Therefore we contradict the statement that, $5+3\sqrt{2}$ is rational.

Hence proved that $5 + 3\sqrt{2}$ is irrational.

$\\$

Question 11:

In Fig. 1, $ABCD$ is a rectangle. Find the values of $x$ and $y$.

Given:

$x+y = 30$   … … … … … i)

$x-y = 14$   … … … … … ii)

Adding i) and ii) we get

$2x = 44 \Rightarrow x = 22$

$\therefore y = 30 - 22 = 8$

$\\$

Question 12: Find the sum of first $8$ multiples of $3$.

The first $8$ multiples of $3$ are : $3, 6, 9, 12, 15, 18, 21, 24$

Number of terms $(n) = 8$

The first term $(a) = 3$

Common difference $(d) = 3$

$S_n =$ $\frac{n}{2}$ $[2a + (n-1)d ]$

$S_8 =$ $\frac{8}{2}$ $[2 \times 3 + (8-1)\times 3]$

$S_8 = 108$

Hence the sum of the first $8$ multiples of $3$ is $108$.

$\\$

Section – C

Question number 13 to 22 carry 3 mark each.

Question 13: A plane left $30$ minutes late than its scheduled time and in order to reach the destination $1500$ km away in time, it had to increase its speed by $100$ km/h from the usual speed. Find its usual speed.

Distance to travel $= 1500$ km

Let the usual speed $= x$ km/hr

Increased speed $= (x+100)$ km/hr

$\therefore$ $\frac{1500}{x+100}$ $+$ $\frac{1}{2}$ $=$ $\frac{1500}{x}$

$\frac{3000 + x + 100}{2(x+100)}$ $=$ $\frac{1500}{x}$

$3000x + x^2 + 100x = 3000x + 300000$

$x^2 + 100x - 300000 = 0$

$x^2 - 500x + 600x - 300000 = 0$

$x( x- 500) + 600(x-500) = 0$

$(x-500)(x+600) = 0$

$x = 500$ km/hr or $x = -600$ km/hr (this is not possible as speed cannot be negative)

Therefore the usual speed $= 500$ km/hr

$\\$

Question 14: Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

Or

If the area of two similar triangles are equal, prove that they are congruent.

Let the side of the square $= a$

Therefore the sides of the equilateral $\triangle ABE = a$ as well

Now $BD = \sqrt{a^2 + a^2} = \sqrt{2} a$

Therefore the sides of the equilateral $\triangle BDF = \sqrt{2} a$

Since the triangles are equilateral, each of the angles are $60^o$. Hence by $AAA$ criterion, the two triangles are similar.

Therefore

$\frac{Ar (\triangle ABE}{Ar (\triangle BDF)}$ $= \Big($ $\frac{AB}{BD}$ $\Big)^2 = \Big($ $\frac{a}{\sqrt{2}a}$ $\Big)^2 =$ $\frac{1}{2}$

Hence proven.

Or

Given: $\triangle ABC$ and $\triangle DEF$

Also $\triangle ABC \sim \triangle DEF$ and $Ar(\triangle ABC) = Ar(\triangle DEF)$

To Prove: $\triangle ABC \cong \triangle DEF$

Since $\triangle ABC \sim \triangle DEF$ we know

$\frac{Ar (\triangle ABC )}{Ar (\triangle DEF)}$ $= \Big($ $\frac{BC}{EF}$ $\Big)^2 = \Big($ $\frac{AB}{DE}$ $\Big)^2 =\Big($ $\frac{AC}{DF}$ $\Big)^2$

$1 = \Big($ $\frac{BC}{EF}$ $\Big)^2 = \Big($ $\frac{AB}{DE}$ $\Big)^2 =\Big($ $\frac{AC}{DF}$ $\Big)^2$

Therefore we get

$1 = \Big($ $\frac{BC}{EF}$ $\Big)^2 \Rightarrow$ $\frac{BC}{EF}$ $= 1 \Rightarrow BC = EF$

$1 = \Big($ $\frac{AB}{DE}$ $\Big)^2 \Rightarrow$ $\frac{AB}{DE}$ $= 1 \Rightarrow AB = DE$

$1 = \Big($ $\frac{AC}{DF}$ $\Big)^2 \Rightarrow$ $\frac{AC}{DF}$ $= 1 \Rightarrow AC = DF$

Therefore $\triangle ABC \cong \triangle DEF$ by SSS criterion.

