\displaystyle \textbf{MATHEMATICS (STANDARD)}

\displaystyle \textbf{Series LRH/1}                                                                      \displaystyle \textbf{Code No. }30/1/1
\displaystyle \text{Roll No.}

\displaystyle \text{Candidates must write the Code on the title page of the answer-book.}
\displaystyle \text{Please check that this question paper contains 15 printed pages.}
\displaystyle \text{Code number given on the right hand side of the question paper should be written on}
\displaystyle \text{the title page of the answer-book by the candidate.}
\displaystyle \text{Please check that this question paper contains 30 questions.}
\displaystyle \text{Please write down the serial number of the question before attempting it.}
\displaystyle \text{15 minutes time has been allotted to read this question paper. The question paper}
\displaystyle \text{will be distributed at 10.15 a.m. From 10.15 a.m. to 10.30 a.m. the student will read}
\displaystyle \text{the question paper only and will not write any answer on the answer script during}
\displaystyle \text{this period.}

\displaystyle \text{Time allowed : 3 hours}                                          \displaystyle \text{Maximum marks : 80}

\displaystyle \textbf{General Instructions :}
\displaystyle \text{(i) All questions are compulsory.}
\displaystyle \text{(ii) The question paper consists of }30\text{ questions divided into four sections }A,B,C\text{ and }D.
\displaystyle \text{Section A comprises ten questions of }1\text{ mark each, Section B five questions of }2\text{ marks each,}
\displaystyle \text{Section C ten questions of }3\text{ marks each and Section D five questions of }6\text{ marks each.}
\displaystyle \text{(iii) All questions in Section A are to be answered in one word, one sentence or as per the exact}
\displaystyle \text{requirement of the question.}
\displaystyle \text{(iv) There is no overall choice. However, an internal choice has been provided in one question of }2\text{ marks,}
\displaystyle \text{three questions of }3\text{ marks and two questions of }6\text{ marks. Attempt only one alternative in each such question.}
\displaystyle \text{(v) In questions on construction, the drawings should be neat and exactly as per the given measurements.}
\displaystyle \text{(vi) Use of calculators is not permitted.}

\displaystyle \textbf{Section A}
\displaystyle \text{Question Numbers 1 to 10 carry 1 mark each.}

\displaystyle \textbf{Question 1. }\text{Has the rational number }\frac{441}{2^{2}\cdot5^{7}\cdot7^{2}}\text{ a terminating or a non-terminating}
\displaystyle \text{decimal representation ?}
\displaystyle \text{Answer:}
\displaystyle \frac{441}{2^{2}\cdot 5^{7}\cdot 7^{2}}=\frac{3^{2}\cdot 7^{2}}{2^{2}\cdot 5^{7}\cdot 7^{2}} 
\displaystyle =\frac{9}{2^{2}\cdot 5^{7}}  
\displaystyle \text{Since the denominator has only the prime factors }2\text{ and }5,  
\displaystyle \text{therefore, the given rational number has a terminating decimal representation.}  
\displaystyle \text{Also, }\frac{9}{2^{2}\cdot 5^{7}}=\frac{9}{312500}=0.0000288  
\\

\displaystyle \textbf{Question 2. }\text{If }\alpha,\ \beta\text{ are the zeroes of a polynomial, such that }\alpha+\beta=6\text{ and }
\displaystyle \alpha\beta=4\text{, then }\text{write the polynomial.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the polynomial be }x^{2}-(\alpha+\beta)x+\alpha\beta.  
\displaystyle \text{Given, }\alpha+\beta=6\text{ and }\alpha\beta=4.  
\displaystyle \therefore\ \text{Polynomial}=x^{2}-6x+4.  
\\

\displaystyle \textbf{Question 3. }\text{If the sum of first }p\text{ terms of an A.P., is }ap^{2}+bp\text{, find its common}
\displaystyle \text{difference.}
\displaystyle \text{Answer:}
\displaystyle S_{p}=ap^{2}+bp  
\displaystyle T_{p}=S_{p}-S_{p-1}  
\displaystyle =\left(ap^{2}+bp\right)-\left[a(p-1)^{2}+b(p-1)\right]  
\displaystyle =ap^{2}+bp-a(p^{2}-2p+1)-bp+b  
\displaystyle =2ap-a+b  
\displaystyle \text{Thus, the first term }A=T_{1}=a+b.  
\displaystyle \text{Also, }T_{p}=A+(p-1)d.  
\displaystyle 2ap-a+b=(a+b)+(p-1)d  
\displaystyle 2ap-2a=(p-1)d  
\displaystyle 2a(p-1)=(p-1)d   
\displaystyle \therefore\ d=2a.  
\\

