2023 CBSE (Class 10) Board Paper Solution Mathematics : 30/4/1

$\textbf{Series: WX1YZ/4} \ \ \ \ \ \textbf{SET } \sim 1 \ \ \ \ \ \textbf{Q.P. Code } 30/4/1$

$\textbf{MATHEMATICS (STANDARD) - Theory}$

Time: 3 Hours Max. Marks:80

General Instructions:

Read the following instructions carefully follow them:

(i) This question paper contains 38 questions. All questions are compulsory.

(ii) Question paper is divided into FIVE sections – Section A, B, C, D and E.

(iii) In section A, question number 1 to 18 are multiple choice questions (MCQs) and question number 19 and 20 are Assertion – Reason based questions of 1 mark each.

(iii) In section B, question number 21 to 25 are very short answer (VSA) type questions of 2 marks each.

(iv) In section C, question number 26 to 31 are short answer (SA) type questions carrying 3 marks each.

(v) In section D, question number 32 to 35 are long answer (LA) type questions carrying 5 marks each.

(vi) ln section E, question number 36 to 38 are case based integrated units of assessment questions carrying 4 marks each. Internal choice is provided in 2 marks question in each case study.

(viii) There is no overall choice. However, an internal choice has been provided in 2 questions in Section B, 2 questions in Section C, 2 questions in Section D and 3 questions in Section E.

$\text{(ix) Draw neat figures wherever required. Take } \pi = \frac{22}{7} \text{ wherever required if not stated.}$

(x) Use of calculators is not allowed.

————————————————————————————————————————————-

SECTION A

Section-A consists of Multiple Choice

Type questions of 1 mark each

————————————————————————————————————————————-

$\displaystyle \text{Question 1: The ratio of HCF to LMC of the least composite number and the} \\ \text{ least prime number is:}$

$\displaystyle \text{(a)} {1:2 } \hspace{1.0cm} \text{(b)} {2:1 } \hspace{1.0cm} \text{(c)} {1:1} \hspace{1.0cm} \text{(d)} {1:3 }$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle \text{Least composite number is 4 and the least prime number is 2.}$

$\displaystyle \text{HCF (4,2) : LCM(4, 2) = 2:4: 1:2 }$

$\\$

$\displaystyle \text{Question 2: The roots of the equation } x^2 + 3x-10 = 0 \text{ are}$

$\displaystyle \text{(a)} {2, -5 } \hspace{1.0cm} \text{(b)} {-2. 5} \hspace{1.0cm} \text{(c)} { 2, 5} \hspace{1.0cm} \text{(d)} {-2, -5}$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle x^2 + 3x-10 = 0$

$\displaystyle \Rightarrow x^2 + 5x - 2x - 10 = 0$

$\displaystyle \Rightarrow x(x+5) -2 ( x+ 5) = 0$

$\displaystyle \therefore x = 2 \text{ and } x = - 5$

$\\$

$\displaystyle \text{Question 3: The next term of the A.P.} \sqrt{6}, \sqrt{24}, \sqrt{54} \text{ is}$

$\displaystyle \text{(a)} {\sqrt{60} } \hspace{1.0cm} \text{(b)} {\sqrt{96}} \hspace{1.0cm} \text{(c)} {\sqrt{72}} \hspace{1.0cm} \text{(d)} {\sqrt{216}}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{First term, } a_1 = \sqrt{6}$

$\displaystyle \text{Second term, }a_2 = \sqrt{24} = 2\sqrt{6}$

$\displaystyle \text{Common difference } = 2\sqrt{6}-\sqrt{6} = \sqrt{6}$

$\displaystyle \text{Next term of A.P = Third term + common difference }$

$= \sqrt{54}+ \sqrt{6}$

$= 3\sqrt{6} + \sqrt{6}$

$= 4\sqrt{6} = \sqrt{96}$

$\\$

$\displaystyle \text{Question 4: The distance of the point} (-1, 7) \text{from x-axis is:}$

$\displaystyle \text{(a)} {-1} \hspace{1.0cm} \text{(b)} {7} \hspace{1.0cm} \text{(c)} {6} \hspace{1.0cm} \text{(d)} {\sqrt{50}}$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{The distance of } (-1, 7) \text{from the x-axis is } 7 \displaystyle \text{ units. }$

$\\$

$\displaystyle \text{Question 5: What is the area of semi-circle of diameter } 'd'$

$\displaystyle \text{(a)} \ {\frac{1}{16} \pi d^2 } \hspace{1.0cm} \text{(b)} \ {\frac{1}{4} \pi d^2 } \hspace{1.0cm} \text{(c)} \ {\frac{1}{8} \pi d^2 } \hspace{1.0cm} \text{(d)} \ {\frac{1}{2} \pi d^2 }$

$\displaystyle \text{Answer:}$

$\fbox{c }$

$\displaystyle \text{Given, diameter of semi-circle } = d$

$\displaystyle \therefore \text{radius of semi-circle }= \frac{d}{2}$

$\displaystyle \therefore \text{Area of semi-circle }= \frac{\pi \Big(\frac{d}{2}\Big)^2}{2} = \frac{\pi d^2}{8}$

$\\$

$\displaystyle \text{Question 6: The empirical relation between the mode, median and mean of a } \\ \text{distribution is:}$

$\displaystyle \text{(a) } \text{{Mode : 3Median - 2Mean }} \hspace{1.0cm} \text{(b) } \text{{Mode = 3Mean - 2Median }} \hspace{1.0cm} \\ \\ \text{(c) } \text{{Mode : 2Median - 3Mean}} \hspace{1.0cm} \text{(d) } \text{{Mode : 2Mean - 3Median }}$

$\displaystyle \text{Answer:}$

$\fbox{a }$

$\displaystyle \text{Empirical formula: Mode = 3Median - 2Mean }$

$\\$

$\displaystyle \text{Question 7: The pair of linear equations} 2x=5y+6 \text{ and } 15y = 6x-18 \\ \text{ represents two lines which are:}$

$\displaystyle \text{(a)} \text{{Intersecting}} \hspace{1.0cm} \text{(b)} \text{{parallel }} \hspace{1.0cm} \\ \\ \text{(c)} \text{{coincident }} \hspace{1.0cm} \text{(d)} \text{{either intersecting or parallel }}$

$\displaystyle \text{Answer:}$

$\fbox{c}$

Given equations can be re-written as:

$\displaystyle 2x-5y-6=0$

$\displaystyle 6x-15y-18=0$

$\displaystyle \frac{a_1}{a_2}= \frac{2}{6} = \frac{1}{3}$

$\displaystyle \frac{b_1}{b_2} =\frac{-5}{-15} = \frac{1}{3}$

$\displaystyle \frac{c_1}{c_2} = \frac{-6}{-18} = \frac{1}{3}$

Therefore the pair of equations has infinitely many solutions. Graphically pair of linear equations represent coincident.

