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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:  In Fig. 1, $PQ$  is a tangent at point $C$ to a circle with center $O$. If $AB$ is the diameter and $\angle CAB = 30^o$, find $\angle PCA$.

Construction: Join $OC$

Now in $\triangle AOC$

$AO = OC$ (radius of the same circle)

Therefore , $\angle OAC= \angle OCA = 30^o$

Also, $\angle OCP= 90^o$

Therefore, $\angle PCA= 90^o - 30^o = 60^o$

$\\$

Question 2: For what value of $k$ will $k+9, 2k-1$, and $2k+7$ are the consecutive terms of an A.P.?

If three terms $x$, $y$ and $z$ are in A.P. then, $2 y = x + z$

Since $k + 9, 2 k - 1$ and $2 k + 7$ are in A.P.

$\therefore 2(2k-1) = (k+9) + (2k+7)$   $\Rightarrow 4k - 2 = 3k + 16$   $\Rightarrow k = 18$

$\\$

Question 3: A ladder, leaning against a wall, makes an angle of $60^o$ with the horizontal. If the foot of the ladder is $2.5$ m away from the wall, find the length of the ladder.

Let $AC$ be a ladder, $AB$ is the wall and $BC$ is the ground. The situation given in the question is shown in the figure.

Therefore $\frac{BC}{AC}$ $=$ $\cos 60^o =$ $\frac{1}{2}$

$\Rightarrow$ $\frac{1}{2}$ $=$ $\frac{2.5}{AC}$

$\Rightarrow AC = 5$ m.   Therefore, the length of ladder is 5 m.

$\\$

Question 4: A card is drawn at random from a well shuffled pack of $52$ playing cards. Find the probability of getting neither a Red card or the queen.

There are $26$ red cards in a deck of $52$ cards

Also there are $4$ queens in a deck of $52$ cards, $2$ queens are black and $2$ are red.

Now, numbers of card which are neither red nor queen are $= 52 - (26 + 2) = 24$

Therefore Probability of getting neither a red card nor a queen is $=$ $\frac{24}{52}$ $=$ $\frac{6}{13}$

$\\$

Section – B

Question number 5 to 10 carry 2 mark each.

Question 5: If $-5$ is the root of the quadratic equation $2x^2 +px - 15 = 0$ and the quadratic equation $p(x^2+x)+ k = 0$ has equal roots, find the value of $k$.

If $-5$ is root of $2x^2 +px - 15 = 0$

$\Rightarrow 2(-5)^2 + p (-5) - 15 = 0$

$\Rightarrow 50 - 5p - 15 = 0$   $\Rightarrow 5p = 35$   $\Rightarrow p = 7$

For $p(x^2+x)+ k = 0$ to have equal roots,

$b^2 - 4ac = 0$     (Determinant should be zero)

$\Rightarrow p^2 - 4pk = 0$ $\Rightarrow k =$ $\frac{p}{2}$ $=$ $\frac{7}{2}$

$\\$

Question 6: Let $P$ and $Q$ be points of trisection of the line segment joining the points $A(2, -2)$ and $B(-7, 4)$ such that $P$ is nearer to $A$. Find the coordinates of $P$ and $Q$.

Since $P$ and $Q$ be the points of trisection of the line segment joining the points $A (2, -2)$ and $B (-7, 4)$ such that $P$ is nearer to $A$. Therefore $P$ divides line segment in the rail $1:2$ and $Q$ divides in $2:1$

By section formula Coordinates of $P$ are

$=$ $\Bigg($ $\frac{1 \times (-7) + 2 \times (2) }{1 + 2} , \frac{1 \times (4) + 2 \times (-2) }{1 + 2}$ $\Bigg)= \Big($ $\frac{-3}{3}, \frac{0}{3}$ $\Big) = (-1, 0)$

By section formula Coordinates of Q are

$=$ $\Bigg($ $\frac{2 \times (-7) + 1 \times (2) }{2 + 1} , \frac{2 \times (4) + 1 \times (-2) }{2 + 1}$ $\Bigg) = \Big($ $\frac{-12}{3}, \frac{6}{3}$ $\Big) = (-4, 2)$

$\\$

Question 7: In Fig. 2, a quadrilateral $ABCD$  is drawn to circumscribe a circle, with center $O$, in such a way that the sides $AB, BC, CD$, and $DA$ touch the circle at the points $P, Q, R$ and $S$ respectively. Prove that $AB + CD = BC + DA$.