Hence proved.

$\\$

Question 15: Prove that the lengths of tangents drawn from an external point to a circle are equal.

Given: Let the circle with center $O$

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

To prove: $PQ = PR$

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

Proof: As $PQ$ is tangent $OQ \perp PQ$. Therefore $\angle OQP = 90^o$

Similarly, As $PR$ is tangent $OR \perp PR$. Therefore $\angle ORP = 90^o$

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

In $\triangle OQP$ and $\triangle ORP$

$\angle OQP = \angle ORP = 90^o$

$OP$ is common

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

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

$\Rightarrow PQ = PR$

Therefore both tangents are equal in length.

$\\$

Question 16: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 2. If the height of the cylinder is $10$ cm and its base is of radius $3.5$ cm. Find the total surface area of the article.

Or

A heap of rice is in the form of a cone of base diameter $24$ m and height $3.5$ m. Find the volume of the rice. How much canvas cloth is required to just cover the heap ?

Radius of the cylinder $(r) = 3.5$ cm Height $(h) = 10$ cm

Curved surface area of cylinder $= 2 \pi r h = 2 \pi \times 3.5 \times 10 = 70 \pi \ cm^2$

Surface area of two hemispheres $= 2 [ \frac{1}{2} \times 4 \pi r^2 ]$ $= 4\pi \times 3.5 \times 3.5 = 49 \pi \ cm^2$

Therefore total surface area $= 70 \pi + 49 \pi = 119 \times$ $\frac{22}{7}$ $= 374 \ cm^2$

Or

Diameter $= 24$ m $\Rightarrow$ Radius $(r) = 12$ m Height $(h) = 3.5$ m

Volume of Rice $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\times$ $\frac{22}{7}$ $\times 12 \times 12 \times 3.5 = 22 \times 2 \times 12 = 528 \ m^3$

Slant height $(l) = \sqrt{12^2 + 3.5^2} = \sqrt{156.25} = 12.5$ m

Curved surface area of the heap $= \pi r l =$ $\frac{22}{7}$ $\times 12 \times 12.5 = 471.43 \ m^2$

Hence Volume of Rice $= 528 \ m^3$ and Canvas required to cover the heap $= 471.43 \ m^2$

$\\$

Question 17: The table below shows the salaries of 280 persons:

 Salary (in thousand Rs.) No.  of persons 5-10 49 10-15 133 15-20 63 20-25 15 25-30 6 30-35 7 35-40 4 40-45 2 45-50 1

 Class Interval Frequency $(f)$$(f)$ Cumulative Frequency $(cf)$$(cf)$ 5 – 10 49 49 10 – 15 133 182 15 – 20 63 245 20 – 25 15 260 25 – 30 6 266 30 – 35 7 273 35 – 40 4 277 40 – 45 2 279 45 – 50 1 280

$cf = 133, f = 133, L = 10, N = 280 \Rightarrow$ $\frac{N}{2}$ $= 140, C = 5$

Median $(M) = L +$ $\frac{\frac{N}{2} - cf}{f}$ $\times C = 10 + \Big($ $\frac{140 - 49}{133}$ $\Big) \times 5 = 10 + 3.42 = 13.42$

Therefore median salary $= 13.42 \times 1000 =13420$ Rs.

$\\$

Question 18: If $4 \tan \theta = 3$, evaluate $\frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1}$

Or

If $\tan 2A = \cot (A - 18^o)$, where $2A$ is an acute angle, find the value of $A$.

Given $4 \tan \theta = 3 \Rightarrow \tan \theta =$ $\frac{3}{4}$ $\Rightarrow \sin \theta =$ $\frac{3}{5}$ $\Rightarrow \cos \theta =$ $\frac{4}{5}$

Therefore

$\frac{4 \sin \theta - \cos \theta + 1}{4 \sin \theta + \cos \theta - 1}$ $=$ $\frac{4 \times \frac{3}{5} - \frac{4}{5} + 1}{4 \times \frac{3}{5} + \frac{4}{5} - 1}$ $=$ $\frac{12 - 4 + 5}{12 + 4 -5}$ $=$ $\frac{13}{11}$

