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...
Read More →Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time.
Question: Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars. Solution: Let x and y be two cars starting from points A and B respectively. Let the speed of the car X be x km/hr and that of the car Y be y km/hr. Case I: When two cars move in the same directions: Suppose two ...
Read More →In figure, ABCD and AEFD are two parallelograms. Prove that
Question: In figure, ABCD and AEFD are two parallelograms. Prove that (i) PE = FQ (ii)ar(ΔAPE) : ar(ΔPFA) = ar(ΔQFD): ar(ΔPFD) (iii) ar(ΔPEA) = ar(ΔQFD) Solution: Given that, ABCD and AEFD are two parallelograms (i) In triangles, EPA and FQD PEA = QFD [corresponding angles] EPA=FQD[corresponding angles] PA = QD [opposite sides of parallelogram] Then,ΔEPA ΔFQD[By AAS condition] Therefore, EP = FQ [C.P.C.T] (ii) Since triangles, PEA and QFD stand on equal bases PE and FQ lies between the same para...
Read More →In figure, X and Y are the mid points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines.
Question: In figure, X and Y are the mid points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(ΔABP) = ar(ΔACQ). Solution: Since X and Y are the mid points of AC and AB respectively. Therefore, XY ∥ BC Clearly, triangles BYC and BXC are on the same base BC and between the same parallels XY and BC ar(ΔBYC) = ar(ΔBXC) ⇒ ar(ΔBYC) ar(ΔBOC) = ar(ΔBXC) ar(ΔBOC) ⇒ ar(ΔBOY) = ar(ΔCOX) ⇒ ar(ΔBOY) + ar(ΔXOY) = ar(ΔCOX) + ar(ΔXOY) ⇒ ar(ΔBXY) = ar(ΔCXY).... (2) We obser...
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Question: $\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}$ Solution: Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4 \sin ^{2} x} d x$ $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4\left(1-\cos ^{2} x\right)} d x$ $\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x}{\cos ^{2} x+4-4 \cos ^{2} x} d x$ $\Rightarrow I=\frac{-1}{3} \int_{0}^{\frac{\pi}{2}} \frac{4-3 \cos ^{2} x-4}{4-3 \cos ^{2} x} d x$ $\Rightarrow I=\frac{-...
Read More →A wheel makes 360 revolutions per minute.
Question: A wheel makes 360 revolutions per minute. Through how many radians does it turn in 1 second? Solution: Number of revolutions taken by the wheel in 1 minute $=360$ Number of revolutions taken by the wheel in 1 second $=\frac{360}{60}=6$ We know, 1 revolution $=2 \pi$ radians $\therefore$ Number of radians the wheel will turn in 1 second $=6 \times 2 \pi=12 \pi$...
Read More →A wheel makes 360 revolutions per minute.
Question: A wheel makes 360 revolutions per minute. Through how many radians does it turn in 1 second? Solution: Number of revolutions taken by the wheel in 1 minute $=360$ Number of revolutions taken by the wheel in 1 second $=\frac{360}{60}=6$ We know, 1 revolution $=2 \pi$ radians $\therefore$ Number of radians the wheel will turn in 1 second $=6 \times 2 \pi=12 \pi$...
Read More →A train covered a certain distance at a uniform speed.
Question: A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr. faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train. Solution: Let the actual speed of the train be Xkm\h mathrm{~km} / \mathrm{hr}$ and the actual time taken by y hours. Then, Dis $\tan c e=$ Speed $\times$ Time Distance covered $=(x y) \mat...
Read More →D is the midpoint of side BC of ΔABC and E is the midpoint of BD.
Question: D is the midpoint of side BC of ΔABC and E is the midpoint of BD. If O is the midpoint of AE, Prove that ar(ΔBOE) = (1/8) ar(ΔABC). Solution: Given that D is the midpoint of sides BC of triangle ABC E is the midpoint of BD and O is the midpoint of AE Since AD and AE are the medians of triangles, ABC and ABD respectively ar(ΔABD) = (1/2) ar(ΔABC) (1) ar(ΔABE) = (1/2) ar(ΔABD) (2) OB is the median of triangle ABE Therefore, ar(ΔBOE) = (1/2) ar(ΔABE) From 1, 2 and 3, we have ar(ΔBOE) = (1...
Read More →Find the length which at a distance of 5280 m will subtend an angle of 1' at the eye.
Question: Find the length which at a distance of 5280 m will subtend an angle of 1' at the eye. Solution: We have: Radius = 5280 m Now, $\theta=1^{\prime}=\left(\frac{1}{60}\right)^{\circ}=\left(\frac{1}{60} \times \frac{\pi}{180}\right)$ radian We know, $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{1}{60} \times \frac{\pi}{180}=\frac{\text { Arc }}{5280}$ $\Rightarrow$ Arc $=\frac{5280 \times 22}{60 \times 180 \times 7}=1.5365 \mathrm{~m}$...