Since tangents drawn from the exterior point to a circle are equal in length.

As, $DR$ and $DS$ are tangents from exterior point $D$ so, $DR = DS$ … … i)

Similarly,

$AP$ and $AS$ are tangents from exterior point $A$ so, $AP = AS$… … … ii)

$BP$ and $BQ$ are tangents from exterior point $B$ so, $BP = BQ$… … … iii)

$CR$ and $CQ$ are tangents from exterior point $C$ so, $CR = CQ$… … … iv)

Adding i) , ii), iii) and iv) , we get

$DR + AP + BP + CR = DS + AS + BQ + CQ$

$(DR + CR) + (AP + BP) = (DS + AS) + (BQ + CQ)$

$CD + AB = DA + BC$

$AB + CD = BC + DA$

Hence proved.

$\\$

Question 8: Prove that the points $(3, 0), (6, 4)$ and $( -1, 3)$ are the vertices of a right angled isosceles triangle.

If $A (3, 0), B (6, 4)$ and $C (-1, 3)$ be the vertices of the triangle. From distance formula,

$AB = \sqrt{(6-3)^2 + (4-0)^2} = \sqrt{9+16} = \sqrt{25} = 5$

$BC = \sqrt{(-1--6)^2 + (3-4)^2} = \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$

$CA = \sqrt{(-1-3)^2 + (3-0)^2} = \sqrt{16+9} = \sqrt{25} = 5$

Therefore we see that $AB^2 + CA^2 = BC^2$

This implies that $\triangle ABC$ is a right angled isosceles triangle.

$\\$

Question 9: The $4^{th}$ term of an AP is zero. Prove that the $25^{th}$ term  of the AP is three times  its $11^{th}$ term.

Let $a$ be the first term and $d$ be the common difference of an A.P. having $n$ terms.

Its $n^{th}$ term is given by, $a_n = a+ (n-1)d$

Therefore $a_4 = a + 3d = 0 \Rightarrow a = - 3d$

$a_{25} = a + 24d \Rightarrow a_{25} = -3d + 24 d = 21d$   … … … … … i)

$a_{11} = a+ 10 d \Rightarrow a_{11} = -3d + 10d = 7 d$  … … … … … ii)

From equation i) and ii)

$a_{25} = 3 a_{11}$  Hence proved.

$\\$

Question 10: In Fig. 3, from an external point $P$, two tangents $PT$ and $PS$ are drawn to a circle with center $O$ and radius$r$. If $OP=2r$, show that $\angle OTS = \angle OST = 30^o$.

In $\triangle OTP, \angle OTP = 90^o$ (tangent to a circle perpendicular to the radius through the point of contact)

$\sin \angle OTP =$ $\frac{OT}{OP}$

$\Rightarrow \sin \angle OTP = \frac{r}{2r}$   $\Rightarrow \sin \angle OTP = \frac{1}{2}$   $\Rightarrow \angle OTP = 30^o$

We know $\angle OTP + \angle OPT + \angle TOP = 180^o$

$\Rightarrow 90^o + 30^o + \angle TOP = 180^o$   $\Rightarrow \angle TOP = 60^o$

Consider $\triangle OTP$ and $\triangle OSP$

$OT = OS$ (radius of the same circle)

$OP = OS$ (tangents from the same external point)

$OP$ is common

$\therefore \triangle OTP \cong \triangle OSP$ (by SSS criterion)

$\Rightarrow \angle TPO = \angle SPO$

$OP$ is bisector of $\angle TPS$

Also $\triangle TQP \cong \triangle SQP$  (by SAS criterion)

Therefore $\angle OQT = OQS = 90^o$

In $\triangle OTQ$,   $\angle OQT + \angle OTQ + \angle TOQ = 180^o$ (angles of a triangle)

$\Rightarrow 90^o + \angle OTQ + \angle 60^o = 180^o$   $\Rightarrow \angle OTQ = 30^o$

Similarly, $\angle OST = 30^o$

Section – C

Question number 11 to 20 carry 3 mark each.