\displaystyle \textbf{Question 4. }\text{In Fig. 1, S and T are points on the sides PQ and PR respectively of}
\displaystyle \triangle PQR\text{, such that PT = }2\text{ cm, TR = }4\text{ cm and ST is parallel to QR. Find the}
\displaystyle \text{ratio of the areas of }\triangle PST\text{ and }\triangle PQR\text{.} \displaystyle \text{Answer:}
\displaystyle \text{Since }ST\parallel QR,\ \triangle PST\sim\triangle PQR.  
\displaystyle PR=PT+TR=2+4=6\text{ cm}  
\displaystyle \frac{PT}{PR}=\frac{2}{6}=\frac{1}{3}  
\displaystyle \text{For similar triangles, the ratio of their areas is the square of the ratio of their}  
\displaystyle \text{corresponding sides.}  
\displaystyle \therefore \frac{\text{Area of }\triangle PST}{\text{Area of }\triangle PQR}=\left(\frac{PT}{PR}\right)^{2}=\left(\frac{1}{3}\right)^{2}=\frac{1}{9}  
\displaystyle \therefore \text{Required ratio }=\text{Area of }\triangle PST:\text{Area of }\triangle PQR=1:9.  
\\

\displaystyle \textbf{Question 5. }\text{In Fig. 2, }\triangle AHK\text{ is similar to }\triangle ABC\text{. If AK = }10\text{ cm, BC = }3.5\text{ cm and}
\displaystyle \text{HK = }7\text{ cm, find AC.} \displaystyle \text{Answer:}
\displaystyle \triangle AHK\sim\triangle ABC  
\displaystyle \therefore \frac{AK}{AC}=\frac{HK}{BC}  
\displaystyle \frac{10}{AC}=\frac{7}{3.5}=2  
\displaystyle AC=\frac{10}{2}=5\text{ cm}  
\displaystyle \therefore AC=5\text{ cm}.  
\\

\displaystyle \textbf{Question 6. }\text{If }3x=\mathrm{cosec}\ \theta\text{ and }\frac{3}{x}=\cot\theta\text{, find the value of }3\left(x^{2}-\frac{1}{x^{2}}\right)\text{.}
\displaystyle \text{Answer:}
\displaystyle 3x=\mathrm{cosec}\ \theta
\displaystyle \frac{3}{x}=\cot\theta
\displaystyle \mathrm{cosec}^{2}\theta-\cot^{2}\theta=1
\displaystyle (3x)^{2}-\left(\frac{3}{x}\right)^{2}=1
\displaystyle 9x^{2}-\frac{9}{x^{2}}=1
\displaystyle 9\left(x^{2}-\frac{1}{x^{2}}\right)=1
\displaystyle 3\left(x^{2}-\frac{1}{x^{2}}\right)=\frac{1}{3}
\\

\displaystyle \textbf{Question 7. }\text{If P}(2,p)\text{ is the mid-point of the line segment joining the points A}(6,-5)
\displaystyle \text{and B}(-2,11)\text{, find the value of }p\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Using mid-point formula,}
\displaystyle \left(2,p\right)=\left(\frac{6+(-2)}{2},\frac{-5+11}{2}\right)
\displaystyle \left(2,p\right)=\left(\frac{4}{2},\frac{6}{2}\right)
\displaystyle \left(2,p\right)=\left(2,3\right)
\displaystyle \therefore\ p=3.
\\

\displaystyle \textbf{Question 8. }\text{If A}(1,2)\text{, B}(4,3)\text{ and C}(6,6)\text{ are the three vertices of a parallelogram}
\displaystyle \text{ABCD, find the coordinates of the fourth vertex D.}
\displaystyle \text{Answer:}
\displaystyle \text{In parallelogram ABCD, diagonals AC and BD bisect each other.}
\displaystyle \text{Let D}=(x,y).
\displaystyle \text{Mid-point of AC}=\left(\frac{1+6}{2},\frac{2+6}{2}\right)=\left(\frac{7}{2},4\right)
\displaystyle \text{Mid-point of BD}=\left(\frac{4+x}{2},\frac{3+y}{2}\right)
\displaystyle \therefore\ \left(\frac{4+x}{2},\frac{3+y}{2}\right)=\left(\frac{7}{2},4\right)
\displaystyle \frac{4+x}{2}=\frac{7}{2}\text{ and }\frac{3+y}{2}=4
\displaystyle 4+x=7\text{ and }3+y=8
\displaystyle x=3\text{ and }y=5
\displaystyle \therefore\ D=(3,5).
\\

\displaystyle \textbf{Question 9. }\text{The slant height of a frustum of a cone is }4\text{ cm and the perimeters}
\displaystyle \text{(circumferences) of its circular ends are }18\text{ cm and }6\text{ cm. Find the curved}
\displaystyle \text{surface area of the frustum.}
\displaystyle \text{[ Use }\pi=\frac{22}{7}\text{ ]}
\displaystyle \text{Answer:}
\displaystyle \text{Curved surface area of frustum}=\pi(R+r)l
\displaystyle 2\pi R=18\text{ cm and }2\pi r=6\text{ cm}
\displaystyle 2\pi(R+r)=18+6=24
\displaystyle \pi(R+r)=12
\displaystyle l=4\text{ cm}
\displaystyle \therefore\ \text{Curved surface area}=12\times 4=48\text{ cm}^{2}
\\

\displaystyle \textbf{Question 10. }\text{A card is drawn at random from a well shuffled pack of }52\text{ playing cards.}
\displaystyle \text{Find the probability of getting a red face card.}
\displaystyle \text{Answer:}
\displaystyle \text{Total number of cards}=52
\displaystyle \text{Red face cards are J, Q and K of hearts and diamonds.}
\displaystyle \text{Number of red face cards}=3+3=6
\displaystyle \therefore\ P(\text{getting a red face card})=\frac{6}{52}=\frac{3}{26}
\\