$\\$

$\displaystyle \text{Question 8: If } \alpha, \beta \text{ are zeros of the polynomial } x^2 - 1, \text{ then value of } (\alpha + \beta) \text{ is: }$

$\displaystyle \text{(a) } {2 } \hspace{1.0cm} \text{(b)} { 1} \hspace{1.0cm} \text{(c) } {-1 } \hspace{1.0cm} \text{(d) } {0 }$

$\displaystyle \text{Answer:}$

$\fbox{d }$

$\displaystyle \text{ Given polynomial:}$

$\displaystyle x^2- 1 = (x-1)(x+1)$

$\displaystyle \text{For zeros, } (x-1)(x+1) = 0$

$\displaystyle \Rightarrow x = 1 \text{ and } x = - 1$

$\displaystyle \text{ Let } \alpha = 1 \text{ and } \beta = - 1$

$\displaystyle \text{Sum of } (\alpha + \beta) = 1 + (-1) = 0$

$\\$

$\displaystyle \text{Question 9: If a pole 6 m high casts a shadow } 2\sqrt{3} \text{ m long on the ground, then sun's elevation is:}$

$\displaystyle \text{(a) } {60^\circ} \hspace{1.0cm} \text{(b) } {45^\circ } \hspace{1.0cm} \text{(c) } {30^\circ } \hspace{1.0cm} \text{(d) } {90^\circ }$

$\displaystyle \text{Answer:}$

$\fbox{ a}$

$\displaystyle \tan \theta = \frac{AB}{AC}$

$\displaystyle \Rightarrow \tan \theta = \frac{6}{2\sqrt{3}}$

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

$\displaystyle \Rightarrow \tan \theta = \sqrt{3}$

$\displaystyle \Rightarrow \tan \theta = \tan 60^\circ$

$\displaystyle \therefore \theta = 60^\circ$

$\\$

$\displaystyle \text{Question 10:} \sec \theta \text{ when expressed in terms of } \cot \theta, \text{ is equal to: }$

$\displaystyle \text{(a) } { \frac{1+\cot^2 \theta}{\cot \theta} } \hspace{1.0cm} \text{(b) } {\sqrt{1+ \cot^2 \theta} } \hspace{1.0cm} \text{(c) } {\frac{\sqrt{1+\cot^2 \theta}}{\cot \theta} } \hspace{1.0cm} \text{(d)} {\frac{\sqrt{1-\cot^2 \theta}}{\cot \theta} }$

$\displaystyle \text{Answer:}$

$\fbox{c}$

We know that

$\displaystyle \sec \theta = \frac{1}{\cos \theta }= \frac{\sin \theta }{\cos \theta } \cdot \frac{1}{\sin \theta } =\frac{1}{\cot \theta } \cdot \mathrm{cosec} \theta =\frac{\sqrt{1+\cot^2 \theta}}{\cot \theta}$

$\\$

$\displaystyle \text{Question 11: Two dice are thrown together. The probability of getting the difference of} \\ \text{ numbers on their upper faces equals to 3 is: }$

$\displaystyle \text{(a) } { \frac{1}{9}} \hspace{1.0cm} \text{(b) } {\frac{2}{9} } \hspace{1.0cm} \text{(c) } {\frac{1}{6} } \hspace{1.0cm} \text{(d) } {\frac{1}{12} }$

$\displaystyle \text{Answer:}$

$\fbox{c}$

$\displaystyle \text{Total number of possible outcomes } = 36 = n(s)$

$\displaystyle \text{Favourable outcomes to get difference of number on the dice as 3 are: } \\ (1, 4), (2, 5), (3, 6), (4, 1), (5, 2), (6, 3)$

$\displaystyle \therefore n(E) = 6$

$\displaystyle \text{Therefore required probability } = \frac{n(E)}{n(s)} = \frac{6}{36} = \frac{1}{6}$

$\\$

$\displaystyle \text{Question 12: In the given figure } \triangle ABC \sim \triangle QPR. \text{ If } AC=6 \text{ cm }, \\ BC = 5 \text{ cm } \text{ and } PR= x; \text { then the value of } x \text{ is }$

$\displaystyle \text{(a) } {3.6 \text{cm}} \hspace{1.0cm} \text{(b) } {2.5 \text{cm} } \hspace{1.0cm} \text{(c) } {10 \text{cm} } \hspace{1.0cm} \text{(d) } {3.2 \text{cm} }$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{Given, } \triangle ABC \sim \triangle QPR$

$\displaystyle \therefore \frac{AB}{QP} = \frac{BC}{PR}= \frac{AC}{QR}$

$\displaystyle \Rightarrow \frac{AB}{QP} = \frac{5}{x} = \frac{6}{3}$

Equating the last two, we get

$\displaystyle x = \frac{5 \times 3}{6} = \frac{5}{2} = 2.5 \text{ cm }$

$\\$

$\displaystyle \text{Question 13: The distance of the point} (-6, 8) \text{ from origin is:}$

$\displaystyle \text{(a) } {6 } \hspace{1.0cm} \text{(b) } { -6} \hspace{1.0cm} \text{(c) } {8 } \hspace{1.0cm} \text{(d) } { 10}$

$\displaystyle \text{Answer:}$

$\fbox{d}$

$\displaystyle \text{Distance between ( -6, 8) and (0, 0) is } \\ = \sqrt{(-6-0)^2+ (8-0)^2} = \sqrt{36+64} =\sqrt{100}= 10$

