Prove
Question: Solution: Let $I=\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x$ ...(1) $I=\int_{0}^{\pi}\left\{\frac{(\pi-x) \tan (\pi-x)}{\sec (\pi-x)+\tan (\pi-x)}\right\} d x$ $\left(\int_{0}^{0} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$ $\Rightarrow I=\int_{0}^{\pi}\left\{\frac{-(\pi-x) \tan x}{-(\sec x+\tan x)}\right\} d x$ $\Rightarrow I=\int_{0}^{\pi} \frac{(\pi-x) \tan x}{\sec x+\tan x} d x$ ...(2) Adding (1) and (2), we obtain $2 I=\int_{0}^{\pi} \frac{\pi \tan x}{\sec x+\tan x} d x$ $\Rig...
Read More →If D, G and R denote respectively the number of degrees, grades and radians in an angle, then
Question: IfD,GandRdenote respectively the number of degrees, grades and radians in an angle, then (a) $\frac{D}{100}=\frac{G}{90}=\frac{2 R}{\pi}$ (b) $\frac{D}{90}=\frac{G}{100}=\frac{R}{\pi}$ (c) $\frac{D}{100}=\frac{G}{100}=\frac{2 R}{\pi}$ (d) $\frac{D}{90}=\frac{G}{100}=\frac{R}{2 \pi}$ Solution: (c) $\frac{D}{100}=\frac{G}{100}=\frac{2 R}{\pi}$ It is the relation between degree, grade and radian....
Read More →If D, G and R denote respectively the number of degrees, grades and radians in an angle, then
Question: IfD,GandRdenote respectively the number of degrees, grades and radians in an angle, then (a) $\frac{D}{100}=\frac{G}{90}=\frac{2 R}{\pi}$ (b) $\frac{D}{90}=\frac{G}{100}=\frac{R}{\pi}$ (c) $\frac{D}{100}=\frac{G}{100}=\frac{2 R}{\pi}$ (d) $\frac{D}{90}=\frac{G}{100}=\frac{R}{2 \pi}$ Solution: (c) $\frac{D}{100}=\frac{G}{100}=\frac{2 R}{\pi}$ It is the relation between degree, grade and radian....
Read More →A man goes 15 meters due west and then 8 meters due north. How far is he from the starting point?
Question: A man goes 15 meters due west and then 8 meters due north. How far is he from the starting point? Solution: Let us draw the diagram. Let A be the starting point. From point B he goes to the north. Therefore, we obtained the following drawing. Now we have to find how far is he from the starting point that is we have to find $l(A C)$. Now we will use Pythagoras theorem to find the length of $A C$. $A C^{2}=A B^{2}+B C^{2}$....(1) Let us substituting the values of AB and BC in equation (1...
Read More →Find the degree measure of the angle subtended at the centre of a circle of radius
Question: Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm. Solution: Length of the arc = 22 cm Radius = 100 cm Now, $\theta=\frac{\text { Arc }}{\text { Radius }}$ $=\frac{22}{100}$ $=\frac{11}{50}$ radian $\therefore$ Angle subtended at the centre by the arc $=\left(\frac{11}{50} \times \frac{180}{\pi}\right)^{\circ}=\left(\frac{11}{5} \times \frac{18}{22} \times 7\right)^{\circ}=\left(\frac{63}{5}\right)^{\circ}=12^{\circ} 36^...
Read More →Write the truth value (T/F) of the following with suitable reasons:
Question: Write the truth value (T/F) of the following with suitable reasons: (i) A circle is a plane figure. (ii) Line segment joining the center to any point on the circle is a radius of the circle, (iii) If a circle is divided into three equal arcs each is a major arc. (iv) A circle has only finite number of equal chords. (v) A chord of a circle, which is twice as long as its radius is the diameter of the circle. (vi) Sector is the region between the chord and its corresponding arc. (vii) The...