\displaystyle \textbf{Section B}
\displaystyle \text{Question Numbers 11 to 15 carry 2 marks each.}

\displaystyle \textbf{Question 11. }\text{If two zeroes of the polynomial }x^{3}-4x^{2}-3x+12\text{ are }\sqrt{3}
\displaystyle \text{ and }-\sqrt{3}\text{, then} \text{ find its third zero.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the third zero be }\alpha.
\displaystyle \text{Sum of zeroes}=\sqrt{3}+(-\sqrt{3})+\alpha=\alpha
\displaystyle \text{For }x^{3}-4x^{2}-3x+12,\text{ sum of zeroes}=\frac{-(-4)}{1}=4
\displaystyle \therefore\ \alpha=4
\displaystyle \text{Hence, the third zero is }4.
\\

\displaystyle \textbf{Question 12. }\text{Find the value of }k\text{ for which the following pair of linear equations have}
\displaystyle \text{infinitely many solutions :}
\displaystyle 2x+3y=7;\qquad (k-1)x+(k+2)y=3k
\displaystyle \text{Answer:}
\displaystyle 2x+3y-7=0
\displaystyle (k-1)x+(k+2)y-3k=0
\displaystyle \text{For infinitely many solutions,}
\displaystyle \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}
\displaystyle \therefore\ \frac{2}{k-1}=\frac{3}{k+2}=\frac{-7}{-3k}
\displaystyle \frac{2}{k-1}=\frac{3}{k+2}
\displaystyle 2(k+2)=3(k-1)
\displaystyle 2k+4=3k-3
\displaystyle k=7
\displaystyle \text{Also, }\frac{2}{k-1}=\frac{7}{3k}
\displaystyle \text{Putting }k=7,\ \frac{2}{6}=\frac{7}{21}=\frac{1}{3}
\displaystyle \therefore\ k=7.
\\

\displaystyle \textbf{Question 13. }\text{In an A.P., the first term is }2\text{, the last term is }29\text{ and sum of the terms is}
\displaystyle 155\text{. Find the common difference of the A.P.}
\displaystyle \text{Answer:}
\displaystyle a=2,\ l=29,\ S_{n}=155
\displaystyle S_{n}=\frac{n}{2}(a+l)
\displaystyle 155=\frac{n}{2}(2+29)
\displaystyle 155=\frac{31n}{2}
\displaystyle n=10
\displaystyle l=a+(n-1)d
\displaystyle 29=2+(10-1)d
\displaystyle 27=9d
\displaystyle d=3
\displaystyle \therefore\ \text{Common difference}=3.
\\

\displaystyle \textbf{Question 14. }\text{If all the sides of a parallelogram touch a circle, show that the}
\displaystyle \text{parallelogram is a rhombus.}
\displaystyle \text{Answer:}
\displaystyle \text{Let ABCD be a parallelogram whose sides touch a circle.}
\displaystyle \text{For a quadrilateral circumscribing a circle, the sums of lengths of opposite sides are equal.}
\displaystyle AB+CD=BC+AD
\displaystyle \text{But in a parallelogram, }AB=CD\text{ and }BC=AD.
\displaystyle \therefore\ AB+AB=BC+BC
\displaystyle 2AB=2BC
\displaystyle AB=BC
\displaystyle \text{Also, }AB=CD\text{ and }BC=AD.
\displaystyle \therefore\ AB=BC=CD=AD
\displaystyle \text{Hence, all the sides of the parallelogram are equal.}
\displaystyle \therefore\ \text{The parallelogram is a rhombus.}
\\

\displaystyle \textbf{Question 15. }\text{Without using trigonometric tables, find the value of the following}
\displaystyle \text{expression :}
\displaystyle \frac{\sec(90^\circ-\theta)\cdot\mathrm{cosec}\ \theta-\tan(90^\circ-\theta)\cot\theta+\cos^{2}25^\circ+\cos^{2}65^\circ}{3\tan27^\circ\cdot\tan63^\circ}

\displaystyle \textbf{Or}

\displaystyle \text{Find the value of cosec }30^\circ\text{, geometrically.}
\displaystyle \text{Answer:}
\displaystyle \text{Using }\sec(90^{\circ}-\theta)=\mathrm{cosec}\ \theta\text{ and }\tan(90^{\circ}-\theta)=\cot\theta,
\displaystyle \frac{\mathrm{cosec}^{2}\theta-\cot^{2}\theta+\cos^{2}25^{\circ}+\cos^{2}65^{\circ}}{3\tan27^{\circ}\cdot\tan63^{\circ}}
\displaystyle =\frac{1+\cos^{2}25^{\circ}+\sin^{2}25^{\circ}}{3\tan27^{\circ}\cdot\cot27^{\circ}}
\displaystyle =\frac{1+1}{3\times 1}
\displaystyle =\frac{2}{3}
\displaystyle \therefore\ \text{Required value}=\frac{2}{3}.