Or

$\tan 2A = \cot (A - 18^o)$

$\Rightarrow \tan 2A = \tan [ 90- (A - 18^o) ]$

$\Rightarrow \tan 2A = \tan ( 108^o - A)$

$\Rightarrow 2A = 108^o - A$

$\Rightarrow A = 36^o$

$\\$

Question 19: Find the area of the shaded region in Fig. 3, where arcs drawn with centres $A, B, C$and $D$ intersect in pairs at mid-points $P, Q, R$ and $S$ of the sides $AB, BC, CD$ and $DA$ respectively of a square $ABCD$ of side $12$ cm. [Use $\pi = 3.14$]

Given square $ABCD$ of side $12$ cm

Area of square $= 12 \times 12 = 144 \ cm^2$

Area of $APS = \frac{1}{4} (\pi \times 6 \times 6) = 9\pi \ cm^2$

Therefore total area enclosed in arc’s $= 4 \times 9\pi = 36 \pi = 36 \times 3.14 = 113.04 \ cm^2$

Therefore shaded area $= 144 - 113.04 = 30.96 \ cm^2$

$\\$

Question 20: If $A(-2, 1), B(a, 0), C(4, b)$ and $D(1, 2)$ are the vertices of a parallelogram $ABCD$, find the values of a and b. Hence find the lengths of its sides.

Or

If $A(-5, 7), B(-4, -5), C(-1, -6)$ and $D(4, 5)$ are the vertices of a quadrilateral, find the area of the quadrilateral $ABCD$.

Mid point of $AB = \Big($ $\frac{-2+4}{2}$ $,$ $\frac{1+b}{2}$ $\Big) = \Big( 1,$ $\frac{1+b}{2}$ $\Big)$

Mid point of $BD = \Big($ $\frac{a+1}{2}$ $,$ $\frac{0+2}{2}$ $\Big) = \Big($ $\frac{a+1}{2}$ $, 1 \Big)$

Since the mid point of $AB$ and $BD$ are the same

$\frac{a+1}{2}$ $=1 \Rightarrow a + 1 = 2 \Rightarrow a = 1$

Similarly, $\frac{1+b}{2}$ $=1 \Rightarrow 1 + b = 2 \Rightarrow b = 1$

$\therefore a =1 , b = 1$

$\therefore AB = \sqrt{(1-(-2))^2+(0-1)^2} = \sqrt{9+1} = \sqrt{10}$

$BC = \sqrt{(4-1)^2+(1-0)^2} = \sqrt{9+1} = \sqrt{10}$

Therefore $DC = \sqrt{10}$ and $DA = \sqrt{10}$

Hence $AB = BC = CD = DA \Rightarrow ABCD$ is a Rhombus

Or

Join $AC$

$\therefore Ar (Quadrilateral ABCD) = Ar (\triangle ABC) + Ar (\triangle ACD)$

Note: For a $\triangle$ with vertices $(x_1, y_1), (x_2, y_2) \ \ \& \ \ (x_3, y_3)$ the area of the triangle is $\Big|$ $\frac{1}{2}$ $[ x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) ] \Big|$

$Ar(\triangle ABC) = \Big|$ $\frac{1}{2}$ $[ -5(-5-(-6)) + (-4)(-6-7) + (-1)(7-9-5)) ] \Big|$

$= \Big|$ $\frac{1}{2}$ $[ -5(1) - 4(-13) + (-1)(12) ] \Big|$

$= \Big|$ $\frac{1}{2}$ $( -5+ 52-12 ) \Big|$

$= \Big|$ $\frac{1}{2}$ $\times 35 \Big|$

$= 17.5$ sq. units

$Ar(\triangle ACD) = \Big|$ $\frac{1}{2}$ $[ -5(5-(-6)) + 4(-6-7) + (-1)(7-5)) ] \Big|$

$= \Big|$ $\frac{1}{2}$ $[ -5(11) + 4(-13) + (-1)(2) ] \Big|$

$= \Big|$ $\frac{1}{2}$ $( -55- 52-2 ) \Big|$

$= \Big|$ $\frac{1}{2}$ $\times (-109) \Big|$

$= 54.5$ sq. units

$\therefore Ar (Quadrilateral ABCD) = 17.5 + 54.5 = 77$ sq. units

$\\$

Question 21: Find HCF and LCM of $404$ and $96$ and verify that HCF $\times$ LCM $=$ Product of the two given numbers.