$\\$

$\displaystyle \text{Question 14: In the given figure,} PQ \text{ is a tangent to the circle with center O. If } \\ \angle OPQ = x. \angle POQ = y , \text{ then } x+y \text{ is }$

$\displaystyle \text{(a) } {45^\circ } \hspace{1.0cm} \text{(b) } {90^\circ } \hspace{1.0cm} \text{(c) } { 60^\circ} \hspace{1.0cm} \text{(d) } {180^\circ }$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{Here, } \angle OQP = 90^\circ \text{ (angle between radius and tangent) }$

$\displaystyle \text{Now, in } \triangle OQP,$

$\displaystyle \angle OQP + \angle QOP + \angle OPQ = 180^\circ$

$\displaystyle 90^\circ + y + x = 180^\circ$

$\displaystyle \Rightarrow x + y = 90^\circ$

$\\$

$\displaystyle \text{Question 15: In the given figure, TA is a tangent to the circle with center O such that OT= 4 cm, } \\ \angle OTA = 30^\circ, \text{ then length of TA is:}$

$\displaystyle \text{(a) } {2\sqrt{3} \text{ cm}} \hspace{1.0cm} \text{(b) } {2 \text{ cm}} \hspace{1.0cm} \text{(c) } {2\sqrt{2} \text{ cm} } \hspace{1.0cm} \text{(d) } {\sqrt{3} \text{ cm} }$

$\displaystyle \text{Answer:}$

$\fbox{a}$

$\displaystyle \text{Here, } \angle OAT = 90^\circ \text{ (angle between radius and tangent)}$

$\displaystyle \text{In } \triangle OAT,$

$\displaystyle \cos 30^\circ = \frac{TA}{OT}$

$\displaystyle \Rightarrow \frac{\sqrt{3}}{2} = \frac{TA}{4}$

$\displaystyle \Rightarrow TA = \frac{4\sqrt{3}}{2} = 2\sqrt{3} \text{cm}$

$\\$

$\displaystyle \text{Question 16: In} \triangle ABC, PQ \parallel BC. \\ \text{ If, PB=6 cm, AP = 4 cm, AQ = 8 cm, find the length of AC. }$

$\displaystyle \text{(a) } {12\text{ cm} } \hspace{1.0cm} \text{(b) } {20\text{ cm} } \hspace{1.0cm} \text{(c) } {6\text{ cm} } \hspace{1.0cm} \text{(d) } {14\text{ cm} }$

$\displaystyle \text{Answer:}$

$\fbox{b}$

$\displaystyle \text{If } PQ \parallel BC \text{ by using basic proportionality theorem, }$

$\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}$

$\displaystyle \Rightarrow \frac{4}{6} = \frac{8}{QC}$

$\displaystyle \Rightarrow QC = \frac{8 \times 6}{4}$

$\displaystyle \Rightarrow QC = 12 \text{ cm }$

$\displaystyle \text{Now, } AC = AQ + QC =8 + 12 = 20 \text{ cm }$

$\\$

$\displaystyle \text{Question 17: If } \alpha , \beta \text{ are zeros of the polynomial } p(x) = 4x^2 - 3x - 7, \\ \text{ then } \Big(\frac{1}{\alpha}+\frac{1}{\beta} \Big) \text{ is equal to:}$

$\displaystyle \text{(a) } {\frac{7}{3}} \hspace{1.0cm} \text{(b) } {\frac{-7}{3} } \hspace{1.0cm} \text{(c) } {\frac{3}{7}} \hspace{1.0cm} \text{(d) } {\frac{-3}{7}}$

$\displaystyle \text{Answer:}$

$\fbox{d}$

$\displaystyle \text{For zeros of polynomial, put } p(x) = 0$

$\displaystyle 4x^2-3x-7 = 0$

$\displaystyle 4x^2 - 7x + 4x - 7 = 0$

$\displaystyle x(4x-7) + ( 4x-7) = 0$

$\displaystyle (4x-7)(x+1) = 0$

$\displaystyle \therefore x = \frac{7}{4} \text{ and } x = -1$

$\displaystyle \text{Let } \alpha = \frac{7}{4} \text{ and } \beta = - 1$

$\displaystyle \therefore \frac{1}{\alpha}+\frac{1}{\beta} = \frac{4}{7} + \frac{1}{-1} =\frac{4}{7} - 1 = \frac{-3}{7}$

$\\$

$\displaystyle \text{Question 18: A card is drawn at random from a well-shuffled pack of 52 cards.} \\ \text{The probability that the card drawn is not an ace is: }$

$\displaystyle \text{(a) } {\frac{1}{13} } \hspace{1.0cm} \text{(b) } {\frac{9}{13} } \hspace{1.0cm} \text{(c) } {\frac{4}{13} } \hspace{1.0cm} \text{(d) } {\frac{12}{13} }$

$\displaystyle \text{Answer:}$

$\fbox{d}$

$\displaystyle \text{Number of ace cards in a pack of 52 = 4 }$

$\displaystyle \text{Number of non-ace cards in a pack of 52 = 48 }$

$\displaystyle \text{Required probability }= \frac{48}{52} = \frac{12}{13}$

$\\$

DIRECTIONS: In the question number 19 and 20 a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option out of the following:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

(d) Assertion (A) is false but Reason (R) is true.

$\displaystyle \text{Question 19: } \\ \text{Assertion (A): The probability that a leap year has 53 Sundays is } \frac{5}{7} \\ \text{Reason (R): The probability that a non-leap year has 53 Sundays is } \frac{5}{7}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{A week has 7 days and total days are 366. }$

$\displaystyle \text{Number of Sundays is a leap year = 52 Sundays + 2 days }$

$\displaystyle \text{Therefore, probability of leap year with 53 Sundays } = \frac{2}{7}$

$\displaystyle \text{Reason: There are 52 Sundays in a non-leap year. But one left over days apart} \\ \text{from those 52 weeks can be either a Monday. Tuesday, Wednesday, Thursday,} \\ \text{Friday, Saturday or Sunday. }$

$\displaystyle \text{Therefore Required probability } = \frac{1}{7}$

$\\$

$\displaystyle \text{Question 20: Assertion (A): a, b, c are in A.P if only if 2b =a+c. } \\ \text{Reason (R): The sum of first n odd natural numbers is } n^2$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Assertion is true because }$

$\displaystyle b-a= c-b \ \ \ \ (a, b, c \text{ are in A.P)}$

$\displaystyle \Rightarrow 2b=a+c$

$\displaystyle \text{Reason: Let } 1+ 3 + 5 + 7 + 9 + \ldots + n, \text{ are sum of n odd natural numbers. }$

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

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

$\displaystyle S_n = \frac{n}{2} (2n)$

$\displaystyle S_n = n^2$

$\displaystyle \text{Hence, the sum of the first } n \text{ odd natural number is } n^2$

$\\$

————————————————————————————————————————————-

SECTION – B

Section – B consists of Very Short Answer (VSA) type questions of 2 marks each.