\displaystyle \textbf{Or}

\displaystyle \text{Consider an equilateral }\triangle ABC\text{ of side }2a.
\displaystyle \text{Let AD be the perpendicular from A on BC.}
\displaystyle \text{Then }BD=DC=a.
\displaystyle \text{In right }\triangle ABD,
\displaystyle AB=2a,\ BD=a.
\displaystyle AD=\sqrt{AB^{2}-BD^{2}}=\sqrt{(2a)^{2}-a^{2}}=\sqrt{3a^{2}}=\sqrt{3}a.
\displaystyle \angle BAD=30^{\circ}.
\displaystyle \sin30^{\circ}=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}.
\displaystyle \therefore\ \mathrm{cosec}\ 30^{\circ}=\frac{1}{\sin30^{\circ}}=\frac{1}{1/2}=2.
\displaystyle \therefore\ \mathrm{cosec}\ 30^{\circ}=2.
\\

\displaystyle \textbf{Section C}
\displaystyle \text{Question Numbers 16 to 25 carry 3 marks each.}

\displaystyle \textbf{Question 16. }\text{Prove that }2-3\sqrt{5}\text{ is an irrational number.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }2-3\sqrt{5}\text{ be a rational number.}
\displaystyle \text{Then }2-3\sqrt{5}=r,\text{ where }r\text{ is rational.}
\displaystyle \therefore\ 3\sqrt{5}=2-r
\displaystyle \sqrt{5}=\frac{2-r}{3}
\displaystyle \text{Since }r\text{ is rational, }\frac{2-r}{3}\text{ is also rational.}
\displaystyle \therefore\ \sqrt{5}\text{ is rational.}
\displaystyle \text{But }\sqrt{5}\text{ is irrational, which is a contradiction.}
\displaystyle \therefore\ 2-3\sqrt{5}\text{ is an irrational number.}
\\

\displaystyle \textbf{Question 17. }\text{The sum of numerator and denominator of a fraction is }3\text{ less than twice the}
\displaystyle \text{denominator. If each of the numerator and denominator is decreased by }1\text{,}
\displaystyle \text{the fraction becomes }\frac{1}{2}\text{. Find the fraction.}

\displaystyle \textbf{Or}

\displaystyle \text{Solve the following pair of equations :}
\displaystyle \frac{4}{x}+3y=8;\qquad \frac{6}{x}-4y=-5

\displaystyle \text{Answer:}
\displaystyle \text{Let the numerator be }x\text{ and denominator be }y.
\displaystyle x+y=2y-3
\displaystyle x-y=-3 \qquad (1)
\displaystyle \frac{x-1}{y-1}=\frac{1}{2}
\displaystyle 2x-2=y-1
\displaystyle 2x-y=1 \qquad (2)
\displaystyle \text{From (1), }x=y-3.
\displaystyle \text{Substituting in (2),}
\displaystyle 2(y-3)-y=1
\displaystyle y-6=1
\displaystyle y=7
\displaystyle x=7-3=4
\displaystyle \therefore\ \text{The required fraction is }\frac{4}{7}.

\displaystyle \textbf{OR}

\displaystyle \text{Let }\frac{1}{x}=t.
\displaystyle 4t+3y=8 \qquad (1)
\displaystyle 6t-4y=-5 \qquad (2)
\displaystyle \text{Multiplying (1) by }4,
\displaystyle 16t+12y=32 \qquad (3)
\displaystyle \text{Multiplying (2) by }3,
\displaystyle 18t-12y=-15 \qquad (4)
\displaystyle \text{Adding (3) and (4),}
\displaystyle 34t=17
\displaystyle t=\frac{1}{2}
\displaystyle \therefore\ \frac{1}{x}=\frac{1}{2}
\displaystyle x=2
\displaystyle \text{Substituting in (1),}
\displaystyle 4\left(\frac{1}{2}\right)+3y=8
\displaystyle 2+3y=8
\displaystyle 3y=6
\displaystyle y=2
\displaystyle \therefore\ x=2,\ y=2.
\\

\displaystyle \textbf{Question 18. }\text{In an A.P., the sum of first ten terms is }-150\text{ and the sum of its next ten}
\displaystyle \text{terms is }-550\text{. Find the A.P.}
\displaystyle \text{Answer:}
\displaystyle S_{10}=-150
\displaystyle \frac{10}{2}[2a+(10-1)d]=-150
\displaystyle 5(2a+9d)=-150
\displaystyle 2a+9d=-30 \qquad (1)
\displaystyle \text{Sum of first }20\text{ terms}=S_{20}=-150-550=-700
\displaystyle \frac{20}{2}[2a+(20-1)d]=-700
\displaystyle 10(2a+19d)=-700
\displaystyle 2a+19d=-70 \qquad (2)
\displaystyle \text{Subtracting (1) from (2),}
\displaystyle 10d=-40
\displaystyle d=-4
\displaystyle \text{Putting }d=-4\text{ in (1),}
\displaystyle 2a+9(-4)=-30
\displaystyle 2a-36=-30
\displaystyle 2a=6
\displaystyle a=3
\displaystyle \therefore\ \text{A.P. is }3,-1,-5,-9,\ldots
\\