$96 = 2^5 \times 3$

$404 = 2^2 \times 101$

Therefore HCF of $96$ and $404 = 2^2 = 4$

Also LCM of $96$ and $404 = 2^5 \times 3 \times 101 = 9696$

Product of two numbers $= 96 \times 404 = 38784$

HCF $\times$ LCM $= 4 \times 9696 = 38784$

Hence Product of two numbers $=$ HCF $\times$ LCM

$\\$

Question 22: Find all zeroes of the polynomial $(2x^4 - 9x^3 + 5x^2 + 3x - 1)$ if two of its zeroes are $(2 + \sqrt{3})$ and $(2 - \sqrt{3})$.

Given $(2+\sqrt{3})$  and $(2-\sqrt{3})$ are zeros of the polynomial $p(x) = 2x^4 - 9x^3+5x^2+3x-1$

Therefore $(x - 2-\sqrt{3})(x - 2+\sqrt{3}) = x^2 -4x+1$ is a factor of the polynomial $p(x)$ Hence

$x^2 -4x+1 ) \overline{2x^4 - 9x^3+5x^2+3x-1} ( 2x^2-x-1 \\ \hspace*{1.6cm} (-) \underline{2x^4-8x^3+2x^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{2cm}-x^3+3x^2+3x-1 \\ \hspace*{1.5cm} (-) \underline{-x^3+4x^2-x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \hspace*{3.5cm}-x^2+4x-1 \\ \hspace*{3cm} (-) \underline{-x^2+4x-1 \ \ \ \ \ } \\ \hspace*{5cm}\times$

Therefore $2x^2-x-1$ is a factor of $p(x)$

$2x^2-x-1 = 2x^2 - 2x + x - 1$

$= 2x(x-1) + (x-1)$

$= (x-1)(2x+1)$

$\Rightarrow (x-1)$ and $(2x+1)$ are factors of $p(x)$

Hence $x = 1$ and $x = -$ $\frac{1}{2}$ are the other two zeros of the given polynomial $2x^4 - 9x^3+5x^2+3x-1$

$\\$

Section – D

Question number 23 to 30 carry 4 mark each.

Question 23: Draw a triangle $ABC$ with $BC = 6$ cm, $AB = 5$ cm and $\angle ABC = 60^o$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the $\triangle ABC$.

$\\$

Question 24: The sum of four consecutive numbers in an AP is $32$ and the ratio of the product of the first and the last term to the product of two middle terms is $7 : 15$. Find the numbers.

Let the four consecutive terms of the AP be $a, a+d, a+2d$ and $a+3d$

Given: $a + ( a+d) + ( a+2d) + ( a+3d) = 32$

$\Rightarrow 4a + 6d = 32$

$\Rightarrow 2a + 3d = 16$

$\Rightarrow d =$ $\frac{16-2a}{3}$   … … … … … i)

Also $\frac{a(a+3d)}{(a+d)(a+2d)}$ $=$ $\frac{7}{15}$

$\Rightarrow 15a^2 + 45ad = 7a^2 + 21 ad + 14d^2$

$\Rightarrow 8a^2 + 24ad - 14 d^2 = 0$

$\Rightarrow 4a^2 + 12 ad - 7d^2 = 0$

$\Rightarrow 4a^2 + 14ad - 2ad - 7d^2 = 0$

$\Rightarrow 2a(2a-d) + 7d(2a-d) = 0$

$(2a-d)(2a+7d) = 0$

Substituting from i)

$\Big(2a-$ $\frac{16-2a}{3}$ $\Big)\Big(2a+7\times$ $\frac{16-2a}{3}$ $\Big) = 0$

$\Rightarrow (6a - 16 + 2a) (6a + 112 - 14a) = 0$

$\Rightarrow (8a-16)(-8a+112) = 0$

$\Rightarrow a = 2$ or $a = 14$

When $a = 2, d =$ $\frac{16-2 \times 2}{3}$ $=$ $\frac{12}{3}$ $= 4$

The the first four terms are $2, 6, 10, 14$

When $a = 14, d =$ $\frac{16-2 \times 14}{3}$ $=$ $\frac{-12}{3}$ $= -4$

The the first four terms are $14, 10, 6, 2$

$\\$

Question 25: In an equilateral $\triangle ABC$, $D$ is a point on side $BC$ such that $BD =$ $\frac{1}{3}$ $BC$. Prove that: $9(AD)^2 = 7(AB)^2$

Or

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

Given: $\triangle ABC$ is an equilateral triangle.

