For the principal values, evaluate the following:
Question: For the principal values, evaluate the following: (i) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right)$ (ii) $\sec ^{-1}(\sqrt{2})+2 \operatorname{cosec}^{-1}(-\sqrt{2})$ (iii) $\sin ^{-1}\left[\cos \left\{2 \operatorname{cosec}^{-1}(-2)\right\}\right]$ (iv) $\operatorname{cosec}^{-1}\left(2 \tan \frac{11 \pi}{6}\right)$ Solution: (i) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)+\operatorname{cosec}^{-1}\left(-\frac{2}{\sqrt{3}}\right...
Read More →If a, b, c are in A.P., then show that:
Question: Ifa,b,care in A.P., then show that: (i)a2(b+c),b2(c+a),c2(a+b) are also in A.P. (ii)b+ca,c+ab,a+bcare in A.P. (iii)bca2,cab2,abc2are in A.P. Solution: Since $a, b, c$ are in $A$. P., we have: $2 b=a+c$ (i) We have to prove the following: $2 b^{2}(a+c)=a^{2}(b+c)+c^{2}(a+b)$ LHS : $2 b^{2} \times 2 b \quad$ (Given) $=4 b^{3}$ RHS : $a^{2} b+a^{2} c+a c^{2}+c^{2} b$ $=a c(a+c)+b\left(a^{2}+c^{2}\right)$ $=a c(a+c)+b\left[(a+c)^{2}-2 a c\right]$ $=a c(2 b)+b\left[(2 b)^{2}-2 a c\right]$ $...
Read More →Construct a triangle similar to Δ ABC in which AB = 4.6 cm,
Question: Construct a triangle similar to ΔABCin whichAB= 4.6 cm,BC= 5.1 cm, A= 60 with scale factor 4 : 5. Solution: Given that Construct a $\triangle A B C$ of given data, $A B=4.6 \mathrm{~cm}, B C=5.1 \mathrm{~cm}$ and $\angle A=60^{\circ}$ and then a triangle similar to it whose sides are $(4: 5=4 / 5)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction Step: I- First of all we draw a line segment $A B=4.6...
Read More →Find the set of values
Question: Find the set of values of $\operatorname{cosec}^{-1}\left(\frac{\sqrt{3}}{2}\right)$ Solution: The value of $\operatorname{cosec}^{-1}\left(\frac{\sqrt{3}}{2}\right)$ is undefined as it is outside the range i.e., $R-(-1,1)$....
Read More →Draw a ΔABC with side BC = 6 cm. AB = 5 cm and ∠ ABC = 60°
Question: Draw a ΔABCwith sideBC= 6 cm.AB= 5 cm and ABC= 60. Then, construct a triangle whose sides are (3/4)thof the corresponding sides of the ΔABC. Solution: Given that Construct a $\triangle A B C$ of given data, $A B=5 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $\angle A B C=60^{\circ}$ and then a triangle similar to it whose sides are $(3 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction Step: I- First ...
Read More →Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.
Question: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus. Solution: LetABCDbe the rectangle andP, Q, RandSbe the midpoints ofAB, BC, CDandDA, respectively.JoinAC, a diagonal of the rectangle.In∆ABC,we have: $\therefore P Q \| A C$ and $P Q=\frac{1}{2} A C$ [By midpoint theorem] Again, in∆DAC,the pointsSandRare the mid points ofADandDC,respectively. $\therefore S R \| A C$ and $S R=\frac{1}{2} A C$ [By midpoint theorem] Now,P...
Read More →Find the principal values of each of the following:
Question: Find the principal values of each of the following: (i) $\operatorname{cosec}^{-1}(-\sqrt{2})$ (ii) $\operatorname{cosec}^{-1}(-2)$ (iii) $\operatorname{cosec}^{-1}\left(\frac{2}{\sqrt{3}}\right)$ (iv) $\operatorname{cosec}^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ Solution: (i) Let $\operatorname{cosec}^{-1}(-\sqrt{2})=y$ Then, cosec $y=-\sqrt{2}$ We know that the range of the principal value branch is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}$. Thus, $\operatorname{cosec} y=-...
Read More →Construct an isosceles triangle whose base is 8 cm and altitude
Question: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 3/2 times the corresponding sides of the isosceles triangle. Solution: Given that Construct an isosceles triangle $\mathrm{ABC}$ in which $A B=B C=6 \mathrm{~cm}$ and altitude $=4 \mathrm{~cm}$ then another triangle similar to it whose sides are $\frac{3}{2}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construct...
Read More →Construct a ΔABC in which AB = 5 cm. ∠B = 60° altitude CD = 3 cm
Question: Construct a ΔABCin which AB = 5 cm. B = 60 altitudeCD= 3 cm. Construct a ΔAQR similar to ΔABCsuch that side ΔAQRis 1.5 times that of the corresponding sides of ΔACB Solution: Given that Construct a triangle $\triangle A B C$ in which $\operatorname{let} A B=5 \mathrm{~cm}, \angle B=90^{\circ}$ and altitutde $C D=3 \mathrm{~cm}$, and then a triangle $\triangle A Q R$ similar to it whose sides are $(1.5$ times $=3 / 2)$ of the corresponding sides of $\triangle A C B$. We follow the follo...