Read More →A rail road curve is to be laid out on a circle.
Question: A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25 in a distance of 40 metres? Solution: Length of the arc = 40 m $\theta=25^{\circ}=\left(25 \times \frac{\pi}{180}\right)=\frac{5 \pi}{36}$ radian We know, $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{5 \pi}{36}=\frac{40}{\text { Radius }}$ $\Rightarrow$ Radius $=\frac{40}{\frac{5 \pi}{36}}$ $=\frac{40 \times 36 \times 7}{5 \times 22}$ $=91.64 \math...
Read More →A rail road curve is to be laid out on a circle.
Question: A rail road curve is to be laid out on a circle. What radius should be used if the track is to change direction by 25 in a distance of 40 metres? Solution: Length of the arc = 40 m $\theta=25^{\circ}=\left(25 \times \frac{\pi}{180}\right)=\frac{5 \pi}{36}$ radian We know, $\theta=\frac{\text { Arc }}{\text { Radius }}$ $\Rightarrow \frac{5 \pi}{36}=\frac{40}{\text { Radius }}$ $\Rightarrow$ Radius $=\frac{40}{\frac{5 \pi}{36}}$ $=\frac{40 \times 36 \times 7}{5 \times 22}$ $=91.64 \math...
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Question: $\int_{0}^{\frac{x}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$ Solution: Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$ $\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \frac{\frac{(\sin x \cos x)}{\cos ^{4} x}}{\frac{\left(\cos ^{4} x+\sin ^{4} x\right)}{\cos ^{4} x}} d x$ $\Rightarrow I=\int_{0}^{\frac{\pi}{4}} \frac{\tan x \sec ^{2} x}{1+\tan ^{4} x} d x$ Let $\tan ^{2} x=t \Rightarrow 2 \tan x \sec ^{2} x d x=d t$ When $x=0, t=0$ and when $x=\...
Read More →The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9°.
Question: The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9. Find the number of sides of the polygons. Solution: Let the number of sides in the first polygon be 5xand the number of sides in the second polygon be 4x. We know: Angle of an $n$-sided regular polygon $=\left(\frac{n-2}{n}\right) 180^{\circ}$ Thus, we have: Angle of the first polygon $=\left(\frac{5 x-2}{5 x}\right) 180^{\circ}$ Angle of the second polygon $=\left(\frac{4 x-2}{4 x}\r...
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Question: $\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$ Solution: $I=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$ $=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin ^{2} \frac{x}{2}}\right) d x$ $=\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{\operatorname{cosec}^{2} \frac{x}{2}}{2}-\cot \frac{x}{2}\right) d x$ Let $f(x)=-\cot \frac{x}{2}$ $\Rightarrow f^{\prime}(x)=-\left(-\frac{1}{2} \operatorn...
Read More →The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9°.
Question: The number of sides of two regular polygons are as 5 : 4 and the difference between their angles is 9. Find the number of sides of the polygons. Solution: Let the number of sides in the first polygon be 5xand the number of sides in the second polygon be 4x. We know: Angle of an $n$-sided regular polygon $=\left(\frac{n-2}{n}\right) 180^{\circ}$ Thus, we have: Angle of the first polygon $=\left(\frac{5 x-2}{5 x}\right) 180^{\circ}$ Angle of the second polygon $=\left(\frac{4 x-2}{4 x}\r...
Read More →Abdul travelled 300 km by train and 200 km by taxi,
Question: Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi he takes 6 minutes longer. Find the speed of the train and that of the taxi. Solution: Let the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases Case I: When Abdul travels 300 Km by train and the 200 Km by taxi Time taken by Abdul to travel $300 \mathrm{Km}$ by train $=\frac{300}{x} h r s$ Time taken by Abdul to ...
Read More →In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC.
Question: In figure ABC and BDE are two equilateral triangles such that D is the midpoint of BC. AE intersects BC in F. Prove that: (i) ar(ΔBDE) = (1/4) ar(ΔABC) (ii)ar(ΔBDE) =(1/2) ar(ΔBAE) (iii)ar(ΔBFE) =ar(ΔAFD) (iv)ar(ΔABC) = 2ar(ΔBEC) (v) ar(ΔFED) =1/8 ar(ΔAFC) (vi) ar(ΔBFE) = 2 ar(ΔEFD) Solution: Given that ABC and BDE are two equilateral triangles. Let AB = BC = CA = x. Then, BD =x/2= DE = BE (i) We have, $\operatorname{ar}(\Delta \mathrm{ABC})=\frac{\sqrt{3}}{4} \mathrm{x}^{2}$ and $\ope...