$\Rightarrow AB = BC = CA$

Also $BD =$ $\frac{1}{3}$ $BC$

To prove: $9 AD^2 = 7 AB^2$

Construction: Draw $AE \perp BC$

Consider $\triangle AEB$ and $\triangle AEC$

$AE$ is common

$\angle AEB = \angle AEC = 90^o$

$AB = AC$ (equilateral triangle)

$\triangle AEB \cong \triangle AEC$ (By RHS criterion)

$\therefore BE = EC$

$\Rightarrow BE =$ $\frac{BC}{2}$

$\Rightarrow BD + DE =$ $\frac{BC}{2}$

$\Rightarrow$ $\frac{BC}{3} + DE =$ $\frac{BC}{2}$

$\Rightarrow DE =$ $\frac{BD}{6}$

Using Pythagoras theorem,

In $\triangle AEB$

$AB^2 = AE^2 + BE^2 \Rightarrow AE^2 = AB^2 - BE^2$   … … … … … i)

In $\triangle AED$

$AD^2 = AE^2 + DE^2 \Rightarrow AE^2 = AD^2 - DE^2$   … … … … … ii)

From i) and ii)

$AB^2 - BE^2 = AD^2 - DE^2$

$\Rightarrow AB^2 - \Big($ $\frac{BC}{2}$ $\Big)^2 = AD^2 - \Big($ $\frac{BC}{6}$ $\Big)^2$

$\Rightarrow AB^2 -$ $\frac{BC^2}{4}$ $= AD^2 -$ $\frac{BC^2}{36}$

$\Rightarrow AB^2 = AD^2 +$ $\frac{8}{36}$ $BC^2$

$AB^2 = AD^2 +$ $\frac{2}{9}$ $BC^2$

But $BC = AB$

$\therefore AB^2 -$ $\frac{2}{9}$ $AB^2 = AD^2$

$\Rightarrow 7AB^2 = 9AD^2$

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)

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 26: A motor boat whose speed is $18$ km/hr in still water takes $1$ hr more to go $24$ km upstream than to return downstream to the same spot. Find the speed of the stream.

Or

A train travels at a certain average speed for a distance of $63$ km and then travels at a distance of $72$ km at an average speed of $6$ km/hr more than its original speed. If it takes $3$ hours to complete total journey, what is the original average speed ?

Distance traveled by the boat $= 24$ km

Speed of the boat $= 18$ km/hr

Let the speed of the stream $= x$ km/hr

Therefore speed of boat upstream $= (18 - x)$ km/hr

Also speed of boat downstream $= (18 + x)$ km/hr

Hence: $\frac{24}{18-x}$ $-$ $\frac{24}{18+x}$ $= 1$

$\Rightarrow 432 + 24 x - (432 - 24x) = 324 - x^2$

$\Rightarrow 48 x = 324 - x^2$

$\Rightarrow x^2 + 48x - 324 = 0$

$\Rightarrow x^2 + 54x - 6x - 324 = 0$

$\Rightarrow x(x+54)-6(x+54) = 0$

$\Rightarrow (x+54)(x-6) = 0$

$\Rightarrow x = 6$ km/hr or $x = -54$ km/hr (not possible as speed cannot be negative)

Therefore the speed of the stream is $6$ km/hr

Or

Let the original speed is $x$ km/hr

Time taken to travel $63$ km $=$ $\frac{63}{x}$ hrs

New speed $= (x+6)$ km/hr

Time taken to travel $72$ km  $=$ $\frac{72}{x+6}$

$\therefore$ $\frac{63}{x}$ $+$ $\frac{72}{x+6}$ $= 3$

$\Rightarrow 63x + 378 + 72x = 3x^2 + 18x$

$\Rightarrow 3x^2 - 117 x - 378 = 0$

$\Rightarrow x^2 - 39x - 126 = 0$

$\Rightarrow x^2 - 42 x + 3x - 126 =0$

$\Rightarrow x(x-42)+3(x-42) = 0$

$\Rightarrow (x-42)(x+3) = 0$

$\Rightarrow x = 42$ km/hr or $x = -3$ km/hr (not possible as speed cannot be negative)