Read More →If the arcs of the same length in two circles subtend angles 65° and 110° at the centre,
Question: If the arcs of the same length in two circles subtend angles 65 and 110 at the centre, find the ratio of their radii. Solution: Let the angles subtended at the centres by the arcs and radii of the first and second circles be $\theta_{1}$ and $r_{1}$ and $\theta_{2}$ and $r_{2} \quad \theta 1$ and $r 1$ and $\theta 2$ and $r 2$, respectively. Thus, we have: $\theta_{1}=65^{\circ}=\left(65 \times \frac{\pi}{180}\right) \operatorname{radian}$ $\theta_{2}=65^{\circ}=\left(110 \times \frac{...
Read More →Fill in the blanks:
Question: Fill in the blanks: (i) All points lying inside/outside a circle are called ______ points/_______ points. (ii) Circle having the same centre and different radii are called _____ circles. (iii) A point whose distance from the center of a circle is greater than its radius lies in _________ of the circle. (iv) A continuous piece of a circle is _______ of the circle. (v) The longest chord of a circle is a ____________ of the circle. (vi) An arc is a __________ when its ends are the ends of...
Read More →The sides of certain triangles are given below. Determine which of them are right triangles.
Question: The sides of certain triangles are given below. Determine which of them are right triangles. (i)a= 7 cm,b= 24 cm andc= 25 cm(ii)a= 9 cm,b= 16 cm andc= 18 cm(iii)a= 1.6 cm,b= 3.8 cm andc= 4 cm(iv)a= 8 cm,b= 10 cm andc= 6 cm Solution: (i) Let $a=7 \mathrm{~cm}$ $b=24 \mathrm{~cm}$ $c=25 \mathrm{~cm}$ In order to prove that the given sides of a certain triangle forms a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two...
Read More →Show that
Question: $\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x=\int_{0}^{\frac{\pi}{2}} 2 \sin x \cos x \tan ^{-1}(\sin x) d x$ Also, let $\sin x=t \Rightarrow \cos x d x=d t$ When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=1$ $\Rightarrow I=2 \int_{0}^{1} t \tan ^{-1}(t) d t$ ...(1) Consider $\int t \cdot \tan ^{-1} t d t=\tan ^{-1} t \cdot \int t d t-\int\left\{\frac{d}{d t}\left(\tan ^{-1} t\right) \int t d t\right\}...
Read More →Find the diameter of the sun in km supposing that it subtends an angle of 32' at the eye of an observer.
Question: Find the diameter of the sun in km supposing that it subtends an angle of 32' at the eye of an observer. Given that the distance of the sun is 91 106km. Solution: LetPQbe the diameter of the Sun andEbe the eye of the observer. Because the distance between the Sun and the Earth is quite large, we will takePQas arcPQ. Now, $r=91 \times 10^{6} \mathrm{~km}$ $\theta=32^{\prime}=\left(\frac{32}{60}\right)^{\circ}=\left(\frac{32}{60} \times \frac{\pi}{180}\right)$ radians $\theta=\frac{\text...
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Question: $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$ Solution: Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$ Also, let $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$ When $x=0, t=-1$ and when $x=\frac{\pi}{4}, t=0$ $\Rightarrow(\sin x-\cos x)^{2}=t^{2}$ $\Rightarrow \sin ^{2} x+\cos ^{2} x-2 \sin x \cos x=t^{2}$ $\Rightarrow 1-\sin 2 x=t^{2}$ $\Rightarrow \sin 2 x=1-t^{2}$ $\therefore I=\int_{-1}^{0} \frac{d t}{9+16\left(1-t^{2}\right)}$ $...
Read More →Find the distance from the eye at which a coin of 2 cm
Question: Find the distance from the eye at which a coin of 2 cm diameter should be held so as to conceal the full moon whose angular diameter is 31'. Solution: LetPQbe the diameter of the coin andEbe the eye of the observer. Also, let the coin be kept at a distancerfrom the eye of the observer to hide the moon completely. Now, $\theta=31^{\prime}=\left(\frac{31}{60}\right)^{\circ}=\left(\frac{31}{60} \times \frac{\pi}{180}\right)$ radians $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Righta...
