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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 $\displaystyle OC$

Now in $\displaystyle \triangle AOC$

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

Therefore , $\displaystyle \angle OAC= \angle OCA = 30^{\circ}$

Also, $\displaystyle \angle OCP= 90^{\circ}$

Therefore, $\displaystyle \angle PCA= 90^{\circ} - 30^{\circ} = 60^{\circ}$

$\\$

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

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

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

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

$\displaystyle \\$

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

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

$\displaystyle \text{Therefore } \frac{BC}{AC} = \cos 60^{\circ} = \frac{1}{2}$

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

$\displaystyle \Rightarrow AC = 5 \text { m }$. Therefore, the length of ladder is 5 m.

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \\$

Section – B

Question number 5 to 10 carry 2 mark each.

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

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

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

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

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

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

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

$\displaystyle \\$

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

By section formula Coordinates of $\displaystyle P$ are

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

By section formula Coordinates of Q are

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

$\displaystyle \\$

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

Similarly,

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

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

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

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

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

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

$\displaystyle CD + AB = DA + BC$

$\displaystyle AB + CD = BC + DA$

Hence proved.

$\displaystyle \\$

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

From equation i) and ii)

$\displaystyle a_{25} = 3 a_{11}$ Hence proved.

$\displaystyle \\$

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 $\displaystyle \triangle OTP, \angle OTP = 90^{\circ}$ (tangent to a circle perpendicular to the radius through the point of contact)

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

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

We know $\displaystyle \angle OTP + \angle OPT + \angle TOP = 180^{\circ}$

$\displaystyle \Rightarrow 90^{\circ} + 30^{\circ} + \angle TOP = 180^{\circ}$ $\displaystyle \Rightarrow \angle TOP = 60^{\circ}$

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

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

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

$\displaystyle OP$ is common

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

$\displaystyle \Rightarrow \angle TPO = \angle SPO$

$\displaystyle OP$ is bisector of $\displaystyle \angle TPS$

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

Therefore $\displaystyle \angle OQT = OQS = 90^{\circ}$

In $\displaystyle \triangle OTQ$, $\displaystyle \angle OQT + \angle OTQ + \angle TOQ = 180^{\circ}$ (angles of a triangle)

$\displaystyle \Rightarrow 90^{\circ} + \angle OTQ + \angle 60^{\circ} = 180^{\circ}$ $\displaystyle \Rightarrow \angle OTQ = 30^{\circ}$

Similarly, $\displaystyle \angle OST = 30^{\circ}$

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 $\displaystyle \triangle ABC \angle C = 90^{\circ}$ (angle in a semicircle)

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

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

$\displaystyle \Rightarrow BC^2 = 25 \Rightarrow BC = 5 \text { cm }$

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

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

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

$\displaystyle \\$

$\\$

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 $\displaystyle =$ Curved surface area of the conical part $\displaystyle +$ curved surface area of cylindrical part.

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

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

Height of cylindrical part is $\displaystyle h = 2.1 \text { m }$

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

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

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

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

$\displaystyle \\$

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

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

Therefore $\displaystyle PA = PB$

Using distance formula

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

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

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

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

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

$\displaystyle \Rightarrow -4bx= - 4ay$

$\displaystyle \Rightarrow bx = ay$

$\displaystyle \\$

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 $\displaystyle =$ Area of circular ring $\displaystyle -$Area of region ABDC

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

Since volume will remain same

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

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

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

$\displaystyle \Rightarrow H = 2 \text { cm }$

$\displaystyle \\$

Question 18: A sphere of diameter $\displaystyle 12 \text { cm }$, 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 $\displaystyle 3\frac{5}{9} \text { cm }$. Find the diameter of the cylindrical vessel.

Diameter of sphere $\displaystyle = 12 \ cm \Rightarrow$ radius $\displaystyle (r) = 6 \text { cm }$

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

Let the radius of cylinder is $\displaystyle R$

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

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

$\displaystyle \\$

Question 19: A man standing on the deck of the ship, which is $\displaystyle 10 \text { m }$eter above the water level, observes the angle of elevation of the top of the hill as $\displaystyle 60^{\circ}$ and the angle of depression of the base of the hill as $\displaystyle 30^{\circ}$. 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 $\displaystyle A$.