Question 11: In Fig. 4, $O$ is the center of the circle such that the diameter $AB = 13$ cm and $AC = 12$ cm. $BC$ is joined. Find the area of shaded region. (Take $\pi = 3.14$)

In $\triangle ABC \angle C = 90^o$ (angle in a semicircle)

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

$\Rightarrow BC^2 + 12^2 = 13^2$

$\Rightarrow BC^2 = 25 \Rightarrow BC = 5$ cm

Area of shaded region $=$ Area of semicircle  $-$ Area of triangle

$=$ $\frac{1}{2}$ $\times 3.14 \times \Big($ $\frac{13}{2}$ $\Big)^2 -$ $\frac{1}{2}$ $\times 5 \times 12$

$= 66.3325 - 30 = 36.3325 \ cm^2$

$\\$

Question 12: In Fig. 5, a tent is an a shape of cylinder surmounted by a conical top of same diameter. If the height of the diameter of the cylindrical part are $2.$1 m and $3$ m respectively and the slant height of the conical part is $2.8$ m, find the cost of the canvas needed to make the tent if the canvas is available at the rate of $500$ Rs sq. meter. (Use $\pi =$ $\frac{22}{7}$)

Canvas needed to make the tent $=$ Curved surface area of the conical part $+$ curved surface area of cylindrical part.

Radius of the conical part $=$ radius of cylindrical part $= r =$ $\frac{3}{2}$ m

Slant height of the conical part is $= l = 2.$8 m.

Height of cylindrical part is  $h = 2.1$ m

Curved surface area of conical part $= \pi r l =$ $\frac{22}{7}$ $\times$ $\frac{3}{2}$ $\times 2.8 = 13.2 m^2$

Curved surface area of cylindrical part $= 2 \pi r h = 2 \times$ $\frac{22}{7}$ $\times$ $\frac{3}{2}$ $\times 2.1 = 19.8 \ m^2$

Total surface area $= 13.2 + 19.8 = 33 \ m^2$

Therefore cost of canvas used $= 33 \times 500 = Rs. \ 16500$

$\\$

Question 13: If the point $P(x, y)$ is equidistant from the points $A(a+b, b-a)$ and $B(a-b, a+b)$. Prove that $bx = ay$.

Since the point $P (x, y)$ is equidistant from the points $A (a + b, b - a)$ and $B (a - b, a + b)$

Therefore $PA = PB$

Using distance formula

$\sqrt{ (x - (a+b))^2 + (y - (b-a))^2 } = \sqrt{ (x - (a-b))^2 + (y - (a+b))^2 }$

$\Rightarrow (x - (a+b))^2 + (y - (b-a))^2 = (x - (a-b))^2 + (y - (a+b))^2$

$\Rightarrow x^2 + (a+b)^2 - 2x(a+b) + y^2 + (b-a)^2 - 2y(b-a) = x^2 + (a-b)^2 -2x(a-b) + y^2 + (a+b)^2 - 2y (a+b)$

$\Rightarrow - 2x(a+b) - 2y(b-a) = -2x(a-b) - 2y (a+b)$

$\Rightarrow -2xa -2xb-2yb+2ya = - 2xa+2xb-2ya-2yb$

$\Rightarrow -4bx= - 4ay$

$\Rightarrow bx = ay$

$\\$

Question 14: In Fig. 6, find the area of the shaded region, enclosed between two concentric circles of radii $7$ cm and $14$ cm  where $\angle AOC = 40^o$. (Use $\pi =$ $\frac{22}{7}$)