————————————————————————————————————————————-

$\displaystyle \text{Question 21: Two number are in the ratio 2 : 3 and their LCM is 180.} \\ \text{What is the HCF of these numbers?}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We know that, }$

$\displaystyle LCM \times HCF = a \times b \ \ \ (a,b \text{ are two numbers } ) \ldots \ldots (i)$

$\displaystyle \text{Let numbers } = 2x \text{ and } 3x$

$\displaystyle \therefore LCM =2 \times 3 \times x= 6 x$

$\displaystyle \therefore 6x = 180$

$\displaystyle \Rightarrow x :30$

$\displaystyle \text{Numbers are: }$

$\displaystyle 2 \times 30 = 60 \text{ and } 3 \times 30=90$

$\displaystyle \text{From equation } (i), \ 180 \times HCF = 60 \times 90$

$\displaystyle \Rightarrow HCF = \frac{60 \times 90}{180}$

$\displaystyle \therefore HCF = 30$

$\\$

$\displaystyle \text{Question 22: If one zero of the polynomial } p(x) = 6x^2+37x-(k-2) \\ \text{is reciprocal of the other, then find the value of } k$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the zeros of the polynomial are } \alpha \text{ and } \frac{1}{\alpha}$

$\displaystyle \text{Product of the zeros } = \frac{-(k-2)}{6}$

$\displaystyle \Rightarrow \alpha \times \frac{1}{\alpha} = \frac{-(k-2)}{6}$

$\displaystyle \Rightarrow 6 = - (k-2)$

$\displaystyle \Rightarrow k = 2-6$

$\displaystyle \Rightarrow k = -4$

$\displaystyle \text{Therefore, value of } k \text{ is } -4$

$\\$

$\displaystyle \text{Question 23:} \\ \text{(A) Find the sum and product of the roots of the quadratic equation } \\ 2x^2-9x + 4=0 \\ {\hspace{5.0cm} \text{OR}} \\ \text{(B) Find the discriminant of the quadratic equation } 4x^2-5=0 \\ \text{ and hence comment on the nature of roots of the equation. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) Given quadratic equation is } 2x^2 - 9x + 4 = 0$

$\displaystyle \text{Sum of the roots } = \frac{-(-9)}{2} = \frac{9}{2}$

$\displaystyle \text{Product of the roots } = \frac{4}{2} = 2$

$\displaystyle \text{Note: For quadratic equation } ax^2 + bx + c = 0, \text{sum of the roots is } \frac{-b}{2} \\ \text{ and product of the roots} = \frac{c}{a}$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{(B) Given quadratic equation is } 4x^2 - 5=0$

$\displaystyle Q \text{ discriminant, } D = b^2 - 4ac$

$\displaystyle \Rightarrow D = 0 - 4(4) (- 5)$

$\displaystyle \Rightarrow D = 80$

$\displaystyle \text{Thus determinant } D = 80$

$\displaystyle \text{Since } D> 0, \text{ then the roots are real and distinct. }$

$\\$

$\displaystyle \text{Question 24: If a fair coin is tossed twice, find the probability of getting atmost one head.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{When a coin is tossed two times.}$

$\displaystyle \text{The possible outcomes are \{TT, HH, TH, HT\} }$

$\displaystyle \therefore n(S) = 4$

$\displaystyle \text{Favourable outcomes : \{HH, HT TH \} }$

$\displaystyle \therefore n(E) = 3$

$\displaystyle \text{Required probability }= \frac{n(E)}{n(s)} = \frac{3}{4}$

$\\$

$\displaystyle \text{Question 25:} \\ \text{(A) Evaluate: } \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^ 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ} \\ \\ {\hspace{5.0cm} \text{OR}} \\ \\ \text{(B) If A and B are acute angles such that } \sin(A-B) = 0 \text{ and } \\ 2 \cos (A+B) - 1 = 0, \text{ then find the angles A and B. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) We have } \frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^ 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

$\displaystyle = \frac{5\Big( \frac{1}{2} \Big)^2 + 4 \Big( \frac{2}{\sqrt{3}} \Big) - (1)^2}{\Big( \frac{1}{2} \Big)^2 + \Big( \frac{\sqrt{3}}{2} \Big)^2}$

$\displaystyle = \frac{\frac{5}{4} + \frac{16}{3} - 1}{\frac{1}{4}+ \frac{3}{4}}$

$\displaystyle = \frac{5}{4}+ \frac{16}{3} - 1 = \frac{15+64-12}{12} = \frac{67}{12}$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{(B) Given } \sin(A-B) = 0 \text{ and } 2 \cos (A+B) - 1 = 0$

$\displaystyle \Rightarrow \sin(A-B) = \sin 0^\circ$

$\displaystyle \text{and } \cos(A+B) = \frac{1}{2}$

$\displaystyle \Rightarrow A-B = 0^\circ \ldots \ldots (i)$

$\displaystyle \text{and } \cos (A+B) = \cos 60^\circ$

$\displaystyle \text{and } A+B = 60^\circ \ldots \ldots (ii)$

$\displaystyle \text{On solving equations (i) and (ii) we get }$

$\displaystyle A = 30^\circ \text{ and } B = 30^\circ$

$\\$

————————————————————————————————————————————-

SECTION – C

Section – C consists of Short Answer (SA) type questions of 3 marks each.

