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SUMMATIVE ASSESSMENT – II

MATHEMATICS

Time allowed: 3 hours                                                             Maximum Marks: 90

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 6 carry 1 mark each.

Question 1: If the quadratic equation $px^2 - 2 \sqrt{5} px + 15 = 0$ has two equal roots, then find the value of $p$.

The given quadratic equation is $px^2 - 2 \sqrt{5} px + 15 = 0$

For real roots, Discriminant $= 0$

$\Rightarrow b^2 - 4ac = 0$

$\Rightarrow ( - 2 \sqrt{5})^2 - 4 (p) \times 15 = 0$

$\Rightarrow 20p^2 - 60 p = 0$

$\Rightarrow 20p(p - 3) = 0$

$\Rightarrow p = 0 or p = 3$

$p = 0$ is not possible as whole equation will become $0$ then. Hence $p = 3$.

$\\$

Question 2: In Figure 1, a tower $AB$ is $20$ m high and $BC$, its shadow on the ground, is $20 \sqrt{3}$ m long. Find the Sun’s altitude.

Let $AB$ be the tower and $BC$ be it’s shadow.

$\tan \theta =$ $\frac{AB}{BC}$

$\Rightarrow \tan \theta =$ $\frac{20}{20\sqrt{3}}$

$\Rightarrow \tan \theta =$ $\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \theta = \tan 30^o$

$\Rightarrow \theta = 30^o$

Therefore sunset is at an altitude  of $30^o$.

$\\$

Question 3: Two different dice are tossed together. Find the probability that the product of the two numbers on the top of the dice is $6$.

Two dice are tossed. Therefore

$S = \begin{bmatrix} (1, 1), (1, 2), (1, 3), (1 4), (1, 5), (1, 6) \\ (2, 1), (2, 2), (2, 3), (2 4), (2, 5), (2, 6) \\ (3, 1), (3, 2), (3, 3), (3 4), (3, 5), (3, 6) \\ (4, 1), (4, 2), (4, 3), (4 4), (4, 5), (4, 6) \\ (5, 1), (5, 2), (5, 3), (5 4), (5, 5), (5, 6) \\ (6, 1), (6, 2), (6, 3), (6 4), (6, 5), (6, 6) \end{bmatrix}$

Therefore the total number of outcomes when two dices are tossed $= 6 \times 6 = 36$

Favorable events of getting the product as $6$ are $( 1, 6), (2, 3), (3, 2), (6, 1) = 4$

$P($ getting products as $6) =$ $\frac{4}{36}$ $=$ $\frac{1}{9}$

$\\$

Question 4: In Figure 2, $PQ$ is a chord of a circle with centre $O$ and $PT$ is a tangent. If $\angle QPT = 60^o$, find $\angle PRQ$.

Given: $PT$ is a tangent so $\angle OPT=90^o$

$\angle QPT =60^o$

$\angle OPQ = \angle OPT- \angle QPT = 90^o-60^o = 30^o$

$\angle OPQ+\angle OQP+ \angle POQ=180^o$       (sum of the angles in a triangle is $180^o$)

$\angle POQ=180^o-30^o-30^o$

$\angle POQ=120^o$

$\angle OPQ= \angle OQP = 30^o$

$\angle POQ= 120^o$.  Also, $\angle PRQ=$ $\frac{1}{2}$ reflex $\angle POQ$

$\angle PRQ=$ $\frac{1}{2} (360-120)$ $=$ $\frac{240}{2}$ $= 120^o$

$\\$

Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: In Figure 3, two tangents $RQ$ and $RP$ are drawn from an external point $R$ to the circle with centre $O$. If  $\angle PRQ = 120^o$, then prove that $OR = PR + RQ$.

