In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN || PQ meets QR at N.
Question: In the adjoining figure,PQRSis a trapezium in whichPQ||SRandMis the midpoint ofPS. A line segmentMN||PQmeetsQRatN. Show thatNis the midpoint ofQR. Solution: Given: In trapeziumPQRS,PQ||SR,MisthemidpointofPSandMN||PQ.Toprove:NisthemidpointofQR.Construction:JoinQS.Proof:In∆SPQ,Since,Misthemid-pointofSPandMO||PQ.Therefore,Oisthemid-pointofSQ. (By Mid-point theorem)Similarly, in∆SRQ,Since,Oisthemid-pointofSQandON||SR (SR||PQandMN||PQ)Therefore,Nisthemid-pointofQR. (By Mid-point theorem)...
Read More →Solve the following
Question: Thenthterm of a sequence is given byan= 2n+ 7. Show that it is an A.P. Also, find its 7th term. Solution: $a_{n}=2 n+7$ $\therefore a_{1}=2 \times 1+7=9$ $a_{2}=2 \times 2+7=11$ $a_{3}=2 \times 3+7=13$ $a_{4}=2 \times 4+7=15$ and so on So, common difference $(d)=11-9=2$ Thus, the above sequence is an $A . P$. with the common difference as 2 $a_{7}=2 \times 7+7=21$...
Read More →Find the principal values of each of the following:
Question: Find the principal values of each of the following: (i) $\sec ^{-1}(-\sqrt{2})$ (ii) $\sec ^{-1}(2)$ (iii) $\sec ^{-1}\left(2 \sin \frac{3 \pi}{4}\right)$ (iv) $\sec ^{-1}\left(2 \tan \frac{3 \pi}{4}\right)$ Solution: (i) Let $\sec ^{-1}(-\sqrt{2})=y$ Then, $\sec y=-\sqrt{2}$ We know that the range of the principal value branch is $[0, \pi]-\left\{\frac{\pi}{2}\right\}$. Thus, $\sec y=-\sqrt{2}=\sec \left(\frac{3 \pi}{4}\right)$ $\Rightarrow y=\frac{3 \pi}{4} \in[0, \pi], y \neq \frac{...
Read More →M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O
Question: MandNare points on opposite sidesADandBCof a parallelogramABCDsuch thatMNpasses through the point of intersectionOof its diagonalsACandBD. Show thatMNis bisected atO. Solution: Given: AparallelogramABCDToprove:MNisbisectedatOProof:InΔ∆OAMandΔ∆OCN,OA=OC (Diagonals ofparallelogrambisecteachother)AOM=CON (Verticallyoppositeangles)MAO=OCN (Alternate interior angles) By ASA congruence criteria, $\Delta O A M \cong \Delta O C N$ $\Rightarrow O M=O N(\mathrm{CPCT})$ Hence,MNisbisectedatO....
Read More →Construct a triangle similar to a given ΔABC such that each of its sides is (5/7)th of the corresponding sides of Δ ABC.
Question: Construct a triangle similar to a given ΔABCsuch that each of its sides is (5/7)th of the corresponding sides of ΔABC.It is given that AB - 5 cm, BC = 7 cm and ABC= 50. Solution: Given that $A B=5 \mathrm{~cm}, B C=7 \mathrm{~cm}$ and $\angle A B C=50^{0}$ Construct a triangle similar to a triangle $A B C$ such that each of sides is $(5 / 7)^{\text {th }}$ of the corresponding sides of triangle $A B C$. We follow the following steps to construct the given Step of construction Step: I- ...
Read More →In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively.
Question: In the adjoining figure,ABCDis a || gm in whichEandFare the midpoints ofABandCDrespectively. IfGHis a line segment that cutsAD,EFandBCatG,PandHrespectively, prove thatGP=PH. Solution: In parallelogramABCD, we have:AD || BCandAB || DC AD = BCandAB = DCAB = AE+BEandDC =DF+FC AE = BE=DF = FC Now,DF = AEandDF || AE. i.e.,AEFDis a parallelogram.AD || EFSimilarly, BEFCis also a parallelogram.EF || BCAD || EF|| BCThus,AD, EFandBCare three parallel lines cut by the transversal lineDCat...