————————————————————————————————————————————-

$\displaystyle \text{Question 26:} \\ \text{(A) How many terms are there in an A.P, whose first and fifth terms are -14 and 2, } \\ \text{respectively and the last term is 62. }\\ {\hspace{5.0cm} \text{OR}} \\ \text{(B) Which term of the A.P: 65, 61,57,53, ... is the first negative term? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) Given first term } (a) = - 14, \text{ fifth term } (a_5) = 2 \\ \text{ and the last term } (a_n) = 62$

$\displaystyle \text{Let common difference be } d.$

$\displaystyle \text{Since } a_5 = a + 4d$

$\displaystyle \Rightarrow 2 = - 14 + 4d$

$\displaystyle \Rightarrow d = 4$

$\displaystyle \Rightarrow a_n = a + (n-1) 4$

$\displaystyle \Rightarrow 62 = -14+(n-14) 4$

$\displaystyle \Rightarrow n-1 = 19$

$\displaystyle \Rightarrow n=20$

$\displaystyle \text{Thus the number of terms in the A.P. is 20. }$

$\displaystyle {\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{Given A.P. is 65, 61 57, 53, ... }$

$\displaystyle \text{Here, first term, } a= 65$

$\displaystyle \text{Common difference, } d = -4$

$\displaystyle \text{Let the } n^{th} \text{ term of the given A.P. be the first negative term.}$

$\displaystyle \because a+n < 0$

$\displaystyle \Rightarrow a+ (n-1) d < 0$

$\displaystyle \Rightarrow 65 + (n-1)(-4) < 0$

$\displaystyle \Rightarrow 69-4n < 0$

$\displaystyle \Rightarrow -4n < -69$

$\displaystyle \Rightarrow n > \frac{69}{4}$

$\displaystyle \Rightarrow n > 17\frac{1}{4}$

$\displaystyle \text{Since, 18 is the natural number just greater that } 17\frac{1}{4}$

$\displaystyle \text{Hence, } n = 18$

$\displaystyle \text{Hence, }18^{th} \text{ term is the first negative term.}$

$\\$

$\displaystyle \text{Question 27: Prove that } \sqrt{5} \text{ is an irrational number.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Assume that } \sqrt{5} \text{ is a rational number. }$

$\displaystyle \text{Then } \sqrt{5} = \frac{a}{b} \text{ where HCF of } a, b \text{ is } 1. \ldots (i)$

$\displaystyle \Rightarrow a = \sqrt{5} b$

$\displaystyle \Rightarrow a^2 = 5b^2 \text{ where } b \neq 0$

$\displaystyle \text{Since, } a^2 \text{ is a multiple of 5, so } a \text{ is also a multiple of 5. }$

$\displaystyle \text{Let } a = 5m$

$\displaystyle \Rightarrow (5m)^2 = 5b^2$

$\displaystyle \Rightarrow 25m^2 = 5b^2$

$\displaystyle \Rightarrow b^2 = 5m^2$

$\displaystyle \text{Since } b^2 \text{ is a multiple of 5, so, } b \text{ is also a multiple of 5. }$

$\displaystyle \text{Let } b = 5n$

$\displaystyle \text{Thus, HCF of } (a, b) = 5 \ldots (ii)$

$\displaystyle \text{From equation (i) and (ii), we get that our assumption is wrong. }$

$\displaystyle \text{Therefore } \sqrt{5} \text{ is not a rational number. It is a irrational number. }$

$\\$

$\displaystyle \text{Question 28: Prove that the angle between the two tangents drawn from an external to} \\ \text{circle is supplementary to the angle subtended by the line joining the points of } \\\text{contact at the center.}$

$\displaystyle \text{Answer:}$

$\displaystyle \textbf{Given: } \text{PA and PB are the tangent drawn from a point P to a circle with center O. }$

$\displaystyle \text{Also, the line segments OA and OB are drawn. }$

$\displaystyle \textbf{To Prove: }\angle APB+ \angle AOB= 180^\circ$

$\displaystyle \textbf{Proof: } \text{We know that the tangents to a circle is perpendicular to the radius } \\ \text{through the points of contact. }$

$\displaystyle \therefore PA \perp OA \Rightarrow \angle OAP = 90^\circ$

$\displaystyle \text{and } PB \perp OB \Rightarrow \angle OBP = 90^\circ$

$\displaystyle \text{Therefore, } \angle OAP + \angle OBP = 180^\circ$

$\displaystyle \text{Hence, }\angle APB + \angle AOB = 180^\circ$

$\displaystyle \text{Note: Sum of all the angles of a quadrilateral is } 360^\circ$

$\\$

$\displaystyle \text{Question 29:} \\ \text{(A) Prove that: } \frac{\sin A - 2\sin^3 A}{2 \cos^3 A - \cos A} = \tan A \\ {\hspace{5.0cm} \text{OR}} \\ \text{(B) Prove that } \sec A (1 - \sin A)(\sec A + \tan A) =1$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) LHS } = \frac{\sin A - 2\sin^3 A}{2 \cos^3 A - \cos A}$

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

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

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

$\displaystyle = \tan A = RHS.$

$\displaystyle \text{Hence proved. }$

${\hspace{5.0cm} \text{OR}}$

$\displaystyle \text{(B) LHS } = \sec A (1 - \sin A)(\sec A + \tan A)$

$\displaystyle = \Big( \sec A - \frac{\sin A}{\cos A} \Big) (\sec A + \tan A)$

$\displaystyle = (\sec A - \tan A) (\sec A + \tan A)$

$\displaystyle = \sec^2 A - \tan^2 A$

$\displaystyle = 1+ \tan^2 A - \tan^2 A$

$\displaystyle = 1 = RHS.$

$\displaystyle \text{Hence proved.}$

$\\$

$\displaystyle \text{Question 30: Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord} \\ \text{ of the larger circle which touches the smaller circle. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{Let the two concentric circles with centers O.}$

$\displaystyle \text{Let AB be the chord of the larger circle which touches the smaller circle at point P. }$

$\displaystyle \text{Therefore, AB is tangent to the smaller circle to the point P.}$

$\displaystyle \therefore OP \perp AB$

$\displaystyle \text{In } \triangle OPA, \ \ \ AO^2 = OP^2 + AP^2$

$\displaystyle \Rightarrow 5^2 = 3^2 + AP^2$

$\displaystyle \Rightarrow AP^2 = 25 - 9 = 16$

$\displaystyle \therefore AP = 4 \text{ cm }$

$\displaystyle \text{Now, in } \triangle OPB \ \ \ \ OP \perp AB$

$\displaystyle \therefore AP = PB$

$\displaystyle \text{Note: Perpendicular form the center of the circle bisects the chord }$