$\angle OPR = \angle OQR = 90^o$   … … … … …  i)

And in $\triangle OPR$ and $\triangle OQR$

$\angle OPR = \angle OQR = 90^o$ (from equation i)

$OP = OQ$ (Radii of same circle)

$OR = OR$ (common side)

$\triangle OPR \cong \triangle OQR$ ( by RHS criterion)

So, $RP = RQ$   … … … … …  ii)

And  $\angle ORP = \angle ORQ$    … … … … …  iii)

$\angle PRQ = \angle ORP + \angle ORQ$

Substitute $\angle PQR = 120^o$ (given) And from equation iii) we get

$\angle ORP + \angle ORP = 120^o$  $\Rightarrow 2 \angle ORP = 120^o$  $\Rightarrow \angle ORP = 60^o$

And we know $\cos 0^o =$ $\frac{Adjacent}{Hypotenuse}$

So in $\triangle OPR$ we get

$\cos \angle ORP =$ $\frac{PR}{OR}$

$\cos 60^o =$ $\frac{PR}{OR}$

$\frac{1}{2}$ $=$ $\frac{PR}{OR}$  (we know $\cos 60^o =$ $\frac{1}{2}$ )

$OR = 2PR$

$OR = PR + PR$ (substitute value from equation ii) we get)

$OR = PR + RQ$

$\\$

Question 6: In Figure 4, a triangle $ABC$ is drawn to circumscribe a circle of radius $3$ cm, such that the segments $BD$ and $DC$ are respectively of lengths $6$ cm and $9$ cm. If the area of $\triangle ABC$ is $54 \ cm^2$, then find the lengths of sides $AB$ and $AC$.

Given $OD = 3$ cm

Construction: Join $OA, OB$ and $OC$

Proof: Area of the $\triangle ABC =$ area of $\triangle OBC +$ area of $\triangle OAC +$ area of $\triangle OBA$

Let $AE = AF = x$

$BD=6$ cm, $BE = 6$ cm (equal tangents)

$DC=9$ cm, $CF=9$ cm (equal tangents)

Area of $\triangle OBC =$ $\frac{1}{2}$ $\times 15 \times 3 =$ $\frac{45}{2}$ $cm^2$

Area of $\triangle OAC =$ $\frac{1}{2}$ $\times (x+9) \times 3 =$ $\frac{3(x+9)}{2}$ $cm^2$

Area of $\triangle OAB =$ $\frac{1}{2}$ $\times (x+6) \times 3 =$ $\frac{3(x+6)}{2}$ $cm^2$

Therefore

$54 =$ $\frac{45}{2}$ $+$ $\frac{3(x+9)}{2}$ $+$ $\frac{3(x+6)}{2}$

$\Rightarrow 54 =$ $\frac{3}{2}$ $\Big( (x+9) + (x+6) + 15 \Big)$

$\Rightarrow 54 =$ $\frac{3}{2}$ $(2x+30)$

$\Rightarrow 36 = 2x+ 30$

$\Rightarrow x = 3$

Therefore sides are $15$ cm, $9$ cm, $12$ cm

$\\$

Question 7: Solve the following quadratic equation for $x : 4x^2 + 4bx - (a^2 - b^2) = 0$

$4x^2 +4bx -(a^2 -b^2) = 0$

$4x^2 +4bx -a^2 + b^2 =0$

$(2x)^2 + 2.(2x).b + b^2 -a^2 =0$

$(2x + b)^2 -a^2 = 0$

Use formula:  $a^2-b^2 = (a -b)(a + b)$

$(2x + b -a)(2x + b +a) =0$

$x =$ $\frac{a - b}{2}$ $, -$ $\frac{a + b}{2}$

$\\$

Question 8: In an AP, if $S_5 + S_7 = 167$ and $S_10 = 235$, then find the AP, where $S_n$ denotes the sum of its first $n$ terms.

Let the first term be $a$ and the common difference be $d$

$S_5 + S_7 = 167$

$\Rightarrow$ $\frac{5}{2}$ $[ 2a + 4d ] +$ $\frac{7}{2}$ $[ 2a + 6d ] = 167$

$\Rightarrow 5(a+ 2d) + 7 ( a + 3d) = 167$

$\Rightarrow 5a + 10 d + 7 a + 21 d = 167$

$\Rightarrow 12 a + 31 d = 167$   … … … … … i)

$S_{10} = 235$

$\Rightarrow$ $\frac{10}{2}$ $[ 2a + 9d ] = 235$

$\Rightarrow 10a + 45 d = 235$

$\Rightarrow 2a + 9 d = 47$   … … … … … ii)