Area of shaded region $=$ Area of circular ring $-$Area of region ABDC

$= \Big( \pi (14^2 - 7^2) -$ $\frac{40}{360}$ $\pi (14^2 - 7^2) \Big)$

$= \pi (14^2 - 7^2) \times \Big($ $\frac{360-40}{360}$ $\Big)$

$=$ $\frac{22}{7}$ $\times 147 \times$ $\frac{8}{9}$

$= 22 \times 21 \times$ $\frac{8}{9}$ $= 410.67 \ cm^2$

$\\$

Question 15: If the ratio of the sum first $n$ terms of two APs  is $(7n+1): (4n+27)$, find the ratio of their $m^{th}$ term.

For first AP, let the first term be $a_1$ and the common difference be $d_1$ and Sum of the first $n$ terms be $S_1$

Similarly, for Second AP, let the first term be $a_2$ and the common difference be $d_2$ and Sum of the first $n$ terms be $S_2$

Therefore $\frac{S_1}{S_2} = \frac{\frac{n}{2} (2a_1+ (n-1)d_1)}{\frac{n}{2} (2a_2+ (n-1)d_2)}$ $=$ $\frac{2a_1+ (n-1)d_1}{ 2a_2+ (n-1)d_2}$

$\Rightarrow \frac{2a_1+ (n-1)d_1}{ 2a_2+ (n-1)d_2}$ $=$ $\frac{7n+1}{4n+27}$

For $m^{th}$ term, replace $n$ by $2m-1$

$\Rightarrow \frac{2a_1+ (2m-1-1)d_1}{ 2a_2+ (2m-1-1)d_2}$ $=$ $\frac{7(2m-1)+1}{4(2m-1)+27}$

$\Rightarrow \frac{a_1+ (m-1)d_1}{ a_2+ (m-1)d_2}$ $=$ $\frac{14m-6}{8m+23}$

Hence the ratio of the $m^{th}$ term is $14m-6 : 8m + 23$

$\\$

Question 16: Solve for $x:$ $\frac{1}{(x-1)(x-2)}$ $+$ $\frac{1}{(x-2)(x-3)}$ $=$ $\frac{2}{3}$ $, x \neq 1, 2, 3$

Given $\frac{1}{(x-1)(x-2)}$ $+$ $\frac{1}{(x-2)(x-3)}$ $=$ $\frac{2}{3}$ $, x \neq 1, 2, 3$

$\Rightarrow$ $\frac{(x-3) + (x-1)}{(x-1)(x-2)(x-3)}$ $=$ $\frac{2}{3}$

$\Rightarrow$ $\frac{2x-4}{(x-1)(x-2)(x-3)}$ $=$ $\frac{2}{3}$

$\Rightarrow$ $\frac{x-2}{(x-1)(x-2)(x-3)}$ $=$ $\frac{1}{3}$

$\Rightarrow$ $\frac{1}{(x-1)(x-3)}$ $=$ $\frac{1}{3}$

$\Rightarrow 3 = (x-1)(x-3)$

$\Rightarrow 3 = x^2 - 4x + 3$

$\Rightarrow x(x-4) = 0$

$\Rightarrow x = 0 \ or \ x = 4$

$\\$

Question 17: A conical vessel with a base of radius $5$ cm and height $24$ cm, is full of water. This water is emptied into a cylindrical vessel of base radius $10$ cm. Find the height to which the water will rise in the cylindrical vessel. (Use $\pi =$ $\frac{22}{7}$)

Let $r, h, l$ be the radius, height and slant height of the cone.

Let $R$ and $H$ be the radius and height of the cylinder.