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

It is given that the angle of elevation of point $\displaystyle E$ from $\displaystyle A$ is $\displaystyle 60^{\circ}$ and the angle of depression of point $\displaystyle C$ from $\displaystyle A$ is $\displaystyle 30^{\circ}$

So, $\displaystyle \angle DAE=60^{\circ}$

$\displaystyle \angle CAD=30^{\circ}$

$\displaystyle \angle CAD= \angle ACB=30^{\circ}$ (alternate angles)

$\displaystyle AB =10 \text { m }$

Suppose $\displaystyle ED = h \text { m }$ and $\displaystyle BC = x \text { m }$

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

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

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

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

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

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

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

$\displaystyle \\$

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 $\displaystyle \{$ TTT, TTH , THT, THH, HTT, HTH, HHT, HHH $\displaystyle \}$

Number of total possible outcomes $\displaystyle = 8$

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

Number of favorable outcomes =3.

$\displaystyle \text{Therefore } P( E ) = \frac{3}{8}$

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

Number of favorable outcomes =4.

$\displaystyle P( E ) \text{Therefore } = \frac{4}{8} = \frac{1}{2}$

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

Number of favorable outcomes =4.

$\displaystyle P( E ) \text{Therefore } = \frac{4}{8} = \frac{1}{2}$

$\displaystyle \\$

Section – D

Question number 21 to 31 carry 4 mark each.

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = 92.4 \ m^2$

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

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

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

$\displaystyle \\$

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 $\displaystyle O$ and two tangents $\displaystyle PQ$ and $\displaystyle PR$ are drawn from a external point $\displaystyle P$ to the circle as shown in the figure given below,

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

$\displaystyle OQ = OR$ (Radius)

$\displaystyle OP$ is common

$\displaystyle \angle OQP = \angle ORP = 90^{\circ}$

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

Hence $\displaystyle PQ = PR$

$\displaystyle \\$

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 $\displaystyle 4 \text { cm }$ with center $\displaystyle O$.

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

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

iv) Draw a radius $\displaystyle OB$, making an angle of $\displaystyle 120^{\circ} (=180^{\circ} - 60^{\circ} )$ with $\displaystyle OA$

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

$\displaystyle \\$

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 $\displaystyle AO' = O'X = XO = r$

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

$\displaystyle \angle ADO' = \angle ACO$

$\displaystyle \angle A$ is common

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

$\displaystyle x = -1$ or $\displaystyle x = 4$

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

$\displaystyle \\$

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

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

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

In $\displaystyle \triangle QPY$

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

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

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

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

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

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

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

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

$\displaystyle \text{Now } \tan 60^{\circ} = \frac{h}{x}$

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

Therefore horizontal distance is $\displaystyle 54.64 \text { m }$

$\displaystyle \\$

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

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

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

Here, $\displaystyle a = 1, d =1$

Given

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

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

Substituting $\displaystyle a = 1, d =1$

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

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

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

$\displaystyle \angle A$ is common

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle (1, 1), (1, 4), (1, 9), (1, 16)$

$\displaystyle (2, 1), (2, 4), (2, 9), (2, 16)$

$\displaystyle (3, 1), (3, 4), (3, 9), (3, 16)$

$\displaystyle (4, 1), (4, 4), (4, 9), (4, 16)$

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

The number of combinations where $\displaystyle xy < 16$ are

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

Therefore the number of possible combinations are $\displaystyle 8$

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

$\displaystyle \\$

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

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

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

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

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

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

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

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

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

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

$\displaystyle \\$

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

Let the speed of stream be $\displaystyle x$.

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

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

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

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

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

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

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

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

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

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

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

Therefore, Speed of boat in upstream is $\displaystyle 24 - 8 = 16 \text { km }$/hr

In downstream, speed of boat is $\displaystyle 24 + 8 = 32 \text { km }$/hr