Solving i) and ii) we get

$\hspace*{1.3cm} 12 a + 31 d = 167 \\ (-) \underline {6 \times [ 2a + 9 d = 47 ] \hspace*{1.3cm}} \\ \hspace*{1.8cm} -23d = - 115$

$\Rightarrow d = 5$

Substituting in ii) we get

$2a = 47 - 9(5) = 2 \Rightarrow a = 1$

Hence the AP is $1, 6, 11, 16, \cdots$

$\\$

Question 9: The points $A(4, 7), B(p, 3)$ and $C(7, 3)$ are the vertices of a right triangle, right-angled at $B$. Find the value of $p$.

$A(4,7), B(p,3)$ and $C(7,3)$

$AB^2+BC^2=AC^2$

$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\Rightarrow AB=\sqrt{(p-4)^2+(3-7)^2}$

$\Rightarrow AB=\sqrt{(p-4)^2+16}$

$AB^2 = (p-4)^2 + 16$

$BC^2 = (7-p)^2 + ( 3- 3)^2 = (7-p)^2$

$AC^2 = (4-7)^2 + (7-3)^2 = 25$

$\therefore (p-4)^2 + 16 + (7-p)^2 = 25$

$\Rightarrow p^2 + 16 - 8p + 16 + 49 + p^2 - 14 p = 25$

$\Rightarrow 2p^2 - 22 p + 56 = 0$

$\Rightarrow p^2 - 11 p + 28 = 0$

$\Rightarrow (p-7)(p-4) = 0$

$\Rightarrow p = 7$ or  $p = 4$

$\\$

Question 10: Find the relation between $x$ and $y$ if the points $A(x, y), B(- 5, 7)$ and $C(- 4, 5)$ are collinear.

It is given that $A(x, y), B(- 5, 7)$ and $C(- 4, 5)$ are collinear.

Therefore the area of $\triangle ABC = 0$

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

$\Rightarrow$ $\frac{1}{2}$ $[ x(7-5) -5(5-y) -4 (y - 7) ] = 0$

$\Rightarrow 2x -25 + 5y - 4 y + 28 = 0$

$\Rightarrow 2x + y + 3 = 0$

$\\$

Section – C

Question number 11 to 20 carry 3 mark each.

Question 11: The $14^{th}$ term of an AP is twice its $8^{th}$ term. If its $6^{th}$ term is $- 8$, then find the sum of its first $20$ terms

Let $a$ be the first term and $d$ be the common difference

$\therefore T_{14} = a + 13 d$

$T_8= a + 7d$

Given, $T_{14} = 2 T_8$

$\Rightarrow a + 13 d = 2a + 14 d$

$\Rightarrow a + d = 0$   … … … … … i)

$T_6 = a + 5d$

Given $T_6 = -8$

$\therefore a + 5d = -8$   … … … … … ii)

Solving i) and ii) we get

$\hspace*{1.0cm} a + d = 0 \\ \underline{(+) - a - 5d = 8} \\ \hspace*{1.0cm} -4d = 8$

$\Rightarrow d = - 2$

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

$\Rightarrow S_{20} =$ $\frac{20}{2}$ $[2(2) + (20-1) (-2) ]$

$\Rightarrow S_{20} = 10(4-38) = - 340$

$\\$

Question 12: Solve for $x : 3 x^2 - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

$\sqrt{3} x^2 - 2 \sqrt{2} x - 2 \sqrt{3} = 0$

On comparing it with $ax^2 + bx + c = 0$  we get

$a = \sqrt{3}, b = - 2 \sqrt{2}$  and $c = - 2 \sqrt{3}$

We know $x =$ $\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$=$ $\frac{-(- 2 \sqrt{2}) \pm \sqrt{(- 2 \sqrt{2})^2 - 4(\sqrt{3})( -2 \sqrt{3})}}{2(\sqrt{3})}$