\displaystyle \textbf{Question 19. }\text{In Fig. 3, ABC is a right triangle, right angled at C and D is the}
\displaystyle \text{mid-point of BC. Prove that AB}^{2}=4\text{AD}^{2}-3\text{AC}^{2}\text{.} \displaystyle \text{Answer:}
\displaystyle \text{In right }\triangle ABC,\text{ by Pythagoras theorem,}
\displaystyle AB^{2}=AC^{2}+BC^{2} \qquad (1)
\displaystyle \text{In right }\triangle ADC,\text{ by Pythagoras theorem,}
\displaystyle AD^{2}=AC^{2}+CD^{2} \qquad (2)
\displaystyle \text{Since D is the mid-point of BC,}
\displaystyle CD=\frac{BC}{2}
\displaystyle BC=2CD
\displaystyle \text{From (1),}
\displaystyle AB^{2}=AC^{2}+(2CD)^{2}
\displaystyle AB^{2}=AC^{2}+4CD^{2} \qquad (3)
\displaystyle \text{From (2), }CD^{2}=AD^{2}-AC^{2}
\displaystyle \text{Substituting in (3),}
\displaystyle AB^{2}=AC^{2}+4(AD^{2}-AC^{2})
\displaystyle AB^{2}=AC^{2}+4AD^{2}-4AC^{2}
\displaystyle AB^{2}=4AD^{2}-3AC^{2}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 20. }\text{Prove the following :}
\displaystyle \frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=1+\tan A+\cot A

\displaystyle \textbf{Or}

\displaystyle \text{Prove the following :}
\displaystyle (\mathrm{cosec}\ A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}

\displaystyle \text{Answer:}
\displaystyle \text{L.H.S.}=\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}
\displaystyle =\frac{\tan A(1-\tan A)+\cot A(1-\cot A)}{(1-\cot A)(1-\tan A)}
\displaystyle =\frac{\tan A-\tan^{2}A+\cot A-\cot^{2}A}{1-\tan A-\cot A+\tan A\cot A}
\displaystyle =\frac{\tan A+\cot A-\tan^{2}A-\cot^{2}A}{2-\tan A-\cot A}
\displaystyle \text{Let }x=\tan A+\cot A.
\displaystyle \text{Then }\tan A\cot A=1
\displaystyle \tan^{2}A+\cot^{2}A=(\tan A+\cot A)^{2}-2\tan A\cot A=x^{2}-2
\displaystyle \therefore\ \text{L.H.S.}=\frac{x-(x^{2}-2)}{2-x}
\displaystyle =\frac{-x^{2}+x+2}{2-x}
\displaystyle =\frac{-(x-2)(x+1)}{-(x-2)}
\displaystyle =x+1
\displaystyle =1+\tan A+\cot A
\displaystyle =\text{R.H.S.}
\displaystyle \text{Hence proved.}

\displaystyle \textbf{OR}

\displaystyle \text{L.H.S.}=(\mathrm{cosec}\ A-\sin A)(\sec A-\cos A)
\displaystyle =\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)
\displaystyle =\left(\frac{1-\sin^{2}A}{\sin A}\right)\left(\frac{1-\cos^{2}A}{\cos A}\right)
\displaystyle =\left(\frac{\cos^{2}A}{\sin A}\right)\left(\frac{\sin^{2}A}{\cos A}\right)
\displaystyle =\sin A\cos A
\displaystyle \text{R.H.S.}=\frac{1}{\tan A+\cot A}
\displaystyle =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}
\displaystyle =\frac{1}{\frac{\sin^{2}A+\cos^{2}A}{\sin A\cos A}}
\displaystyle =\sin A\cos A
\displaystyle \therefore\ \text{L.H.S.}=\text{R.H.S.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 21. }\text{Construct a triangle ABC in which BC = }8\text{ cm, }\angle B=45^\circ\text{ and }
\displaystyle \angle C=30^\circ\text{.}  \text{Construct another triangle similar to }\triangle ABC\text{ such that its sides are }
\displaystyle \frac{3}{4}\text{ of the } \text{corresponding sides of }\triangle ABC\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Steps of construction:}
\displaystyle \text{1. Draw a line segment }BC=8\text{ cm.}
\displaystyle \text{2. At B, draw a ray making an angle of }45^{\circ}\text{ with BC.}
\displaystyle \text{3. At C, draw a ray making an angle of }30^{\circ}\text{ with CB.}
\displaystyle \text{4. Let the two rays intersect at A. Thus, }\triangle ABC\text{ is obtained.}
\displaystyle \text{5. From B, draw a ray }BX\text{ making an acute angle with BC on the opposite side of A.}
\displaystyle \text{6. Mark four equal points }B_{1},B_{2},B_{3},B_{4}\text{ on }BX.
\displaystyle \text{7. Join }B_{4}C.
\displaystyle \text{8. Through }B_{3},\text{ draw a line parallel to }B_{4}C\text{ meeting BC at }C'.
\displaystyle \text{9. Through }C',\text{ draw a line parallel to CA meeting BA at }A'.
\displaystyle \text{Then }\triangle A'BC'\text{ is the required triangle similar to }\triangle ABC.
\\