$\displaystyle \text{Thus, }AB = 2AP = 2 \times 4 = 8 \text{ cm }$

$\displaystyle \text{Hence, length of the chord of the larger circle is 8 cm. }$

$\\$

$\displaystyle \text{Question 31: Find the value of } 'p' \text{ for which the quadratic equation } \\ px(x-2) + 6 =0 \text{ has two equal roots.}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{For equal roots determinant is } = 0$

$\displaystyle \therefore b^2 - 4ac = 0$

$\displaystyle \text{Given equation is } px(x-2)+6 = 0$

$\displaystyle \Rightarrow px^2 - 2px + 6 = 0$

$\displaystyle \text{Here, } a = p, b= - 2p \text{ and } c = 6$

$\displaystyle \text{Therefore } (-2p)^2 - 4(p)(6) = 0$

$\displaystyle \Rightarrow 4p^2 - 24 p = 0$

$\displaystyle \Rightarrow 4p^2 = 24p$

$\displaystyle \Rightarrow p^2 = 6 p$

$\displaystyle \Rightarrow p^2 - 6p = 0$

$\displaystyle \Rightarrow p ( p-6) = 0$

$\displaystyle \therefore p = 0 \text{ or } p = 6$

$\displaystyle \therefore p = 6$

$\displaystyle \text{Note: If } p = 0, \text{ the the given equation is not a quadratic equation. }$

$\\$

————————————————————————————————————————————-

SECTION. D

Section – D consists of Long Answer (LA) type questions of 4 marks each.

————————————————————————————————————————————-

$\displaystyle \text{Question 32:} \\ \text{(A) A straight highway leads to the foot of a tower. A man standing on the top of the} \\ \text{75 m high tower observes two cars at angles of depression of } 30^\circ \text{and }60^\circ, \\ \text{which are approaching the foot of the tower. If one car is exactly behind the other} \\ \text{on the same side of the tower, find the distance between the two cars. (Use } \sqrt{3} = 1.73 ) \\ \text{ \hspace{5.0cm} OR} \\ \text{(B) From the top of a 7 m high building the angle of elevation of the top of a cable } \\ \text{tower is } 60^\circ \text{and the angle of depression of its foot is } 30^\circ. \text{Determine the height of} \\ \text{the tower. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{In right } \triangle ABC,$

$\displaystyle \tan 30^\circ = \frac{AB}{BC}$

$\displaystyle \frac{1}{\sqrt{3}} = \frac{75}{BD + x}$

$\displaystyle \therefore BD + x = 75\sqrt{3} \ldots \ldots (i)$

$\displaystyle \text{In right } \triangle ABD,$

$\displaystyle \tan 60^\circ = \frac{AB}{BD}$

$\displaystyle \sqrt{3} = \frac{AB}{BD}$

$\displaystyle BD = \frac{75}{\sqrt{3}} \ldots \ldots (ii)$

$\displaystyle \text{From equations (i) and (ii), we get }$

$\displaystyle \frac{75}{\sqrt{3}} + x = 75\sqrt{3}$

$\displaystyle \Rightarrow x =75\sqrt{3} - \frac{75}{\sqrt{3}}$

$\displaystyle \Rightarrow x =75\sqrt{3} - \frac{75\sqrt{3}}{3}$

$\displaystyle \Rightarrow x = 75\sqrt{3} \Big( 1 - \frac{1}{3} \Big)$

$\displaystyle \Rightarrow x = 75\sqrt{3} \Big( \frac{2}{3} \Big)$

$\displaystyle \Rightarrow x = \frac{150}{\sqrt{3}}$

$\displaystyle \Rightarrow x = \frac{150}{1.73}$

$\displaystyle \Rightarrow x = 86.705$

$\displaystyle \Rightarrow x = 86.71 \text{ m }$

$\displaystyle \text{(B) Let AB be the building of height 7 m and EC be the height of the tower. }$

$\displaystyle \text{A is the point from where elevation of tower is } 60^\circ \text{ and the angle of depression of } \\ \text{its foot is } 45^\circ$

$\displaystyle EC = DE + CD$

$\displaystyle \text{Also, } CD = AB = 7 \text{ m and } BC = AD$

$\displaystyle \text{In right } \triangle ABC,$

$\displaystyle \tan 45^\circ = \frac{AB}{BC}$

$\displaystyle \Rightarrow 1 = \frac{7}{BC}$

$\displaystyle \Rightarrow BC = 7$

$\displaystyle \text{Since, } BC = AD$

$\displaystyle \text{Hence, } AD = 7 \text{ m }$

$\displaystyle \text{In right } \triangle ADE,$

$\displaystyle \tan 60^\circ = \frac{DE}{AD}$

$\displaystyle \Rightarrow \sqrt{3} = \frac{DE}{7}$

$\displaystyle \Rightarrow DE = 7\sqrt{3} \text{ m }$

$\displaystyle \text{Hence, } EC = CD + ED$

$\displaystyle = 7 + 7\sqrt{3}$

$\displaystyle = 7( 1 + \sqrt{3})$

$\displaystyle = 7 ( 1+ 1.732)$

$\displaystyle = 7 \times 2.732$

$\displaystyle = 19.124 \text{ m } \sim 19 \text{ m }$

$\\$

$\displaystyle \text{Question 33:} \\ \text{(A) D is a point on the side BC of a triangle ABC such that } \angle ADC = \angle BAC, \\ \text{ prove that } CA^2 = CB \cdot CD \\ {\hspace{5.0cm} OR} \\ \text{(B) If AD and PM are medians of triangle ABC and PQR respectively where } \\ \triangle ABC \sim \triangle PQR, \text{ prove that } \frac{AB}{PQ} = \frac{AD}{PM}$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(A) } \textbf{Given: } \text{D is the point on the side BC of } \triangle ABC \text{ such that } \angle ADC = \angle BAC$