Read More →In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively.
Question: In figure, ABC is a right angled triangle at A, BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AXDE meets BC at Y. Show that (i)ΔMBC ΔABD (ii)ar(BYXD) = 2ar(ΔMBC) (iii)ar(BYXD) = ar(ABMN) (iv)ΔFCB ΔACE (v) ar(CYXE) = 2ar(ΔFCB) (vi)ar(CYXE) = ar(ACFG) (vii) ar(BCED) = ar(ABMN) + ar(ACFG) Solution: (i) InΔMBCandΔABD, we have MB = AB BC = BD AndMBC =ABD [since, MBC andABC are obtained by addingABC to a right angle.] So, by SAS congruence criterion, w...
Read More →A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km/hr.
Question: A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km/hr. Through what angle has it turned in 10 seconds? Solution: Time = 10 seconds Speed $=66 \mathrm{~km} / \mathrm{h}=\frac{66 \times 1000}{3600} \mathrm{~m} / \mathrm{s}$ We know, Speed $=\frac{\text { Distance }}{\text { Time }}$ $\Rightarrow \frac{66 \times 1000}{3600}=\frac{\text { Distance }}{\text { Time }}$ $\Rightarrow$ Distance $=\frac{66 \times 1000}{3600} \times 10=\frac{1100}{6} \mat...
Read More →A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km/hr.
Question: A railway train is travelling on a circular curve of 1500 metres radius at the rate of 66 km/hr. Through what angle has it turned in 10 seconds? Solution: Time = 10 seconds Speed $=66 \mathrm{~km} / \mathrm{h}=\frac{66 \times 1000}{3600} \mathrm{~m} / \mathrm{s}$ We know, Speed $=\frac{\text { Distance }}{\text { Time }}$ $\Rightarrow \frac{66 \times 1000}{3600}=\frac{\text { Distance }}{\text { Time }}$ $\Rightarrow$ Distance $=\frac{66 \times 1000}{3600} \times 10=\frac{1100}{6} \mat...
Read More →If the sides of a triangle are 3 cm,
Question: If the sides of a triangle are 3 cm, 4 cm and 6 cm long, determine whether the triangle is a right-angled triangle. Solution: We have, $a=3 \mathrm{~cm}$ $b=4 \mathrm{~cm}$ $c=6 \mathrm{~cm}$ In order to prove that the triangle is a right angled triangle we have to prove that square of the larger side is equal to the sum of the squares of the other two sides. Here, the larger side is $c=6 \mathrm{~cm}$. Hence, we have to prove that $a^{2}+b^{2}=c^{2}$. Let solve the left hand side of t...
Read More →Show that
Question: $\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$ Solution: Let $I=\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$ $I=\int_{0}^{1} \frac{1}{(\sqrt{1+x}-\sqrt{x})} \times \frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})} d x$ $=\int_{0}^{1} \frac{\sqrt{1+x}+\sqrt{x}}{1+x-x} d x$ $=\int_{0}^{1} \sqrt{1+x} d x+\int_{0}^{1} \sqrt{x} d x$ $=\left[\frac{2}{3}(1+x)^{\frac{3}{2}}\right]_{0}^{1}+\left[\frac{2}{3}(x)^{\frac{3}{2}}\right]_{0}^{1}$ $=\frac{2}{3}\left[(2)^{\frac{3}{2}}-1\right]+\frac{...
Read More →In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔABD) = ar(ΔADE) = ar(ΔAEC).
Question: In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ΔABD) = ar(ΔADE) = ar(ΔAEC). Solution: Draw a line l through A parallel to BC. Given that, BD = DE = EC We observed that the triangles ABD and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal. Hence, ar(ΔABD) = ar(ΔADE) = ar(ΔAEC)....