Read More →A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common.
Question: A square is inscribed in an isoscelesright triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse. Solution: Given:In anisoscelesright∆ABC,CMPNis asquare.Toprove:PbisectsthehypotenuseABi.e.,AP=PB.Proof:InsquareCMPN,CM=MP=PN=CN (Allsidesareequal.)Also,∆ABCis an isosceleswithAC=BC.⇒AN+NC=CM+MB⇒AN=MB (∵CN=CM) ...(i)Now,In∆ANPand∆PMB,AN=MB [From (i)]ANP= PMB=90PN=PM (Sides of squa...
Read More →P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
Question: P, Q, RandSare respectively the midpoints of the sidesAB,BC,CDandDAof a quadrilateralABCD. Show that (i) $P Q \| A C$ and $P Q=\frac{1}{2} A C$ (ii)PQ||SR (iii)PQRSis a parallelogram. Solution: Given: In quadrilateralABCD,P,Q,RandSarerespectivelythemidpointsofthesidesAB,BC,CDandDA.To prove: (i) $P Q \| A C$ and $P Q=\frac{1}{2} A C$ (ii) $P Q \| S R$ (iii) $P Q R S$ is a parallelogram Proof: (i) In $\Delta A B C$ Since,PandQare the mid points of sidesABandBC,respectively. (Given) $\Rig...
Read More →Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case.
Question: Show that each of the following sequences is an A.P. Also find the common difference and write 3 more terms in each case. (i) 3, 1, 5, 9 ... (ii) 1, 1/4, 3/2, 11/4, ... (iii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}, \ldots$ (iv) 9, 7, 5, 3, ... Solution: (i) We have: $-1-3=-4$ $-5-(-1)=-4$ $-9-(-5)=-4 \ldots$ Thus, the sequence is an A.P. with the common difference being $-4$. The next three terms are as follows: $-9-4=-13$ $-13-4=-17$ $-17-4=-21$ (ii) We have: $1 / 4-(-1)=5 / 4$...
Read More →Evaluate each of the following:
Question: Evaluate each of the following: (i) $\tan ^{-1} 1+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$ (ii) $\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$ (iii) $\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left\{\cos \left(\frac{13 \pi}{6}\right)\right\}$ Solution: (i) Let $\sin ^{-1}\left(-\frac{1}{2}\right)=y$ Then, $\sin y=-\frac{1}{2}$ We know that the range of the principal v...
Read More →In the adjoining figure, ∆ABC is a triangle and through A, B, C, lines are drawn, parallel respectively to BC,
Question: In the adjoining figure,∆ABCis a triangle and throughA,B,C, lines are drawn, parallel respectively toBC,CAandAB, intersecting atP,QandR. Prove that the perimeter of ∆PQRis double the perimeter of ∆ABC. Solution: Perimeter of∆ABC=AB+BC+CA...(i)Perimeter of∆PQR= PQ+QR+PR ...(ii)BC||QAandCA||QBi.e.,BCQAis a parallelogram.BC =QA ...(iii)Similarly,BC|| ARandAB ||CRi.e.,BCRAis a parallelogram.BC = AR ...(iv)But,QR= QA+ARFrom (iii) and (iv), we get:⇒QR= BC+BC⇒QR=2BC $\therefore B C=\frac{1}...
Read More →The Fibonacci sequence is defined by
Question: The Fibonacci sequence is defined by a1= 1 =a2,an=an 1+an 2forn 2 Find $\frac{a_{n+1}}{a_{n}}$ for $n=1,2,3,4,5$. Solution: $a_{1}=1=a_{2}, a_{n}=a_{n-1}+a_{n-2}$ for $n2$ Then we have: $a_{3}=a_{2}+a_{1}=1+1=2$ $a_{4}=a_{3}+a_{2}=2+1=3$ $a_{5}=a_{4}+a_{3}=3+2=5$ $a_{6}=a_{5}+a_{4}=5+3=8$ For $n=1, \frac{a_{n+1}}{a_{n}}=\frac{a_{2}}{a_{1}}=\frac{1}{1}=1$ For $n=2, \frac{a_{n+1}}{a_{n}}=\frac{a_{3}}{a_{2}}=\frac{2}{1}=2$ For $n=3, \frac{a_{n+1}}{a_{n}}=\frac{a_{4}}{a_{3}}=\frac{3}{2}$ F...