Since volume will remain same

Volume of water in conical vessel $=$ Volume in cylindrical vessel

Therefore  $\frac{1}{3}$ $\pi r^2 h = \pi R^2 H$

$\Rightarrow \frac{1}{3}$ $\pi 5^2 (24) = \pi 10^2 H$

$\Rightarrow H = 2$ cm

$\\$

Question 18: A sphere of diameter $12$ cm, is dropped into a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $3\frac{5}{9}$ cm. Find the diameter of the cylindrical vessel.

Diameter of sphere $= 12 \ cm \Rightarrow$  radius $(r) = 6$ cm

Rise in water level in cylinder $= 3$ $\frac{5}{9}$ $=$ $\frac{32}{9}$ cm

Let the radius of cylinder is $R$

Therefore $\frac{4}{3}$ $\pi (6)^3 = \pi R^2$ $\frac{32}{9}$

$\Rightarrow R^2 =$ $\frac{4}{3}$ $\times 6^3 \times$ $\frac{9}{32}$ $= 3\sqrt{3}$ cm

$\\$

Question 19: A man standing on the deck of the ship, which is $10$ meter above the water level, observes the angle of elevation of the top of the hill as $60^o$ and the angle of depression of the base of the hill as $30^o$. Find the distance of the hill from the ship and the height of the ship.

The man is standing on the deck of the ship at point $A$.

Let $CE$ be the hill with base at $C$.

It is given that the angle of elevation of point $E$ from $A$ is $60^o$ and the angle of depression of point $C$ from $A$ is $30^o$

So, $\angle DAE=60^o$

$\angle CAD=30^o$

$\angle CAD= \angle ACB=30^o$ (alternate angles)

$AB =10$ m

Suppose $ED = h$ m and $BC = x$ m

In $\triangle EAD$, we have $\tan 60^o =$ $\frac{ED}{AD}$

$\Rightarrow \sqrt{3} =$ $\frac{h}{x}$       $(BC = AD = x)$

$\Rightarrow h = \sqrt{3} x$

In $\triangle ABC$, we have $\tan 30^o =$ $\frac{AB}{BC}$

$\Rightarrow$ $\frac{1}{\sqrt{3}}$ $=$ $\frac{10}{x}$   $\Rightarrow x = 10\sqrt{3}$ m

Therefore distance of the hill from the ship is $10 \sqrt{3}$ m

And height of the hill is $h = x\sqrt{3}+ 10 = 10 \sqrt{3} \times \sqrt{3} + 10 = 40$ m

$\\$

Question 20: Three different coins are tossed together. Find the probability of getting (i) exactly two heads (ii) at least two heads (iii) at least two tails

If three different coins tossed together,

Total possible outcomes are $\{$ TTT, TTH , THT, THH, HTT, HTH, HHT, HHH $\}$

Number of total possible outcomes $= 8$

i ) Exactly two heads (favorable outcomes) $= \{$ TTH , THT, HTT $\}$

Number of favorable outcomes $= 3$   Therefore  $P( E ) =$ $\frac{3}{8}$

ii ) At least two heads is (favorable outcomes) $= \{$ THH, HTH, HHT, HHH $\}$

Number of favorable outcomes $= 4$   Therefore $P( E ) =$ $\frac{4}{8}$ $=$ $\frac{1}{2}$

iii ) At least two tails is (favorable outcomes) $=\{$ TTT, TTH , THT, HTT $\}$

Number of favorable outcomes $= 4$   Therefore $P( E ) =$ $\frac{4}{8}$ $=$ $\frac{1}{2}$

$\\$

Section – D

Question number 21 to 31 carry 4 mark each.

Question 21: Due to heavy flood in a  state, thousands were rendered homeless. $50$ schools collectively offered to the state government to provide place and canvas for $1500$ tents to be fixed by government and decided to share the whole expenditure equally. The lower part of the tent is cylindrical of base radius $2.8$ m and height $3.5$ m, with conical upper part of the same base radius but of height $2.1$ m. If the canvas used to make  the tent costs Rs. $120$ sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem.(Use $\pi =$ $\frac{22}{7}$)

Let the radius of base of both cylinder and cone be $r$

Let the height of cone be $H$ and that of cylinder be $h$.