\displaystyle \textbf{Question 22. }\text{Point P divides the line segment joining the points A}(2,1)\text{ and}
\displaystyle \text{B}(5,-8)\text{ such that }\frac{AP}{AB}=\frac{1}{3}\text{. If P lies on the line }2x-y+k=0\text{, find the}
\displaystyle \text{value of }k\text{.}
\displaystyle \text{Answer:}
\displaystyle \frac{AP}{AB}=\frac{1}{3}
\displaystyle \therefore\ AP:PB=1:2
\displaystyle \text{Using section formula,}
\displaystyle P=\left(\frac{1\cdot 5+2\cdot 2}{1+2},\frac{1\cdot(-8)+2\cdot 1}{1+2}\right)
\displaystyle P=\left(\frac{5+4}{3},\frac{-8+2}{3}\right)
\displaystyle P=(3,-2)
\displaystyle \text{Since P lies on }2x-y+k=0,
\displaystyle 2(3)-(-2)+k=0
\displaystyle 6+2+k=0
\displaystyle k=-8
\\

\displaystyle \textbf{Question 23. }\text{If R}(x,y)\text{ is a point on the line segment joining the points P}(a,b)\text{ and}
\displaystyle \text{Q}(b,a)\text{, then prove that }x+y=a+b\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let R divide PQ in the ratio }m:n.
\displaystyle \text{Using section formula,}
\displaystyle x=\frac{mb+na}{m+n}
\displaystyle y=\frac{ma+nb}{m+n}
\displaystyle \therefore\ x+y=\frac{mb+na}{m+n}+\frac{ma+nb}{m+n}
\displaystyle =\frac{mb+na+ma+nb}{m+n}
\displaystyle =\frac{m(a+b)+n(a+b)}{m+n}
\displaystyle =\frac{(m+n)(a+b)}{m+n}
\displaystyle =a+b
\displaystyle \therefore\ x+y=a+b.
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 24. }\text{In Fig. 4, the boundary of shaded region consists of four semicircular arcs,}
\displaystyle \text{two smallest being equal. If diameter of the largest is }14\text{ cm and that of the}
\displaystyle \text{smallest is }3.5\text{ cm, calculate the area of the shaded region.}
\displaystyle \text{[ Use }\pi=\frac{22}{7}\text{ ]}

\displaystyle \textbf{Or}

\displaystyle \text{Find the area of the shaded region in Fig. 5, if AC = }24\text{ cm, BC = }10\text{ cm}
\displaystyle \text{and O is the centre of the circle.}
\displaystyle \text{[Use }\pi=3.14\text{]}

\displaystyle \text{Answer:}
\displaystyle \text{Diameter of largest semicircle}=14\text{ cm}
\displaystyle \text{Diameter of each smallest semicircle}=3.5\text{ cm}
\displaystyle \text{Diameter of middle semicircle}=14-3.5-3.5=7\text{ cm}
\displaystyle \text{Area of shaded region}=\text{Area of largest semicircle}+\text{Area of middle semicircle}
\displaystyle \hspace{0.2cm}-2\times\text{Area of smallest semicircle}
\displaystyle =\frac{1}{2}\pi(7)^{2}+\frac{1}{2}\pi\left(\frac{7}{2}\right)^{2}-2\times\frac{1}{2}\pi\left(\frac{3.5}{2}\right)^{2}
\displaystyle =\frac{49\pi}{2}+\frac{49\pi}{8}-\frac{49\pi}{16}
\displaystyle =\frac{392\pi+98\pi-49\pi}{16}
\displaystyle =\frac{441\pi}{16}
\displaystyle =\frac{441}{16}\times\frac{22}{7}
\displaystyle =86.625\text{ cm}^{2}
\displaystyle \therefore\ \text{Area of shaded region}=86.625\text{ cm}^{2}.

\displaystyle \textbf{OR}

\displaystyle \text{Since O is the centre and lies on AB, AB is the diameter of the circle.}
\displaystyle \text{In right }\triangle ABC,
\displaystyle AB^{2}=AC^{2}+BC^{2}
\displaystyle AB^{2}=24^{2}+10^{2}=576+100=676
\displaystyle AB=26\text{ cm}
\displaystyle \therefore\ r=\frac{26}{2}=13\text{ cm}
\displaystyle \text{Area of semicircle}=\frac{1}{2}\pi r^{2}
\displaystyle =\frac{1}{2}\times 3.14\times 13^{2}
\displaystyle =265.33\text{ cm}^{2}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}\times BC\times AC
\displaystyle =\frac{1}{2}\times 10\times 24=120\text{ cm}^{2}
\displaystyle \text{Area of shaded region}=265.33-120=145.33\text{ cm}^{2}
\displaystyle \therefore\ \text{Area of shaded region}=145.33\text{ cm}^{2}.
\\

\displaystyle \textbf{Question 25. }\text{Cards bearing numbers }1,\ 3,\ 5,\ \_\_\_,\ 35\text{ are kept in a bag. A card is drawn}
\displaystyle \text{at random from the bag. Find the probability of getting a card bearing}
\displaystyle \text{(i) a prime number less than }15\text{.}
\displaystyle \text{(ii) a number divisible by }3\text{ and }5\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Total cards}=\frac{35-1}{2}+1=18
\displaystyle \text{(i) Prime numbers less than }15\text{ among the given cards are }3,5,7,11,13.
\displaystyle \text{Number of favourable outcomes}=5
\displaystyle P(\text{prime number less than }15)=\frac{5}{18}
\displaystyle \text{(ii) Numbers divisible by both }3\text{ and }5\text{ are multiples of }15.
\displaystyle \text{Among the given cards, these are }15\text{ and }30.
\displaystyle \text{But }30\text{ is not present since only odd numbers are in the bag. }
\displaystyle \text{Therefore, only }15\text{ is favourable.}
\displaystyle \text{Number of favourable outcomes}=1
\displaystyle P(\text{number divisible by }3\text{ and }5)=\frac{1}{18}
\\