Read More →Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle
Question: Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are (2/3) of the corresponding sides of it. Solution: Given that Construct a triangle of sides $4 \mathrm{~cm}, 5 \mathrm{~cm}$ and $6 \mathrm{~cm}$ and then a triangle similar to it whose sides are $(2 / 3)$ of the corresponding sides of it. We follow the following steps to construct the given Step of construction Step: I- First of all we draw a line segment. Step: II- WithAas centre and ra...
Read More →Write the first five terms in each of the following:
Question: Let an be a sequence. Write the first five terms in each of the following: (i)a1= 1,an=an 1+ 2,n 2 (ii)a1= 1 =a2,an=an 1+an 2,n 2 (iii)a1=a2= 2,an=an 1 1,n 2 Solution: (i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$ $a_{2}=a_{1}+2=1+2=3$ $a_{3}=a_{2}+2=5$ $a_{4}=a_{3}+2=7$ $a_{5}=a_{4}+2=9$ Hence, the five terms are 1, 3, 5, 7 and 9. (ii)a1= 1 =a2,an=an 1+an 2,n 2 $a_{3}=a_{2}+a_{1}=1+1=2$ $a_{4}=a_{3}+a_{2}=2+1=3$ $a_{5}=a_{4}+a_{3}=3+2=5$ Hence, the five terms are 1, 1, 2, 3 and 5. (iii)a1=...
Read More →Divide a line segment of length 14 cm internally in the ratio 2 : 5.
Question: Divide a line segment of length 14 cm internally in the ratio 2 : 5. Also, justify your construction. Solution: Given that Determine a point which divides a line segment of lengthinternally in the ratio of. We follow the following steps to construct the given Step of construction Step: I-First of all we draw a line segment. Step: II- We draw a raymaking an acute anglewith. Step: III- Draw a rayparallel toAXby making an acute angle. Step IV- Mark of two pointsonand three pointsonin such...
Read More →Write the first five terms in each of the following:
Question: Let an be a sequence. Write the first five terms in each of the following: (i)a1= 1,an=an 1+ 2,n 2 (ii)a1= 1 =a2,an=an 1+an 2,n 2 (iii)a1=a2= 2,an=an 1 1,n 2 Solution: (i) $a_{1}=1, a_{n}=a_{n-1}+2, n \geq 2$ $a_{2}=a_{1}+2=1+2=3$ $a_{3}=a_{2}+2=5$ $a_{4}=a_{3}+2=7$ $a_{5}=a_{4}+2=9$ Hence, the five terms are 1, 3, 5, 7 and 9. (ii)a1= 1 =a2,an=an 1+an 2,n 2 $a_{3}=a_{2}+a_{1}=1+1=2$ $a_{4}=a_{3}+a_{2}=2+1=3$ $a_{5}=a_{4}+a_{3}=3+2=5$ Hence, the five terms are 1, 1, 2, 3 and 5. (iii)a1=...
Read More →A ∆ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides
Question: A $\triangle A B C$ is given. If lines are drawn through $A, B, C$, parallel respectively to the sides $B C, C A$ and $A B$, forming $\triangle P Q R$, as shown in the adjoining figure, show that $B C=\frac{1}{2} Q R$. Solution: BC|| QAandCA|| QBi.e.,BCQAis a parallelogram. BC = QA ...(i)Similarly,BC|| ARandAB ||CR.i.e.,BCRAis a parallelogram.BC = AR...(ii)ButQR = QA+ ARFrom (i) and (ii), we get:QR= BC + BC⇒QR= 2BC $\therefore B C=\frac{1}{2} Q R$...
Read More →Divide a line segment of length 9 cm internally in the ratio 4 : 3.
Question: Divide a line segment of length 9 cm internally in the ratio 4 : 3. Also, give justification of the construction. Solution: Given that Determine a point which divides a line segment of lengthinternally in the ratio of. We follow the following steps to construct the given Step of construction Step: I- First of all we draw a line segment. Step: II- We draw a raymaking an acute anglewith. Step: III- Draw a rayparallel toAXby making an acute angle. Step IV- Mark of two pointsonand three po...