Let the slant height of cone be $l$, Then,

Area of tent $=$ CSA of cylindrical base $+$ CSA of cone

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

$= 2 \pi r h + \pi r \sqrt{H^2 + r^2}$

$= 2 \pi r h + \pi r \sqrt{2.1^2 + 2.8^2}$

$= 2 \pi r h + \pi r \sqrt{12.25}$

$= 2 \pi r h + \pi r \times 3.5$

$=$ $\frac{22}{7}$ $\times 2.8 ( 2 \times 3.5 + 3.5)$

$=$ $\frac{22}{7}$ $\times 2.8 \times 10.5$

$= 92.4 \ m^2$

Cost of canvas used in making one tent $= Rs. \ 92.4 \times 120 = Rs.11088$

Cost of canvas used in making $1500$ tents $= Rs. \ 11088 \times 1500= Rs.16632000$

Share of one school $= Rs. \$ $\frac{16632000}{50}$ $= Rs. 332640$

$\\$

Question 22: Prove that the lengths of the tangents drawn from an external point to the circle are equal.

Let us draw a circle with center $O$ and two tangents $PQ$ and $PR$ are drawn from a external point $P$ to the circle as shown in the figure given below,

Consider $\triangle PQO$ and $\triangle PRO$

$OQ = OR$ (Radius)

$OP$ is common

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

Therefore $\triangle PQO \cong \triangle PRO$   (By SAS criterion)

Hence $PQ = PR$

$\\$

Question 23: Draw a circle of radius $4$ cm. Draw two tangents to the circle inclined at an angle of $60^o$ to each other.

i) Draw a circle of radius $4$ cm with center $O$.

ii) Take a point $A$ on the circumference of circle and join $OA$.

iii) Draw a perpendicular to $OA$ at $A$

iv) Draw a radius $OB$, making an angle of $120^o (=180^o - 60^o)$ with $OA$

v) Draw a perpendicular to $OB$ at $B$. Let these perpendiculars intersect at $P$.

$\\$

Question 24: In Fig. 7, two equal circles with center $O$ and $O'$, touch each other at $X$. $OO'$ produced meets the circle with center $O'$ at $A$. $AC$ is a tangent to the circle with center $O$, at the point $C$. $O'D$ is perpendicular to $AC$. Find the value of $\frac{DO'}{CO}$.

As, circles are equal, so their radius are also equal.

That means $AO' = O'X = XO = r$

Consider $\triangle ADO'$ and $\triangle ACO$

$\angle ADO' = \angle ACO$

$\angle A$ is common

$\triangle ADO' \sim \triangle ACO$   (By AA criterion)

$\therefore$ $\frac{AO'}{AO}$ $=$ $\frac{DO'}{CO}$

$\Rightarrow$ $\frac{r}{3r}$ $=$ $\frac{DO'}{CO}$

$\Rightarrow$ $\frac{DO'}{CO}$ $=$ $\frac{1}{3}$

$\\$

Question 25: Solve for $x:$ $\frac{1}{x+1}$ $+$ $\frac{2}{x+2}$ $=$ $\frac{4}{x+4}$ $, x \neq -1, -2, -4$

Given $\frac{1}{x+1}$ $+$ $\frac{2}{x+2}$ $=$ $\frac{4}{x+4}$ $, x \neq -1, -2, -4$

$\Rightarrow \frac{x+2 + 2 (x+1)}{(x+1)(x+2)}$ $=$ $\frac{4}{x+4}$

$\Rightarrow (3x+3)(x+4) = 4 (x+1)(x+2)$

$\Rightarrow 3x^2 + 3x + 12x + 12 = 4x^2 + 4x + 8x + 8$

$\Rightarrow 3x^2 + 15x + 12 = 4x^2 + 12x+8$

$\Rightarrow x^2 - 3x -4 = 0$

$\Rightarrow (x+1)(x-4) = 0$

$x = -1$ or $x = 4$

Since $x \neq -1, x = 4$

$\\$

Question 26: The angle of elevation of the top $Q$  of a vertical tower $PQ$ from a point $X$ on the ground is $60^o$. From a point $Y, 40$ m vertically above $X$, the angle of elevation  of the top $Q$ of tower is $45^o$. Find the height of the tower $PQ$ and the distance $PX$.