$\displaystyle \textbf{To Prove: } CA^2 = CB \cdot CD$

$\displaystyle \textbf{Proof: } \text{From } \triangle ADC \text{ and } \triangle BAC$

$\displaystyle \angle ADC = \angle BAC \text{ (given) }$

$\displaystyle \angle ACD = \angle BCA \text{ (common angle) }$

$\displaystyle \therefore \triangle ADC \sim \triangle BAC \text{ (By AA similarity criterion) }$

$\displaystyle \text{We know that, the corresponding sides of similar triangles are in proportion. }$

$\displaystyle \therefore \frac{CA}{CB} = \frac{CD}{CA}$

$\displaystyle \Rightarrow CA^2 = CB \cdot CD$

$\displaystyle \text{Hence Proved.}$

$\displaystyle \text{(B) Given, } \triangle ABC \sim \triangle PQR$

$\displaystyle \text{We know that, the corresponding sides of similar triangles are in proportion. }$

$\displaystyle \therefore \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR} \ldots \ldots (i)$

$\displaystyle \text{Also, } \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R \ldots \ldots (ii)$

$\displaystyle \text{Since AD and PM are medians, they will divide opposite sides. }$

$\displaystyle \therefore BD = \frac{BC}{2} \text{ and } QM = \frac{QR}{2} \ldots \ldots (iii)$

$\displaystyle \text{From equation (i) and (ii) we get }$

$\displaystyle \frac{AB}{PQ} = \frac{BD}{QM} \ldots \ldots (iv)$

$\displaystyle \text{In } \triangle ABD \text{ and } \triangle PQM$

$\displaystyle \angle B = \angle Q$

$\displaystyle \frac{AB}{PQ} = \frac{BD}{QM}$

$\displaystyle \therefore \triangle ABD \sim \triangle PQM \text{ (By SAS similarity criterion) }$

$\displaystyle \text{Thus, } \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$

$\displaystyle \text{Hence, } \frac{AB}{PQ} = \frac{AD}{PM}$

$\displaystyle \text{Hence proved. }$

$\\$

$\displaystyle \text{Question 34: A student was asked to make a model shaped like a cylinder with two} \\ \text{cones attached to its ends by using a thin aluminium sheet. The diameter of the model}\\ \text{is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the}\\ \text{volume of air contained in the model. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{We have (the figure) }$

$\displaystyle \text{Height of the cylinder:} = 12 - 4 = 8 \text{ cm }$

$\displaystyle \text{Radius of cone / cylinder } = \frac{3}{2} = 1.5 \text{ cm }$

$\displaystyle \text{Height of the cone } = 2 \text{ cm }$

$\displaystyle \text{Volume of the cylinder } = \pi r^2 h = \pi (1.5)^2 \times 8 = 18 \pi \ cm^3$

$\displaystyle \text{Volume of the cone } = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (1.5)^2 \times 2 = 1.5 \pi \ cm^3$

$\displaystyle \text{Total volume = Volume of cylinder + (volume of cone) } \times 2 \\ {\hspace{3.0cm} = 18\pi + 3\pi = 21\pi = 21 \times \frac{22}{7} = 66 \ cm^3 }$

$\\$

$\displaystyle \text{Question 35: The monthly expenditure on milk in 200 families of a Housing Society is given below:}$

$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Monthly } & 1000 & 1500 & 2000 & 2500 & 3000 & 3500 & 4000 & 4500 \\ \text{Expenditure in Rs. } & - 1500 & - 2000 & - 2500 & - 3000 & - 3500 & - 4000 & - 4500 & - 5000 \\ \hline \text{Number } & 24 & 40 & 33 & x & 30 & 22 & 16 & 7 \\ \text{of Families } & & & & & & & & \\ \hline \end{array}$

$\displaystyle \text{Find the value of } x \text{ and also, find the median and mean expenditure on milk. }$

$\displaystyle \text{Answer:}$

We have,

$\displaystyle 20+40+33+x+30+22+16+7 = 2000 \text{[Note: Total number of families = 200]}$

$\displaystyle \Rightarrow x = 200-172 = 28$

$\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Expenditure} & \text{No. of} & \text{Cumulative} & x_i & d_i = x_1 - 2750 & u_i = \frac{x-2750}{h} & f_i u_i \\ \text{(in Rs.)} & \text{families} (f_i) & \text{frequency (c.f)} & & & & \\ \hline 1000 - 1500 & 24 & 24 & 1250 & - 1500 & -3 & -72 \\ \hline 1500 - 2000 & 40 & 64 & 1750 & -1000 & -2 & -80 \\ \hline 2000 - 2500 & 33 & 97 & 2250 & -500 & -1 & -33 \\ \hline 2500 - 3000 & 28 & 125 & 2750 & 0 & 0 & 0 \\ \hline 3000 - 3500 & 30 & 155 & 3250 & 500 & 1 & 30 \\ \hline 3500 - 4000 & 22 & 177 & 3750 & 1000 & 2 & 44 \\ \hline 4000 - 4500 & 16 & 193 & 4250 & 1500 & 3 & 48 \\ \hline 4500 - 5000 & 7 & 200 & 4750 & 2000 & 4 & 28 \\ \hline \text{Total} & 200 & & & & & -35 \\ \hline \end{array}$

For Mean:

From table above:

$\displaystyle \Sigma f_i = 200, \ \ \ \Sigma f_i u_i = - 35, \ \ \ h = 500, \ \ \ A = 2750$

$\displaystyle \because \ \text{Mean }( \overline{x}) = A + \Big( \frac{\Sigma f_i u_i}{\Sigma f_i} \Big) \times h = 2750 + \Big( \frac{-35}{200} \Big) \times 500 = 2750-87.5 = 2662.5$

$\displaystyle \text{Hence, the mean monthly expenditure is Rs. } 2662.50$

For Median:

From table above:

$\displaystyle \Sigma f_i = N = 200. \text{ then } \frac{N}{2} - \frac{200}{2} = 100, \text{which lies in interval } 2500 - 3000.$

$\displaystyle \text{Median class: } 2500 - 3000$

$\displaystyle \text{Hence, } l = 2500, \ \ \ f = 97, \ \ \ h = 500$

$\displaystyle \because \ \ \ \text{Median} = l + \frac{\Big( \frac{N}{2} - c.f. \Big)}{f} \times h \\ \\ { \hspace{2.0cm} = 2500 + \frac{100-97}{28} \times 500 = 2500 + \frac{3}{28} \times 500 } \\ \\ {\hspace{2.0cm} = 2500 + 53.57 = 2553.57 }$

$\\$

————————————————————————————————————————————-

SECTION. E

Section – E consists of three Case Studies. Based questions of 4 marks each.