Read More →K, L, M and N are points on the sides AB, BC, CD and DA respectively of a square
Question: K, L, MandNare points on the sidesAB,BC,CDandDArespectively of a squareABCDsuch thatAK=BL=CM=DN. Prove thatKLMNis a square. Solution: Given: InsquareABCD,AK=BL=CM=DN.Toprove:KLMNisasquare.Proof:InsquareABCD,AB=BC=CD=DA (Allsidesofasquareareequal.)And,AK=BL=CM=DN (Given)So,AB-AK=BC-BL=CD-CM=DA-DN⇒KB=CL=DM=AN ...(1) In $\Delta N A K$ and $\Delta K B L$, $\angle N A K=\angle K B L=90^{\circ}$ (Each angle of a square is a right angle.) $A K=B L$ (Given) $A N=K B$ [From (1)] So, by SAS con...
Read More →For the principal values, evaluate each of the following:
Question: For the principal values, evaluate each of the following: (i) $\tan ^{-1}(-1)+\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ (ii) $\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$ Solution: (i) $\tan ^{-1}(-1)+\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)$ $=\tan ^{-1}\left\{\tan \left(-\frac{\pi}{4}\right)\right\}+\cos ^{-1}\left(\cos \frac{3 \pi}{4}\right)$ $\left[\because\right.$ Range of $\tan$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) ;-\frac{\pi}{4} \i...
Read More →Determine a point which divides a line segment of length
Question: Determine a point which divides a line segment of length 12 cm internally in the ratio 2 : 3 Also, justify your construction. Solution: Given that Determine a point which divides a line segment of lengthinternally in the ratio of. We follow the following steps to construct the given Step of construction Step: I- First of all we draw a line segment. Step: II- We draw a raymaking an acute anglewith. Step: III- Draw a rayparallel toAXby making an acute angle. Step IV- Mark of two pointson...
Read More →Solve the following
Question: Let an be a sequence defined by a1= 3 and,an= 3an 1+ 2, for alln 1 Find the first four terms of the sequence. Solution: Given: a1= 3 And,an= 3an 1+ 2 for alln 1 $a_{2}=3 a_{2-1}+2=3 a_{1}+2=11$ $a_{3}=3 a_{3-1}+2=3 a_{2}+2=35$ $a_{4}=3 a_{4-1}+2=3 a_{3}+2=107$ Thus, the first four terms of the sequence are $3,11,35,107$....
Read More →A sequence is defined by
Question: A sequence is defined byan=n3 6n2+ 11n 6,nϵN. Show that the first three terms of the sequence are zero and all other terms are positive. Solution: Given: an=n3 6n2+ 11n 6,nϵN For $n=1, a_{1}=1^{3}-6 \times 1^{2}+11 \times 1-6=0$ For $n=2, a_{2}=2^{3}-6 \times 2^{2}+11 \times 2-6=0$ For $n=3, a_{3}=3^{3}-6 \times 3^{2}+11 \times 3-6=0$ For $n=4, a_{4}=4^{3}-6 \times 4^{2}+11 \times 4-6=60$ For $n=5, a_{5}=5^{3}-6 \times 5^{2}+11 \times 5-6=240$ and so on Thus, the first three terms are ...
Read More →Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm
Question: Draw a line segmentABof length 8 cm. Taking A as centre, draw a circle of radius 4 cm and takingBas centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle. Solution: Given that Construct a circle of radius, andextended diameter each at distance of 7cm from its centre. Construct the pair of tangents to the circle from these two points. We follow the following steps to construct the given Step of construction Step: I First of all...
Read More →Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 from its centre.
Question: Draw two tangents to a circle of radius 3.5 cm from a pointPat a distance of 6.2 from its centre. [CBSE 2013] Solution: Steps of ConstructionStep 1. Draw a circle with O as centre and radius 3.5 cm.Step 2. Mark a point P outside the circle such that OP = 6.2 cm.Step 3. Join OP. Draw the perpendicular bisector XY of OP, cutting OP at Q.Step 4. Draw a circle with Q as centre and radius PQ (or OQ), to intersect the given circle at the points T and T'.Step 5. Join PT and PT'. Here, PT and ...
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