Read More →The radius of a circle is 30 cm.
Question: The radius of a circle is 30 cm. Find the length of an arc of this circle, if the length of the chord of the arc is 30 cm. Solution: LetABbe the chord andObe the centre of the circle. Here, AO = BO = AB= 30 cm Therefore, $\Delta A O B$ is an equilateral triangle. Now, Radius = 30 cm $\theta=60^{\circ}=\left(60 \times \frac{\pi}{180}\right)=\frac{\pi}{3} \operatorname{radian}$ $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{\pi}{3}=\frac{\operatorname{Arc}}{30}$ $\Rig...
Read More →The radius of a circle is 30 cm.
Question: The radius of a circle is 30 cm. Find the length of an arc of this circle, if the length of the chord of the arc is 30 cm. Solution: LetABbe the chord andObe the centre of the circle. Here, AO = BO = AB= 30 cm Therefore, $\Delta A O B$ is an equilateral triangle. Now, Radius = 30 cm $\theta=60^{\circ}=\left(60 \times \frac{\pi}{180}\right)=\frac{\pi}{3} \operatorname{radian}$ $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{\pi}{3}=\frac{\operatorname{Arc}}{30}$ $\Rig...
Read More →In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that:
Question: In a triangle ABC, if L and M are points on AB and AC respectively such that LM ∥ BC. Prove that: (i)ar(ΔLCM) = ar(ΔLBM) (ii)ar(ΔLBC) = ar(ΔMBC) (iii)ar(ΔABM) = ar(ΔACL) (iv) ar(ΔLOB) = ar(ΔMOC) Solution: (i) Clearly triangles LMB and LMC are on the same base LM and between the same parallels LM and BC. ar(ΔLMB) = ar(ΔLMC) ... (1) (ii) We observe that triangles LBC and MBC are on the same base BC and between same parallels LM and BC. ar(ΔLBC) = ar(ΔMBC) ... (2) (iii) We have, ar(ΔLMB) ...
Read More →Show that
Question: $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$ Solution: Let $I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$ $\Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{-(-\sin 2 x)}} d x$ $\Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{-(-1+1-2 \sin x \cos x)}} d x$ $\Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{(\sin x+\cos x)}{\sqrt{1-\left(\sin ^{2} x+...
Read More →In figure, ABCD is a parallelogram. O is any point on AC. PQ ∥ AB and LM ∥ AD.
Question: In figure, ABCDis a parallelogram. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that:ar(∥gmDLOP) = ar(∥gmBMOQ). Solution: Since a diagonal of a parallelogram divides it into two triangles of equal area Therefore,ar(ΔADC) = ar(ΔABC) ⇒ ar(ΔAPO) + ar(∥gmDLOP) + ar(ΔOLC) ⇒ ar(ΔAOM) + ar(∥gmBMOQ) + ar(ΔOQC) ... (1) Since AO and Oc are diagonals of parallelograms AMOP and OQCL respectively. ar(ΔAPO) = ar(ΔAMO) .... (2) Andar(ΔOLC) = ar(ΔOQC) .... (3) Subtracting 2 and 3 from 1, we get ar...
Read More →Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
Question: Find the angle in radians through which a pendulum swings if its length is 75 cm and the tip describes an arc of length (i) 10 cm (ii) 15 cm (iii) 21 cm. Solution: We know: Radius = 75 cm(i) Length of the arc = 10 cm Now, $\theta=\frac{\text { Radius }}{\text { Rade }}$ $=\frac{10}{75}$ $=\frac{2}{15} \operatorname{radian}$ (ii) Length of the arc = 15 cm Now, $\theta=\frac{\text { Arc }}{\text { Radius }}$ $=\frac{15}{75}$ $=\frac{1}{5}$ radian (iii) Length of the arc = 21 cm Now, $\th...
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