$=$ $\frac{2\sqrt{2} \pm \sqrt{8 + 24}}{2\sqrt{3}}$

$=$ $\frac{2\sqrt{2} \pm \sqrt{32}}{2\sqrt{3}}$

$=$ $\frac{\sqrt{2} \pm \sqrt{8}}{\sqrt{3}}$

$=$ $\frac{\sqrt{2} \pm 2\sqrt{2}}{\sqrt{3}}$

Hence $x =$ $\frac{\sqrt{2} + 2\sqrt{2}}{\sqrt{3}}$ $=$ $\frac{3\sqrt{2}}{\sqrt{3}}$ $= \sqrt{6}$

or $x =$ $\frac{\sqrt{2} - 2\sqrt{2}}{\sqrt{3}}$ $=$ $\frac{-\sqrt{2}}{\sqrt{3}}$ $= -$ $\sqrt{\frac{2}{3}}$

$\\$

Question 13: The angle of elevation of an aeroplane from a point A on the ground is $60^o$. After a flight of $15$ seconds, the angle of elevation changes to $30^o$. If the aeroplane is flying at a constant height of $1500 \sqrt{3}$ m, find the speed of the plane in km/hr.

$\tan 60^o =$ $\frac{1500\sqrt{3}}{AC}$

$\Rightarrow AC =$ $\frac{1500\sqrt{3}}{\sqrt{3}}$ $= 1500$ m

$\tan 30^o =$ $\frac{1500\sqrt{3}}{AE}$

$\Rightarrow AE = 1500\sqrt{3} \times \sqrt{3} = 4500$ m

$\therefore CE = 4500 - 1500 = 3000$ m

Time taken to travel $CE = 15$ sec

$\therefore Speed =$ $\frac{3000 \ m}{15 \ sec}$ $= 200$ $\frac{m}{sec}$ $= 200 \times$ $\frac{3600}{1000}$ $\frac{km}{hr}$ $= 720$ km/hr

$\\$

Question 14: If the coordinates of points $A$ and $B$ are $(-2, - 2)$ and $(2, - 4)$ respectively, find the coordinates of $P$ such that $AP =$ $\frac{3}{7}$ $AB$, where $P$ lies on the line segment $AB$.

Let the coordinates of point $P$ be $P(x, y)$

Given $AP =$ $\frac{3}{7}$ $AB$

$\Rightarrow AP =$ $\frac{3}{7}$ $(AP + PB)$

$\Rightarrow 7AP = 3AP + 3 PB$

$\Rightarrow 4AP = 3 P$

$\Rightarrow$ $\frac{AP}{PB}$ $=$ $\frac{3}{4}$

Hence $P$ divides $AB$ in the ratio of $3:4$

Now using section’s formula

$x =$ $\frac{3(2) + 4(-2)}{3+4}$ $=$ $\frac{6-8}{7}$ $= -$ $\frac{2}{7}$

$y =$ $\frac{3(-4) + 4(-2)}{3+4}$ $=$ $\frac{-12-8}{7}$ $= -$ $\frac{20}{7}$

$\therefore$ Coordinates of $P = \Big( -$ $\frac{2}{7}$ $, -$ $\frac{20}{7}$ $\Big)$

$\\$

Question 15: The probability of selecting a red ball at random from a jar that contains only red, blue and orange balls is $\frac{1}{4}$. The probability of selecting a blue ball at random from the same jar is $\frac{1}{3}$ . If the jar contains $10$ orange balls, find the total number of balls in the jar.

Let the number of red balls be $x$ and number of Blue balls be $y$

No of Orange balls $= 10$

$\therefore$ Total number of balls $= x + y + 10$

$P(Red \ ball) =$ $\frac{1}{4}$ and $P(Blue \ ball) =$ $\frac{1}{3}$

$\therefore$ $\frac{x}{x+y+10}$ $=$ $\frac{1}{4}$

$\Rightarrow 4x = x + y + 10$

$\Rightarrow 3x - y = 10$   … … … … … i)

Similarly, $\frac{y}{x+y+10}$ $=$ $\frac{1}{3}$

$\Rightarrow 3y = x + y + 10$

$\Rightarrow -x + 2y = 10$  … … … … … ii)

Solving i) and ii) we get

$\hspace*{1.8cm} 3x - y = 10 \\ (+) \underline{ 3 \times [ -x + 2y = 10 ] } \\ \hspace*{2.7cm} 5y = 40$