Read More →In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively,
Question: In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of ∆ABC. Show that EDF = A, DEF = B and DEF = C. Figure Solution: ∆ABCis shown below. D, EandFare the midpoints of sidesBC,CAand AB, respectively.AsFandEare the mid points of sidesABandACof∆ABC.FE∣∣BC (By mid point theorem)Similarly,DE∣∣FBandFD ∣∣AC.Therefore,AFDE, BDEFandDCEFare all parallelograms.In parallelogramAFDE,we have:A=EDF (Opposite angles are equal)In parallelogramBDEF, we have:B=DE...
Read More →Solve the following
Question: If $a^{2}, b^{2}, c^{2}$ are in A.P., prove that $\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}$ are in A.P. Solution: $a^{2}, b^{2}, c^{2}$ are in A.P. $\Rightarrow b^{2}-a^{2}=c^{2}-b^{2}$ $\Rightarrow(b+a)(b-a)=(c-b)(c+b)$ $\Rightarrow \frac{b-a}{c+b}=\frac{c-b}{b+a}$ $\Rightarrow \frac{b-a}{(c+a)(c+b)}=\frac{c-b}{(b+a)(c+a)} \quad\left[\right.$ Multiplying both the sides by $\left.\frac{1}{c+a}\right]$ $\Rightarrow \frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}-\frac{1}{c+a}$ $\therefore ...
Read More →Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm.
Question: Draw a right triangle in which the sides (other than hypotenuse) are of lengths 5 cm and 4 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. Solution: Given that Construct a right triangle of sides let $A B=5 \mathrm{~cm}, A C=4 \mathrm{~cm}$, and $\angle A=90^{\circ}$ and then a triangle similar to it whose sides are $(5 / 3)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to const...
Read More →Prove that the line segments joining the middle points
Question: Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles. Solution: ∆ABCis shown below. D, EandFare the midpoints of sidesAB, BCandCA, respectively. As,DandEare the mid points of sidesAB, andBCof ∆ABC.DE∣∣ AC (By midpoint theorem)Similarly,DF∣∣ BCandEF∣∣ AB.Therefore,ADEF, BDFEandDFCEare all parallelograms.Now,DEis the diagonal of the parallelogramBDFE.∆BDE ∆FEDSimiilarly,DFis thediagonal of the parallelogramADEF. ∆D...
Read More →Solve the following
Question: If $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P., prove that: (i) $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P. (ii)a(b+c),b(c+a),c(a+b) are in A.P. Solution: Given: $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in A.P. $\therefore \frac{2}{b}=\frac{1}{a}+\frac{1}{c}$ $\Rightarrow 2 a c=a b+b c \quad \ldots(1)$ (i) To prove: $\frac{b+c}{a}, \frac{c+a}{b}, \frac{a+b}{c}$ are in A.P. $2\left(\frac{a+c}{b}\right)=\frac{b+c}{a}+\frac{a+b}{c}$ $\Rightarrow 2 a c(a+c)=b c(b+c...
Read More →Draw a right triangle ABC in which AC = AB = 4.5 cm and ∠A = 90°.
Question: Draw a right triangleABCin whichAC=AB= 4.5 cm and A = 90. Draw a triangle similar to ΔABCwith its sides equal to (5/4)th of the corresponding sides of ΔABC. Solution: Given that Construct a right triangle of sides $A B=A C=4.5 \mathrm{~cm}$, and $\angle A=90^{\circ}$ and then a triangle similar to it whose sides are $(5 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction Step: I- First of all we ...
Read More →In the adjoining figure, AD and BE are the medians of ∆ABC
Question: In the adjoining figure, $A D$ and $B E$ are the medians of $\triangle A B C$ and $D F \| B E$. Show that $C F=\frac{1}{4} A C$. Solution: In∆ABC,we have:AC = AE + EC...(i)AE =EC ...(ii) [BEis the median of∆ABC]AC = 2EC ...(iii)In∆BEC,DF||BE.EF = CF (By midpoint theorem, asDis the midpoint ofBC)ButEC = EF+CF⇒EC= 2⨯ CF ...(iv)From (iii) and (iv), we get:AC =2 ⨯(2⨯CF) $\therefore C F=\frac{1}{4} A C$...
Read More →In the adjoining figure, AD is a median of ∆ABC and DE || BA. Show that BE is also a median of ∆ABC.
Question: In the adjoining figure,ADis a median of∆ABCandDE||BA. Show thatBEis also a median of ∆ABC. Solution: ADis a median of ∆ABC.BD = DCWe know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.Here, in∆ABC, Dis the mid point ofBCandDE||BA(given). ThenDEbisectsAC.i.e.,AE = ECEis the midpoint ofAC.⇒BEis the median of∆ABC....