Let $QM = y$,  therefore in $\triangle QMY$

$\tan 45^o =$ $\frac{y}{MY}$   $\Rightarrow$ $\frac{y}{MY}$ $= 1$   $\Rightarrow MY = y$

In $\triangle QPY$

$\tan 60^o =$ $\frac{40+y}{PX}$

$\Rightarrow \sqrt{3} =$ $\frac{40+y}{y}$

$\Rightarrow \sqrt{3}y = 40+y$

$\Rightarrow (\sqrt{3} - 1)y = 40$

$\Rightarrow y =$ $\frac{40}{\sqrt{3} - 1}$

$\Rightarrow y =$ $\frac{40}{\sqrt{3} - 1}$ $\times$ $\frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$\Rightarrow y = 20 (\sqrt{3} + 1)$

Therefore the height of the tower $= 40 + 20 (\sqrt{3} + 1) = 94.64$ m

Now  $\tan 60^o =$ $\frac{h}{x}$

$\Rightarrow x =$ $\frac{94.64}{\sqrt{3}}$ $= 54.64$ m

Therefore horizontal distance is $54.64$ m

$\\$

Question 27: The houses in a row are numbered consecutively from $1$ to $49$. Show that there exists a value of $X$ such that the sum of the numbers of houses preceding the house numbered $X$ is equal to sum of the numbers of houses following $X$.

The number of houses preceding $X$ is $X - 1$

Sum of the number of houses preceding the house numbered, $X = 1 + 2 + 3+.... X - 1.$

Here, $a = 1, d =1$

Given

$S_{X-1} = S_{49} - S_X$

$\Rightarrow$ $\frac{X-1}{2}$ $\Big( 2a + (X-1-1)d \Big) =$ $\frac{49}{2}$ $(2a+48d) -$ $\frac{X}{2}$ $(2a + (X-1)d)$

Substituting $a = 1, d =1$

$\Rightarrow$ $\frac{X-1}{2}$ $\Big( 2 + (X-1-1) \Big) =$ $\frac{49}{2}$ $(2+48) -$ $\frac{X}{2}$ $(2a + (X-1))$

$\Rightarrow ({X-1}) \Big( 2 + (X-2) \Big) =$ $\frac{49}{2}$ $(2 +48 ) - ({X}) (2a + (X-1) )$

$\Rightarrow (X-1) X = 45 \times 50 - X ( 1+X)$

$\Rightarrow 2X^2 = 49 \times 50$

$\Rightarrow X^2 = 49 \times 25$

$\Rightarrow X = 7 \times 5 = 35$

$\\$

Question 28: In Fig. 8, the vertices of $\triangle ABC$ are $A( 4, 6), B(1, 5)$ and $C(7, 2)$. A line segment $DE$ is drawn to intersect the sides $AB$ and $AC$ at $D$  and $E$ respectively such that $\frac{AD}{AB}$ $=$ $\frac{AE}{AC}$ $=$ $\frac{1}{3}$. Calculate the area of $\triangle ADE$ and compare it with the area of $\triangle ABC$.