$\Rightarrow y = 8$

$\therefore x =$ $\frac{10+8}{3}$ $= 6$

Hence the total number of balls $= 6 + 8 + 10 = 24$

$\\$

Question 16: Find the area of the minor segment of a circle of radius $14$ cm, when its central angle is $60^o$. Also find the area of the corresponding major segment. (Use $\pi =$ $\frac{22}{7}$)

Radius $(r) = 14$ cm $\theta = 60^o$

Therefore Area of $OAPB =$ $\frac{60}{360}$ $\times \pi r^2 =$

$\frac{1}{6}$ $\times \frac{22}{7} \times 14^2 = 102.67 \ cm^2$

Area of $\triangle AOB =$ $\frac{1}{2}$ $\times base \times height$

We draw $OM \perp AB$

$\therefore \angle OMB = \angle OMA = 90^o$

In $\triangle OMA$ and $\triangle OMB$

$\angle OMA = \triangle OMB = 90^o$  (by construction)

$OA = OB$   (both are radius of the same circle)

$OM$ is common

$\therefore \triangle OMA \cong \triangle OMB$  (by RHS criterion)

$\Rightarrow \angle AOM = \angle BOM$

$\therefore \angle AOM = \angle BOM =$ $\frac{1}{2}$ $\angle BOA$

$\Rightarrow \angle AOM = \angle BOM =$ $\frac{1}{2}$ $\times 60 = 30^o$

Also, since $\triangle OMB \cong \triangle OMA$

$\therefore BM = AM$

$\Rightarrow BM = AM =$ $\frac{1}{2}$ $AB$

$\Rightarrow AB = 2 BM$  … … … … … i)

In right $\triangle OMB$

$\frac{AM}{AO}$ $= \sin 30^o$

$\Rightarrow$ $\frac{AM}{14}$ $=$ $\frac{1}{2}$

$\Rightarrow AM =$ $\frac{14}{2}$

$\Rightarrow AB = 2 \times$ $\frac{14}{2}$ $= 14$

Similarly, In right $\triangle OMA$

$\frac{OM}{AO}$ $= \cos 30^o$

$\Rightarrow$ $\frac{OM}{14}$ $=$ $\frac{\sqrt{3}}{2}$

$\Rightarrow OM =$ $\frac{\sqrt{3}}{2}$ $\times 14$

$\therefore$ Area of $\triangle AOB =$ $\frac{1}{2}$ $\times 14 \times$ $\frac{\sqrt{3}}{2}$ $\times 14 = 84.87 \ cm^2$

Therefore area of segment $APB = 102.67-84.87 = 17.80 \ cm^2$

Area of major segment $= \pi r^2 - 17.80 = 3.14 \times 14^2 - 17.80 = 598.2 \ cm^2$

$\\$

Question 17: Due to sudden floods, some welfare associations jointly requested the government to get $100$ tents fixed immediately and offered to contribute $50\%$ of the cost. If the lower part of each tent is of the form of a cylinder of diameter $4.2$ m and height $4$ m with the conical upper part of same diameter but of height $2.8$ m, and the canvas to be used costs Rs. $100$ per sq. m, find the amount, the associations will have to pay. What values are shown by these associations ? (Use $\pi =$ $\frac{22}{7}$)

$l = \sqrt{r^2 + h^2} = \sqrt{(2.1)^2 + (2.8)^2} = \sqrt{12.25} = 3.5$ m

Surface Area of canvas $=$ CSA of Cone $+$ CSA of cylinder

$= \pi r l + 2 \pi r H$

$=$ $\frac{22}{7}$ $(2.1) [ 3.5 + 2 \times 4 ]$

$= 22 \times 0.3 \times 11.5 = 75.9 \ m^2$

Total cost of canvas $= 100$ $\frac{Rs}{m^2}$ $\times 75.9 \ m^2 = 7590$ Rs.

Association will pay $=$ $\frac{50}{100}$ $\times 7590 = 3795$ Rs. per tent

Hence for $100$ tents, Association will pay $= 3795 \times 100 = 379500$ Rs.