\displaystyle \textbf{Section D}
\displaystyle \text{Question Numbers 26 to 30 carry 6 marks each.}

\displaystyle \textbf{Question 26. }\text{Three consecutive positive integers are such that the sum of the square of}
\displaystyle \text{the first and the product of the other two is }46\text{, find the integers.}

\displaystyle \textbf{Or}

\displaystyle \text{The difference of squares of two numbers is }88\text{. If the larger number is }5
\displaystyle \text{less than twice the smaller number, then find the two numbers.}

\displaystyle \text{Answer:}
\displaystyle \text{Let the three consecutive positive integers be }x,\ x+1,\ x+2.
\displaystyle x^{2}+(x+1)(x+2)=46
\displaystyle x^{2}+x^{2}+3x+2=46
\displaystyle 2x^{2}+3x-44=0
\displaystyle 2x^{2}+11x-8x-44=0
\displaystyle x(2x+11)-4(2x+11)=0
\displaystyle (x-4)(2x+11)=0
\displaystyle x=4\text{ or }x=-\frac{11}{2}
\displaystyle \text{Since }x\text{ is a positive integer, }x=4.
\displaystyle \therefore\ \text{The integers are }4,\ 5,\ 6.

\displaystyle \textbf{OR}

\displaystyle \text{Let the smaller number be }x.
\displaystyle \text{Then larger number}=2x-5.
\displaystyle (2x-5)^{2}-x^{2}=88
\displaystyle 4x^{2}-20x+25-x^{2}=88
\displaystyle 3x^{2}-20x-63=0
\displaystyle 3x^{2}-27x+7x-63=0
\displaystyle 3x(x-9)+7(x-9)=0
\displaystyle (x-9)(3x+7)=0
\displaystyle x=9\text{ or }x=-\frac{7}{3}
\displaystyle \text{Since }x\text{ is positive, }x=9.
\displaystyle \text{Larger number}=2(9)-5=13
\displaystyle \therefore\ \text{The two numbers are }9\text{ and }13.
\\

\displaystyle \textbf{Question 27. }\text{Prove that the ratio of the areas of two similar triangles is equal to the}
\displaystyle \text{square of the ratio of their corresponding sides.}
\displaystyle \text{Using the above, prove the following :}
\displaystyle \text{If the areas of two similar triangles are equal, then prove that the triangles}
\displaystyle \text{are congruent.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\triangle ABC\sim\triangle DEF.
\displaystyle \therefore\ \frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}
\displaystyle \text{Also, for similar triangles,}
\displaystyle \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEF}=\left(\frac{AB}{DE}\right)^{2}=\left(\frac{BC}{EF}\right)^{2}=\left(\frac{CA}{FD}\right)^{2}
\displaystyle \text{Thus, the ratio of the areas of two similar triangles is equal to the square}
\displaystyle \text{of the ratio of their corresponding sides.}
\displaystyle \text{Now, given that the areas of two similar triangles are equal.}
\displaystyle \therefore\ \frac{\text{Area of }\triangle ABC}{\text{Area of }\triangle DEF}=1
\displaystyle \left(\frac{AB}{DE}\right)^{2}=1
\displaystyle \frac{AB}{DE}=1
\displaystyle AB=DE
\displaystyle \text{Similarly, }BC=EF\text{ and }CA=FD.
\displaystyle \therefore\ \triangle ABC\cong\triangle DEF\text{ by SSS congruence criterion.}
\displaystyle \text{Hence proved.}
\\

\displaystyle \textbf{Question 28. }\text{From the top of a }7\text{ m high building, the angle of elevation of the top of a}
\displaystyle \text{tower is }60^\circ\text{ and the angle of depression of the foot of the tower is }30^\circ\text{. Find}
\displaystyle \text{the height of the tower.}
\displaystyle \text{Answer:}
\displaystyle \text{Let AB be the building of height }7\text{ m and CD be the tower of height }h\text{ m.}
\displaystyle \text{Let BC be the horizontal distance between the building and the tower.}
\displaystyle \tan30^{\circ}=\frac{AB}{BC}
\displaystyle \frac{1}{\sqrt{3}}=\frac{7}{BC}
\displaystyle BC=7\sqrt{3}
\displaystyle \tan60^{\circ}=\frac{CD-AB}{BC}
\displaystyle \sqrt{3}=\frac{h-7}{7\sqrt{3}}
\displaystyle h-7=21
\displaystyle h=28
\displaystyle \therefore\ \text{Height of the tower}=28\text{ m}.
\\

\displaystyle \textbf{Question 29. }\text{A milk container is made of metal sheet in the shape of frustum of a cone}
\displaystyle \text{whose volume is }10459\frac{3}{7}\text{ cm}^{3}\text{. The radii of its lower and upper circular}
\displaystyle \text{ends are }8\text{ cm and }20\text{ cm respectively. Find the cost of metal sheet used in}
\displaystyle \text{making the container at the rate of Rs. }1.40\text{ per square centimeter.}
\displaystyle \text{[ Use }\pi=\frac{22}{7}\text{ ]}