Read More →Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another
Question: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Solution: Given that Construct a triangle of sides $A B=5 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $A C=7 \mathrm{~cm}$ and then a triangle similar to it whose sides are $(7 / 5)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction Step: I- First of all we dra...
Read More →Find the domain of
Question: Find the domain of (i) $\sec ^{-1}(3 x-1)$ (ii) $\sec ^{-1} x-\tan ^{-1} x$ Solution: (ii) Let $f(x)=g(x)-h(x)$, where Therefore, the domain of $f(x)$ is given by the intersection of the domain of $g(x)$ and $h(x)$ The domain of $g(x)$ is $[0, \pi / 2) \cup[\pi, 3 \pi / 2)$ The domain of $h(x)$ is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Therfore, the intersection of $g(x)$ and $h(x)$ is $\mathrm{R}-\{n \pi, n \in Z\}$...
Read More →In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively.
Question: In the adjoining figure,ABCDis a trapezium in whichAB||DCandP,Qare the midpoints ofADandBCrespectively.DQandABwhen produced meet atE. Also,ACandPQintersect atR. Prove that (i)DQ=QE, (ii)PR||AB, (iii)AR=RC. Solution: Given:AB||DC,AP = PDandBQ= CQ(i) In∆QCDand ∆QBE, we have:DQC=BQE (Vertically opposite angles) DCQ=EBQ (Alternate angles, asAE||DC)BQ=CQ (Pis the midpoints)∆QCD∆QBEHence,DQ = QE (CPCT)(ii) Now, in∆ADE,PandQare the midpoints ofADandDE,respectively.PQ||AE⇒PQ||AB||DC⇒ AB||P...
Read More →For the principal values, evaluate the following:
Question: For the principal values, evaluate the following: (i) $\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)$ (ii) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)-2 \sec ^{-1}\left(2 \tan \frac{\pi}{6}\right)$ Solution: (i) $\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)=\tan ^{-1}\left(\tan \frac{\pi}{3}\right)-\sec ^{-1}\left(\sec \frac{2 \pi}{3}\right)$ $=\frac{\pi}{3}-\frac{2 \pi}{3}$ $=-\frac{\pi}{3}$ (ii) $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)-2 \sec ^{-1}\left(2 \tan \frac{\pi}{6}\right)=-\sin ^{-1}\left(\fr...
Read More →Draw a ΔABC in which BC = 6 cm, AB = 4 cm and AC = 5 cm.
Question: Draw a ΔABCin whichBC= 6 cm,AB= 4 cm andAC= 5 cm. Draw a triangle similar to ΔABCwith its sides equal to (3/4)thof the corresponding sides of ΔABC. Solution: Given that Construct a triangle of sides $A B=4 \mathrm{~cm}, B C=6 \mathrm{~cm}$ and $A C=5 \mathrm{~cm}$ and then a triangle similar to it whose sides are $(3 / 4)^{\text {th }}$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction Step: I- First of all we dra...
Read More →The nth term of a sequence is given by an
Question: Thenthterm of a sequence is given byan= 2n2+n+ 1. Show that it is not an A.P. Solution: We have: $a_{n}=2 n^{2}+n+1$ $a_{1}=2 \times 1^{2}+1+1$ = 4 $a_{2}=2 \times 2^{2}+2+1$ =11 $a_{3}=2 \times 3^{2}+3+1$ =22 $a_{2}-a_{1}=11-4$ =7 and $a_{3}-a_{2}=22-11$ =11 Since, $a_{2}-a_{1} \neq a_{3}-a_{2}$ Hence, it is not an AP....
Read More →In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm.
Question: In a parallelogramPQRS,PQ= 12 cm andPS= 9 cm. The bisector of PmeetsSRinM.PMandQRboth when produced meet atT. Find the length ofRT. Solution: Given: In parallelogramPQRS,PQ= 12 cm andPS= 9 cm. The bisector of SPQmeetsSRatM.Let SPQ= 2x.⇒ SRQ= 2xand TPQ=x.Also,PQ∥SR⇒ TMR= TPQ=x.In △TMR,SRQis an exterior angle.⇒ SRQ= TMR+ MTR⇒ 2x=x+ MTR⇒ MTR=x⇒ △TPQis an isosceles triangle.⇒TQ=PQ= 12 cmNow,RT=TQQR= TQPS= 12 9= 3 cm...
Read More →Construct a triangle similar to a given ΔABC such that each of its sides is (2/3)rd
Question: Construct a triangle similar to a given ΔABCsuch that each of its sides is (2/3)rdof the corresponding sides of ΔABC. It is given that BC = 6 cm, B= 50 and C= 60. Solution: Given that Construct a triangle of given data, $B C=6 \mathrm{~cm}, \angle \mathrm{B}=50^{\circ}$ and $\angle \mathrm{C}=60^{\circ}$ and then a triangle similar to it whose sides are $(2 / 3)$ of the corresponding sides of $\triangle A B C$. We follow the following steps to construct the given Step of construction S...
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