This is an example of showing humanity and helpful nature of the association.

$\\$

Question 18: A hemispherical bowl of internal diameter $36$ cm contains liquid. This liquid is filled into $72$ cylindrical bottles of diameter $6$ cm. Find the height of the each bottle, if $10\%$ liquid is wasted in this transfer.

Radius of hemispherical bowl $=$ $\frac{36}{2}$ $= 18$ cm

Volume of hemispherical bowl $=$ $\frac{2}{3}$ $\pi (18)^3 = 3888 \pi \ cm^3$

Volume of liquid to be transferred $= 0.9 \times 3888 \pi = 3499.2 \pi \ cm^3$

Radius of cylindrical bottle $=$ $\frac{6}{2}$ $= 3$ cm

Let the height of Bottle $= h$

$\therefore$ Volume of Bottle $= \pi r^2 h = \pi (3)^2 h$

Total volume of $72$ bottles $= \pi (3)^2 h \times 72 = 648 \pi h \ cm^3$

Therefore $648 \pi h = 3499.2 \pi \Rightarrow h =$ $\frac{3499.2}{648}$ $= 5.4$ cm

Therefore height of bottle is $5.4$ cm

$\\$

Question 19: A cubical block of side $10$ cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs. $5$ per $100$ sq. cm. [ Use $\pi = 3.14$ ]

The greatest diameter of the hemisphere can be $10$ cm

TSA of solid $=$ SA of cube $+$ CSA of hemisphere $-$ Area of base of hemisphere

$= 6 \times 10^2 + 2 \times$ $\frac{22}{7}$ $\times 5^2 -$ $\frac{22}{7}$ $\times 5^2$

$= 600 +$ $\frac{22}{7}$ $\times 25$

$= 687.57 \ cm^2$

Cost of painting $=$ $\frac{5 \ Rs}{100 \ cm^2}$ $= 0.05$ $\frac{Rs}{cm^2}$

$\therefore$ Cost of painting $= 678.57 \times 0.05 = 33.93$ Rs.

$\\$

Question 20: $504$ cones, each of diameter $3.5$ cm and height $3$ cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. (Use $\pi =$ $\frac{22}{7}$)

Diameter of cone $(d) = 3.5$ cm

Height of cone $(h)= 3.0$ cm

Volume of come $=$ $\frac{1}{3}$ $\pi r^2 h =$ $\frac{1}{3}$ $\times \frac{22}{7}$ $\times \Big($ $\frac{3.5}{2}$ $\Big)^2 \times 3 =$ $\frac{77}{8}$ $\ cm^3$

$\therefore$ Volume of $504$ cones $= 504 \times$ $\frac{77}{8}$ $= 4851 \ cm^3$

Let the radius of the sphere $= r$

$\therefore$ $\frac{4}{3}$ $\pi r^3 = 4851$

$\Rightarrow r^3 =$ $\frac{3}{4}$ $\times$ $\frac{7}{22}$ $\times 4851 = 1157.625 \ cm^3$

$\Rightarrow r = 10.5$ cm

$\therefore$ Surface area of sphere $= 4 \pi r^2 = 4 \times$ $\frac{22}{7}$ $\times 10.5^2 = 1386 \ cm^2$

$\\$

Section – D

Question number 21 to 31 carry 4 mark each.

Question 21: The diagonal of a rectangular field is $16$ metres more than the shorter side. If the longer side is $14$ metres more than the shorter side, then find the lengths of the sides of the field.

Let the shorter side $= x$

$\therefore (x+14)^2 + x^2 = (x+16)^2$

$\Rightarrow x^2 + 196 + 28x + x^2 = x^2 + 256 + 32x$

$\Rightarrow x^2 - 4x - 60 = 0$

$\Rightarrow x^2 - 10x + 6 x - 60 = 0$

$\Rightarrow x(x-10) + 6(x-10) = 0$

$\Rightarrow (x-10)(x+6) = 0$

$\Rightarrow x = 10$ or $x = -6$ (not possible)