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Question: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$ Solution: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}=\frac{\sqrt{x^{2}+1}}{x^{4}}\left[\log \left(x^{2}+1\right)-\log x^{2}\right]$ $=\frac{\sqrt{x^{2}+1}}{x^{4}}\left[\log \left(\frac{x^{2}+1}{x^{2}}\right)\right]$ $=\frac{\sqrt{x^{2}+1}}{x^{4}} \log \left(1+\frac{1}{x^{2}}\right)$ $=\frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right)$ $=\frac{1}...
Read More →The angles of a triangle are in A.P. such that the greatest is 5 times the least.
Question: The angles of a triangle are in A.P. such that the greatest is 5 times the least. Find the angles in radians. Solution: Let the angles of the triangle be $(a-d)^{\circ},(a)^{\circ}$ and $(a+d)^{\circ}$. We know, $a-d+a+a+d=180$ $\Rightarrow 3 a=180$ $\Rightarrow a=60$ Given: Greatest angle $=5 \times$ Least angle or, $\frac{\text { Greatest angle }}{\text { Least angle }}=5$ or, $\frac{a+d}{a-d}=5$ or, $\frac{60+d}{60-d}=5$ or, $60+d=300-5 d$ or, $6 d=240$ or, $d=40$ Hence, the angles ...
Read More →In figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm .If X and Y are, respectively,
Question: In figure, ABCDis a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm .If X and Y are, respectively, the mid points of AD and BC, prove that: (i) XY = 50 cm (ii) DCYX is a trapezium (iii) ar(trap. DCYX) = (9/11) ar(XYBA). Solution: (i) Join DY and produce it to meet AB produced at P. In triangles BYP and CYD we have, BYP =CYD [Vertically opposite angles] DCY =PBY [Since, DC ∥ AP] And BY = CY So, by ASA congruence criterion, we have (ΔBYP) (ΔCYD) ⇒DY = Yp and DC = BP ⇒Y is the m...
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Question: $\tan ^{-1} \sqrt{\frac{1-x}{1+x}}$ Solution: $I=\tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$ Let $x=\cos \theta \Rightarrow d x=-\sin \theta d \theta$ $I=\int \tan ^{-1} \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-\sin \theta d \theta)$ $=-\int \tan ^{-1} \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}} \sin \theta d \theta$ $=-\int \tan ^{-1} \tan \frac{\theta}{2} \cdot \sin \theta d \theta$ $=-\frac{1}{2} \int \theta \cdot \sin \theta d \theta$ $=-\frac{1}{2}\left[...
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Question: $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$ Solution: Let $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^{2}}+\frac{C}{(x+2)}$ ...(1) $\Rightarrow x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C\left(x^{2}+2 x+1\right)$ $\Rightarrow x^{2}+x+1=A\left(x^{2}+3 x+2\right)+B(x+2)+C\left(x^{2}+2 x+1\right)$ $\Rightarrow x^{2}+x+1=(A+C) x^{2}+(3 A+B+2 C) x+(2 A+2 B+C)$ Equating the coefficients of $x^{2}, x$, and constant term, we obtain $A+C=1$ $3 A+B+2 C=1$ $2 A+2 B+C=1$ On solving these equations,...
Read More →The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second.
Question: The angle in one regular polygon is to that in another as 3 : 2 and the number of sides in first is twice that in the second. Determine the number of sides of two polygons. Solution: Let the number of sides in the first polygon be 2xand the number of sides in the second polygon isx. We know: Angle of an $n$-sided regular polygon $=\left(\frac{n-2}{n}\right) \pi$ radian $\therefore$ Angle of the first polygon $=\left(\frac{2 x-2}{2 x}\right) \pi=\left(\frac{x-1}{x}\right) \pi$ radian An...
Read More →In figure, PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥ CR. Prove that ar (ΔPQE) = ar(ΔCFD).
Question: In figure, PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥ CR. Prove that ar (ΔPQE) = ar(ΔCFD). Solution: Given that PSDA is a parallelogram Since, AP ∥ BQ ∥ CR ∥ DS and AD ∥ PS Therefore, PQ = CD (equ. 1) In triangle BED, C is the midpoint of BD and CF ∥ BE Therefore, F is the midpoint of ED ⇒EF = PE Similarly, EF = PE Therefore, PE = FD (equ.2) In triangles PQE and CFD, we have PE = FD Therefore,EPQ =FDC [Alternate angles] So, by SAS criterion, we have ΔPQE ΔDCF ⇒ ar(ΔPQE...
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