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

$=$ $\frac{1}{2}$ $| 4(5-2) + 1(2-6) + 7(6-5) |$

$=$ $\frac{15}{2}$ $= 7.5$ sq. units

Now in $\triangle ADE$ and $\triangle ABC$

$\frac{AD}{AB}$ $=$ $\frac{AE}{AC}$ $=$ $\frac{1}{3}$

$\angle A$ is common

$\therefore \triangle ADE \sim \triangle ABC$

Therefore $\frac{ar (\triangle ADE)}{ar (\triangle ABC)}$ $= \Big($ $\frac{AD}{AB}$ $\Big)^2= \Big($ $\frac{1}{3}$ $\Big)^2$

$\Rightarrow ar (\triangle ADE) =$ $\frac{ar (\triangle ABC)}{9}$ $=$ $\frac{7.5}{9}$ $=$ $\frac{5}{6}$ sq. units

$\\$

Question 29: A number $x$ is selected randomly from the numbers $1, 2, 3$ and $4$. Another number $y$ is selected at random from the numbers $1, 4, 9$ and $16$. Find the probability that the product of $x$ and $y$  is less than $16$.

Number $x$ can be selected in four ways. Corresponding to each such way, there are four ways of selecting $y$.

Therefore two numbers can be selected in one of the following ways,

$(1, 1), (1, 4), (1, 9), (1, 16)$
$(2, 1), (2, 4), (2, 9), (2, 16)$
$(3, 1), (3, 4), (3, 9), (3, 16)$
$(4, 1), (4, 4), (4, 9), (4, 16)$

Therefore, the two numbers can be selected in $16$ ways

The number of combinations where $xy < 16$ are

$(1, 1), (1, 4), (1, 9), (2, 1), (2, 4), (3, 1), (3, 4), (4, 1)$

Therefore the number of possible combinations are $8$

Therefore probability that the product $xy$ is less than $16$ is $=$ $\frac{8}{16}$ $=$ $\frac{1}{2}$

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Question 30: In Fig. 9, is shown a sector $OAP$ of a circle with center $O$ containing $\angle \theta$ . $AB$ is perpendicular to the radius $OA$ and meets $OP$ produced at $B$. Prove that the perimeter of shaded region is $r [ \tan \theta + \sec \theta +$ $\frac{\pi \theta}{180}$ $- 1 ]$ .

In $\triangle AOB, \cos \theta =$ $\frac{OA}{OB}$

$\Rightarrow OB =$ $\frac{OA}{\cos \theta}$    $\Rightarrow OB = r \sec \theta$

$PB = OP - OB = r \sec \theta - r$

Also $\tan \theta =$ $\frac{AB}{OA}$

$AB = OA \tan \theta$   $\Rightarrow AB = r \tan \theta$

Length of arc $AP =$ $\frac{\theta}{360^o}$ $(2 \pi r) =$ $\frac{\pi \theta}{180^o}$

Perimeter of shaded region $= PB + AB +$ Length of arc $AP$

$= \Big( r \sec \theta - r + r \tan \theta +$ $\frac{\pi r \theta}{180^o}$ $\Big)$

$= r\Big( \sec \theta + \tan \theta +$ $\frac{\pi \theta}{180^o}$ $- 1 \Big)$

$\\$

Question 31: A motor boat whose speed  is $24$ km/h in still water takes $1$ hour more to go $32$ km upstream than to return down stream  to the same spot. Find the speed of the stream.

Let the speed of stream be $x$.

Therefore, Speed of boat in upstream is $24 - x$

In downstream, speed of boat is $24 + x$

Given, Time taken in the upstream journey ‒ Time taken in the downstream journey $= 1$ hour

$\frac{32}{24-x}$ $-$ $\frac{32}{24+x}$ $=$ $1$

$\Rightarrow \frac{24+x - 24+x}{24^2 - x^2}$ $=$ $\frac{1}{32}$

$\Rightarrow \frac{2x}{576 - x^2}$ $=$ $\frac{1}{32}$

$\Rightarrow x^2 + 64x-576 = 0$

$\Rightarrow x^2 + 72x - 8 x - 576 = 0$

$\Rightarrow x(x+72) -8 (x+72) = 0$

$\Rightarrow (x-8) (x+72) = 0$

$\Rightarrow x = 8, -72$. Since speed cannot be negative, $x = 8$ km/hr.

Therefore, Speed of boat in upstream is $24 - 8 = 16$ km/hr

In downstream, speed of boat is $24 + 8 = 32$ km/hr