————————————————————————————————————————————-

$\displaystyle \text{Question 36:}$

Two schools ‘P’ and ‘Q’ decided to award prizes to their students for two games of Hockey Rs. $x$ per student and Cricket Rs. $y$ per student. School ‘P’ decided to award a total of Rs. 9,500 for the two games to 5 and 4 students respectively; while school ‘Q’ decided to award Rs. 7,370 for the two games to 4 and 3 students respectively.

$\displaystyle \text{Based on the above information answer the following questions: }$

$\displaystyle \text{(i) Represent the following information algebraically (in terms of } x \text{ and } y ).$

$\displaystyle \text{(ii) (a) What is the prize amount for hockey? }$

$\displaystyle \text{ (b) Prize amount on which game is more and by how much? }$

$\displaystyle \text{(iii) What will be the total prize amount if there are 2 students each from two games ? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Given Rs. } x \text{ and Rs.} y \text{ are the price money per student for Hockey and Cricket respectively. }$

$\displaystyle \therefore 5x + 4y = 9500 \ldots \ldots (i)$

$\displaystyle \text{and } 4x + 3y = 7370 \ldots \ldots (ii)$

(ii)

(a) On multiplying equation (i) by 4 and equation (ii) by 5 we get

$\begin{array}{ccccc} 20x & + & 16y & = & 38000 \\ 20x & + & 15y & = & 36850 \\ - & - & - & & \\ \hline & & y & = & 1150 \end{array}$

On substituting value of $y$ in equation (ii) we get

$5x + 4( 1150) = 9500$

$\Rightarrow 5x + 4600 = 9500$

$\Rightarrow 5x = 4900$

$\Rightarrow x = 980$

$\displaystyle \text{Thus, prize money for Hockey is Rs. 980. }$

$\displaystyle \text{(b) From part (a),}$

$\displaystyle \text{Prize money for Hockey = Rs. 980 }$

$\displaystyle \text{Price money for Cricket = Rs. 1150 }$

$\displaystyle \text{Difference between price money = 1150 - 980 = Rs. 170 }$

$\displaystyle \text{Thus the price money is Rs. 170 more for cricket in comparison to hockey }$

$\displaystyle \text{(iii) Total prize money = 2 ( Prize money for Hockey) + (Prize money for Cricket) }$

$= 2 (980 + 1150) = 2 \times 2130 = 4260 \ \text{Rs. }$

$\\$

$\displaystyle \text{Question 37:}$

Jagdhish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field from growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O.

$\displaystyle \text{(i) Taking O as origin, coordinates of P are (-200, 0) and of Q are (200, 0).} \\ \text{PQRS being a square, what are the coordinates of R and S? }$

$\displaystyle \text{(ii) (a) What is the area of square PQRS ? }$

$\displaystyle \text{(b) What is the length of diagonal PR in square PQRS? }$

$\displaystyle \text{(iii) If S divides CA in the ratio K : 1, what is the value of K where point } \\ \text{A is (200, 800)? }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Coordinates of R = (200,400) }$

$\displaystyle \text{Coordinates of S = (- 200,400) }$

$\displaystyle \text{(ii) Since, side of square PQRS = 400 }$

$\displaystyle \text{Thus, area of square PQRS } = (400)^2 = 160000 \text{ units}^2$

$\displaystyle \text{We know that, diagonal of square } = \sqrt{2} \times \text{side }$

$\displaystyle \text{Therefore diagonal PR of square PQRS } = \sqrt{2} \times 400 = 400\sqrt{2} \text{ unit }$

$\displaystyle \text{(iii) Using section formula }$

$\displaystyle 200 = \frac{200K + 1(-600)}{K+1}$

$\displaystyle \Rightarrow - 200 K - 200 = 200 K - 60$

$\displaystyle \Rightarrow - 400 K = - 400$

$\displaystyle \Rightarrow K =$

$\displaystyle \text{Note: Here S is the mid-point of CA, hence S divides CA in ratio 1:1 }$

$\\$

$\displaystyle \text{Question 38:}$

Governing council of a local public development authority of Dehradun decided to build an adventurous playground on the top of a hill, which will have adequate space for parking.

After survey, it was decided to build a rectangular playground, with a semi-circular area allotted for parking at one end of the playground. The length and breadth of the rectangular playground are 14 units and 7 units, respectively. There are two quadrants of radius 2 units on one side for special seats.

$\displaystyle \text{Based on the above information, answer the following questions: }$

$\displaystyle \text{(i) What is the total perimeter of the parking area? }$

$\displaystyle \text{(ii) (a) What is the total area of parking and the two quadrants? }$

$\displaystyle \text{(b) What is the ratio of area of playground to the area of parking area? }$

$\displaystyle \text{(iii) Find the cost of fencing the playground and parking area at the rate of } \\ \text{Rs. 2 per unit. }$

$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Radius of semi-circle } (r) = \frac{7}{2} =3.5 \text{ units}$

$\displaystyle \text{Circumference of semi-circle } = \pi r = \frac{22}{7} \times 3.5 = 11 \text{ units}$

$\displaystyle \therefore \text{Perimeter of parking area = circumference of semi-circle + diameter of semi-circle } = 11 + 7 = 18 \text{ units}$

$\displaystyle \text{(ii) (a) Area of parking } = \frac{\pi r^2}{2} = \frac{22}{7} \times \frac{1}{2} \times (3.5)^2 = 11\times 0.5 \times 3.5 = 19.25 \text{ unit}^2$

$\displaystyle \text{Area of quadrants } = 2 \times \text{(area of one quadrant)} \\ \\ = 2 \times \frac{\pi {r_1}^2 } {4} = 2 \times \frac{22}{7} \times \frac{1}{4} \times 2^2 = 6.285 \text{ unit}^2$