Hence original speed $= 42$ km/hr

$\\$

Question 27: As observed from the top of a $100$ m high light house from the sea-level, the angles of depression of two ships are $30^o$ and $45^o$. If one ship is exactly behind the other on the same side of the light house, find the distance between the two ships. [Use $\sqrt{3} = 1.732$]

In $\triangle ABC:$

$\frac{100}{BC}$ $=\tan 45^o = 1 \Rightarrow BC = 100$ m

In $\triangle ABD:$

$\frac{100}{BD}$ $=\tan 30^o = 1 \Rightarrow BC = 100 \sqrt{3}$ m

$\therefore DC = 100 \sqrt{3} -100 = 173.2 - 100 = 73.2$ m

Therefore the distance between the ships is $73.2$ m.

$\\$

Question 28: The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are $10$ cm and $30$ cm respectively. If its height is $24$ cm, find :
(i) The area of the metal sheet used to make the bucket.
(ii) Why we should avoid the bucket made by ordinary plastic ? [Use $\pi = 3.14$]

Diameter of upper end of bucket $=30$ cm

Radius of the upper end of the frustum of cone $( r_2) = 15$ cm

Diameter of lower end of bucket $= 10$ cm

radius of the lower end of the frustum of cone $( r_1) = 5$ cm

Height of the frustum of Cone $(h) = 24$ cm

Volume of bucket $=$ $\frac{\pi}{3}$ $h ({r_1}^2 + {r_2}^2 + r_1r_2 )$

$=$ $\frac{1}{3}$ $\times$ $3.14$ $\times 24 \times ( 5^2 + 15^2 + 5 \times 15)$

$= 8164 \ cm^3 = 8.164$ liters

$l = \sqrt{h^2 + (r_2 - r_1)^2} = \sqrt{24^2 + (15-5)^2} = \sqrt{24^2 + 10^2} = 26$ cm

Therefore surface area $= \pi {r_1}^2 + \pi (r_1 + r_2)l$

$= 3.14$  $( 5^2 + (5+15)\times 26 )$

$= 1711.3 \ cm^2$

Hence, the Area of metal sheet used to make the bucket is $= 1711.3 \ cm^2$

$\\$

Question 29: The mean of the following distribution is $18$. Find the frequency $f$ of the class $19 - 21$.

 Class: 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Frequency: 3 6 9 13 $f$$f$ 5 4

Or

The following distribution gives the daily income of 50 workers of a factory:

 Daily Income (in Rs) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

 Class Interval Frequency $(f_i)$$(f_i)$ $x_i$$x_i$ $f_ix_i$$f_ix_i$ 11-13 3 12 36 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 $f$$f$ 20 $20f$$20f$ 21-23 5 22 110 23-25 4 24 96 $\Sigma f_i = 40 + f$$\Sigma f_i = 40 + f$ $\Sigma f_ix_i = 704 + 20f$$\Sigma f_ix_i = 704 + 20f$

Mean $= \frac{\Sigma f_ix_i }{40 + f }$ $=$ $\frac{704 + 20f}{40 + f}$

$\Rightarrow 720 + 18f = 704 + 20f$

$\Rightarrow 2f = 16$

$\Rightarrow f = 8$

Or

 Daily wages in Rs (less than) Frequency $(f_i)$$(f_i)$ Cumulative Frequency Less than 120 12 12 Less than 140 14 26 Less than 160 8 34 Less than 180 6 40 Less than 200 10 50

$\\$

Question 30: Prove that: $\frac{\sin A - 2 \sin^3 A }{2 \cos^3 A - \cos A}$ $= \tan A$

LHS $= \frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A}$

$= \frac{\sin A ( 1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)}$

$= \frac{\sin A ( 1 - \sin^2 A - \sin^2 A)}{\cos A ( \cos^2 A + \cos^2 A - 1)}$

$= \frac{\sin A ( \cos^2 A - \sin^2 A)}{\cos A ( \cos^2 A - \sin^2 A)}$

$= \frac{\sin A }{\cos A }$

$= \tan A$

$=$ RHS. Hence Proved.

$\\$