\displaystyle \textbf{Or}

\displaystyle \text{A toy is in the form of a hemisphere surmounted by a right circular cone of}
\displaystyle \text{the same base radius as that of the hemisphere. If the radius of base of the}
\displaystyle \text{cone is }21\text{ cm and its volume is }\frac{2}{3}\text{ of the volume of the hemisphere,}
\displaystyle \text{calculate the height of the cone and the surface area of the toy.}
\displaystyle \text{[ Use }\pi=\frac{22}{7}\text{ ]}

\displaystyle \text{Answer:}
\displaystyle R=20\text{ cm},\ r=8\text{ cm},\ V=10459\frac{3}{7}\text{ cm}^{3}=\frac{73216}{7}\text{ cm}^{3}
\displaystyle V=\frac{1}{3}\pi h(R^{2}+r^{2}+Rr)
\displaystyle \frac{73216}{7}=\frac{1}{3}\times\frac{22}{7}\times h(20^{2}+8^{2}+20\times 8)
\displaystyle \frac{73216}{7}=\frac{1}{3}\times\frac{22}{7}\times h(400+64+160)
\displaystyle \frac{73216}{7}=\frac{1}{3}\times\frac{22}{7}\times h\times 624
\displaystyle h=16\text{ cm}
\displaystyle l=\sqrt{h^{2}+(R-r)^{2}}
\displaystyle l=\sqrt{16^{2}+(20-8)^{2}}=\sqrt{256+144}=20\text{ cm}
\displaystyle \text{Curved surface area}=\pi(R+r)l
\displaystyle =\frac{22}{7}\times(20+8)\times 20=1760\text{ cm}^{2}
\displaystyle \text{Area of lower base}=\pi r^{2}=\frac{22}{7}\times 8^{2}=\frac{1408}{7}\text{ cm}^{2}
\displaystyle \text{Total area of metal sheet}=1760+\frac{1408}{7}=\frac{13728}{7}\text{ cm}^{2}
\displaystyle \text{Cost}=\frac{13728}{7}\times 1.40=\text{Rs. }2745.60
\displaystyle \therefore\ \text{Cost of metal sheet}=\text{Rs. }2745.60

\displaystyle \textbf{OR}

\displaystyle r=21\text{ cm}
\displaystyle \text{Volume of cone}=\frac{2}{3}\times\text{Volume of hemisphere}
\displaystyle \frac{1}{3}\pi r^{2}h=\frac{2}{3}\times\frac{2}{3}\pi r^{3}
\displaystyle \frac{1}{3}\pi r^{2}h=\frac{4}{9}\pi r^{3}
\displaystyle h=\frac{4r}{3}=\frac{4\times 21}{3}=28\text{ cm}
\displaystyle l=\sqrt{r^{2}+h^{2}}=\sqrt{21^{2}+28^{2}}=\sqrt{441+784}=35\text{ cm}
\displaystyle \text{Surface area of toy}=\text{CSA of cone}+\text{CSA of hemisphere}
\displaystyle =\pi rl+2\pi r^{2}
\displaystyle =\frac{22}{7}\times 21\times 35+2\times\frac{22}{7}\times 21^{2}
\displaystyle =2310+2772=5082\text{ cm}^{2}
\displaystyle \therefore\ \text{Height of cone}=28\text{ cm and surface area of toy}=5082\text{ cm}^{2}.
\\

\displaystyle \textbf{Question 30. }\text{Find the mean, mode and median of the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Class :} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 \\ \hline \text{Frequency :} & 4 & 4 & 7 & 10 & 12 & 8 & 5 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class} & f & x & fx \\ \hline 0-10 & 4 & 5 & 20 \\ \hline 10-20 & 4 & 15 & 60 \\ \hline 20-30 & 7 & 25 & 175 \\ \hline 30-40 & 10 & 35 & 350 \\ \hline 40-50 & 12 & 45 & 540 \\ \hline 50-60 & 8 & 55 & 440 \\ \hline 60-70 & 5 & 65 & 325 \\ \hline \end{array}
\displaystyle \sum f=50,\quad \sum fx=1910
\displaystyle \text{Mean}=\frac{\sum fx}{\sum f}=\frac{1910}{50}=38.2
\displaystyle \text{For median, }N=50,\quad \frac{N}{2}=25
\displaystyle \text{Median class is }30-40.
\displaystyle l=30,\ cf=15,\ f=10,\ h=10
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-cf}{f}\right)h
\displaystyle =30+\left(\frac{25-15}{10}\right)10
\displaystyle =40
\displaystyle \text{For mode, modal class is }40-50.
\displaystyle l=40,\ f_{1}=12,\ f_{0}=10,\ f_{2}=8,\ h=10
\displaystyle \text{Mode}=l+\left(\frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}}\right)h
\displaystyle =40+\left(\frac{12-10}{2(12)-10-8}\right)10
\displaystyle =40+\frac{20}{6}
\displaystyle =43.33
\displaystyle \therefore\ \text{Mean}=38.2,\quad \text{Median}=40,\quad \text{Mode}=43.33
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.