Hence the shorter side $= 10$ m, diagonal $= 26$ m and Longer side $= 24$ m

$\\$

Question 22: Find the $60^{th}$ term of the AP $8, 10, 12, \cdots$, if it has a total of $60$ terms and hence find the sum of its last $10$ terms

Given AP: $8, 10, 12, ...$

$\therefore a = 8$ and $d = 2$

$\therefore T_{60} = a + (60-1) d = 8 + 59 \times 2 = 126$

For sum of the last $10$ terms, lets look at the series backwards

$\therefore a = 126, d = -2 n = 10$

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

$\Rightarrow S_{10} =$ $\frac{10}{2}$ $[ 2 \times 126 + (10-1)(-2) ]$

$= 5[ 252 - 18] = 1170$

Therefore sum of the last $10$ terms $= 1170$

$\\$

Question 23: A train travels at a certain average speed for a distance of $54$ km and then travels a distance of $63$ km at an average speed of $6$ km/h more than the first speed. If it takes $3$ hours to complete the total journey, what is its first speed ?

Total time taken $= 3$ hr

$\therefore$ $\frac{54}{x}$ $+$ $\frac{63}{x+6}$ $= 3$

$\Rightarrow$ $\frac{18}{x}$ $+$ $\frac{21}{x+6}$ $= 1$

$\Rightarrow 18x + 108 + 21 x = x^2 + 6x$

$\Rightarrow 39x + 108 = x^2 + 6x$

$\Rightarrow x^2 - 33x - 108 = x$

$\Rightarrow x^2 - 36x + 3x - 108 = 0$

$\Rightarrow x(x-36) + 3 (x - 36) = 0$

$\Rightarrow (x-36)(x+ 3) = 0$

$\Rightarrow x = 36$ or $x = -3$ (not possible)

$\therefore x = 36$ km/hr

$\\$

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

Given: Circle with center $O$

$PQ$ and $PR$ are tangents touching the circle at $Q$ and $R$ respectively.

To Prove: $PQ = PR$

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

Proof: Since $PQ$ is a tangent, $\angle PQO = 90^o$

Similarly, $\angle PRO = 90^o$

Consider $\triangle POQ$ and $\triangle POR$

$OQ = OR$     (radius of the circle)

$\angle PQO = \angle POR = 90^o$

$PO$ is common

$\therefore \triangle POQ \cong \triangle POR$ (By RHS criterion)

$\therefore PQ = PR$. Hence proved.

$\\$

Question 25: Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

Given: $P$ is the mid point of $\widehat{PQ}$, $O$ is the center, $AB$ is the chord

To Prove: $AB \parallel ST$

$\angle OPT = 90^o$ (tangent)

Since $P$ is the mid point of $\widehat{PQ}$,

$\widehat{AP} = \widehat{PB}$

$\Rightarrow \angle AOP = \angle BOP$

$\Rightarrow \angle AOM = \angle BOM$

In $\triangle AOM$ and $\triangle BOM$

$AO = OB$    (radius)

$OM$ is common

$\angle AOM = \angle BOM$

$\therefore \triangle AOM \cong \triangle BOM$    (By SAS criterion)

$\Rightarrow \angle AMO = \angle BMO$

$\angle AMO + \angle BMO = 180^o$

$\Rightarrow \angle AMO = \angle BMO = 90^o$

$\therefore \angle AMO = \angle OPS = 90^o$

$\angle BMO = \angle OPT = 90^o$

Corresponding angles are equal, hence $AB \parallel ST$. Hence proved.

$\\$

Question 26: Construct a $\triangle ABC$ in which $AB = 6$ cm, $\angle A = 30^o$ and $\angle B = 60^o$. Construct another $\triangle AB'C'$ similar to $\triangle ABC$ with base $AB' = 8$ cm.

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Question 27: At a point $A, 20$ metres above the level of water in a lake, the angle of elevation of a cloud is $30^o$. The angle of depression of the reflection of the cloud in the lake, at $A$ is $60^o$. Find the distance of the cloud from $A$.

$\tan 30^o =$ $\frac{h}{AD}$   $\Rightarrow AD = \sqrt{3} h$

$\tan 60^o =$ $\frac{h+40}{AD}$

$\Rightarrow \sqrt{3} =$ $\frac{h+40}{AD}$

$\Rightarrow 3h = h + 40$

$\Rightarrow 2h = 40$   $\Rightarrow h = 20$ m

$\therefore EC = 20 + h = 20 + 20 = 40$ m

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Question 28: A card is drawn at random from a well-shuffled deck of playing cards.
Find the probability that the card drawn is
(i) a card of spade or an ace.
(ii) a black king.
(iii) neither a jack nor a king.
(iv) either a king or a queen.

Total number of cards $= 52$

i) P( a card of spades or an ace) $=$ $\frac{13 + 3}{52}$ $=$ $\frac{16}{52}$ $= \frac{4}{13}$

ii) P(a black king) $=$ $\frac{2}{52}$ $=$ $\frac{1}{26}$

iii) P( neither o jack nor a king) $=$ $\frac{52 - (4 + 4)}{52}$ $=$ $\frac{44}{52}$ $=$ $\frac{11}{13}$

iv) P (either a king or a queen) $=$ $\frac{4+4}{52}$ $=$ $\frac{8}{52}$ $=$ $\frac{2}{13}$

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Question 29: Find the values of $k$ so that the area of the triangle with vertices $(1, - 1), (- 4, 2k)$ and $(- k, - 5)$ is $24$ sq. units.

Vertices: $(1, -1), (-4, 2k), (-k, -5)$

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

$\Rightarrow 24 =$ $\frac{1}{2}$ $[ 1 (2k + 5) + (-4)(-5+1) + (-k)(-1-2k) ]$

$\Rightarrow 48 = 2k + 5 + 16 + k + 2k^2$

$\Rightarrow 2k^2 + 3k - 27 =0$

$\Rightarrow 2k^2 + 9k - 6k - 27 = 0$

$\Rightarrow (2k+9)(k-3) = 0$

$\Rightarrow k = -$ $\frac{9}{2}$ or $k = 3$

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Question 30: In Figure 5, $PQRS$ is a square lawn with side $PQ = 42$ metres. Two circular flower beds are there on the sides $PS$ and $QR$ with centre at $O$, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

Area of square lawn $= 42 \times 42 = 1764 \ m^2$

Let $OP = OS = x$

$\angle POS = 90^o$ (diagonals of square are perpendicular to each other)

$\therefore x^2 + x^2 = 42^2$  $\Rightarrow 2x^2 = 1764$  $\Rightarrow x^2 = 882$

Area of sector $POS =$ $\frac{90}{360}$ $\times \pi (x)^2$  $=$ $\frac{1}{4}$ $\pi x^2$  $=$ $\frac{1}{4}$ $\times$ $\frac{22}{7}$ $\times 882$  $= 693 \ m^2$

Area of $\triangle POS =$ $\frac{1}{2}$ $\times x \times x =$ $\frac{1}{2}$ $x^2 =$ $\frac{1}{2}$ $\times 882 = 441 \ m^2$

$\therefore$ Area of one flower bed $= 693 - 441 = 252 \ m^2$

$\therefore$ Total area of flower bed $= 2 \times 252 = 504 \ m^2$

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Question 31: From each end of a solid metal cylinder, metal was scooped out in hemispherical form of same diameter. The height of the cylinder is $10$ cm and its base is of radius $4.2$ cm. The rest of the cylinder is melted and converted into a cylindrical wire of $1.4$ cm thickness. Find the length of the wire. (Use $\pi =$ $\frac{22}{7}$)

Volume of cylinder $= \pi (4.2)^2 \times 10 = 176.4 \pi \ cm^3$
Volume of $2$ hemisphere $= 2 \times$ $\frac{2}{3}$ $\times \pi (4.2)^3 = 98.784 \pi \ cm^3$
$\therefore$ Volume melted $= 176.4 \pi - 98.78 \pi = 77.616 \pi \ cm^3$
Diameter of cylindrical wire $= 1.4$ cm
$\therefore 77.616 \pi = \pi \Big($ $\frac{1.4}{2}$ $\Big)^2 \times l$
$l =$ $\frac{77.616 \times 2 \times 2}{1.4 \times 1.4}$ $= 158.4$ m