The angles of a triangle are in A.P.
Question: The angles of a triangle are inA.P.and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. 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: $\frac{\text { Number of degrees in the least angle }}{\text { Number of degrees in the mean angle }}=\frac{1}{120}$ or, $\frac{a-d}{a}=\frac{1}{120}$ or, $\f...
Read More →The angles of a triangle are in A.P.
Question: The angles of a triangle are inA.P.and the number of degrees in the least angle is to the number of degrees in the mean angle as 1 : 120. 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: $\frac{\text { Number of degrees in the least angle }}{\text { Number of degrees in the mean angle }}=\frac{1}{120}$ or, $\frac{a-d}{a}=\frac{1}{120}$ or, $\f...
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Question: $\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}$ Solution: $\begin{aligned} I =\int\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) e^{x} \\ =\int\left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right) e^{x} \\ =\int\left(\frac{1+\sin x \cos x}{\cos ^{2} x}\right) e^{x} \\ =\int\left(\sec ^{2} x+\tan x\right) e^{x} \end{aligned}$ Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x$ $\therefore I=\int\left(f(x)+f^{\prime}(x)\right] e^{x} d x$ $=e^{x} f(x)+\mathrm{C}$ $=e^{x} \tan x+\mathrm{C}$...
Read More →A takes 3 hours more than B to walk a distance of 30 km.
Question: A takes 3 hours more than $B$ to walk a distance of $30 \mathrm{~km}$. But, if $A$ doubles his pace (speed) he is ahead of $B$ by $1 \frac{1}{2}$ hours. Find the speeds of $A$ and $B$. Solution: Let the speed of A and B be x Km/hr and y Km/hr respectively. Then, Time taken by A to cover $30 \mathrm{~km}=\frac{30}{x} h r s$, And, Time taken by B to cover $30 \mathrm{~km}=\frac{30}{y} h r s$. By the given conditions, we have $\frac{30}{x}-\frac{30}{y}=3$ $\frac{10}{x}-\frac{10}{y}=1 \cdo...
Read More →In figure, CD ∥ AE and CY ∥ BA.
Question: In figure, CD ∥ AE and CY ∥ BA. (i) Name a triangle equal in area ofΔCBX (ii) Prove thatar(ΔZDE) = ar(ΔCZA) (iii) Prove that ar(BCZY) = ar(ΔEDZ) Solution: Since, triangle BCA and triangle BYA are on the same base BA and between same parallel s BA and CY. Thenar(ΔBCA) = ar(ΔBYA) ⇒ ar(ΔCBX) + ar(ΔBXA) = ar(ΔBXA) + ar(ΔAXY) ⇒ ar(ΔCBX) = ar(ΔAXY) (1) Since, trianglesACE and ADE are on the same base AE and between same parallels CD and AE Then,ar(ΔACE) = ar(ΔADE) ar(ΔCZA) + ar(ΔAZE) = ar(ΔA...
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Question: $\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$ Solution: $I=\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$ Let $x=\cos ^{2} \theta \Rightarrow d x=-2 \sin \theta \cos \theta d \theta$ $\begin{aligned} I =\int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}(-2 \sin \theta \cos \theta) d \theta \\ =-\int \sqrt{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}} \sin 2 \theta d \theta} \end{aligned}$ $=-\int \tan \frac{\theta}{2} \cdot 2 \sin \theta \cos \theta d \theta$ $=-2 \int \frac{\sin ...
Read More →ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point
Question: ABCD is a parallelogram. G is a point on AB such that AG = 2GB and E is point on DC such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that: (i) ar(ADEG) = ar(GBCE) (ii)ar(ΔEGB) = (1/6) ar(ABCD) (iii)ar(ΔEFC) =(1/2) ar(ΔEBF) (iv)ar(ΔEGB) = 3/2 ar(ΔEFC) (v) Find what portion of the area of parallelogram is the area of ΔEFG. Solution: Given: ABCD is a parallelogram AG = 2 GB, CE = 2 DE and BF = 2 FC To prove: (i)ar(ADEG) = ar(GBCE). (ii)ar(ΔEGB) =(1/6) ar(ABCD). (iii) ...
Read More →The angle of a quadrilateral are in A.P. and the greatest angle is 120°.
Question: The angle of a quadrilateral are in A.P. and the greatest angle is 120. Express the angles in radians. Solution: Let the angles of the quadrilateral be $(a-3 d)^{\circ},(a-d)^{\circ},(a+d)^{\circ}$ and $(a+3 d)^{\circ}$. We know: $a-3 d+a-d+a+d+a-2 d=360$ $\Rightarrow 4 a=360$ $\Rightarrow a=90$ We have: Greatest angle $=120^{\circ}$ Now, $a+3 d=120$ $\Rightarrow 90+3 d=120$ $\Rightarrow 3 d=30$ $\Rightarrow d=10$ Hence, $(a-3 d)^{\circ},(a-d)^{\circ},(a+d)^{\circ}$ and $(a+3 d)^{\circ...
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Question: $\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]$ Solution: Let $I=\int \frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} x+\cos ^{-1} x} d x$ It is known that, $\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}=\frac{\pi}{2}$ $\Rightarrow I=\int \frac{\left(\frac{\pi}{2}-\cos ^{-1} \sqrt{x}\right)-\cos ^{-1} \sqrt{x}}{\frac{\pi}{2}} d x$ $=\frac{2}{\pi} \int\left(\frac{\pi}{2}-2 \cos ^{-1} \sqrt{x}\right) d x$ $=\frac{2}{\pi} \cdot \fr...
Read More →Find the magnitude, in radians and degrees, of the interior angle of a regular
Question: Find the magnitude, in radians and degrees, of the interior angle of a regular (i) pentagon (ii) octagon (iii) heptagon (iv) duodecagon. Solution: (i) Sum of the interior angles of the polygon $=(n-2) \pi$ Number of sides in the pentagon $=5$ $\therefore$ Sum of the interior angles of the pentagon $=(5-2) \pi=3 \pi$ Each angle of the pentagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{3 \pi}{5} \mathrm{rad}$ Each angle of the pentagon ...
Read More →In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that:
Question: In a triangle ABC, P and Q are respectively the mid points of AB and BC and R is the midpoint of AP. Prove that: (i) ar(ΔPBQ) = ar(ΔARC). (ii) ar(ΔPRQ) =1/2 ar(ΔARC). (iii) ar(ΔRQC) = 3/8 ar(ΔABC). Solution: We know that each median of a triangle divides it into two triangles of equal area. (i) Since CR is the median ofΔCAP ar(ΔCRA) = (1/2) ar(ΔCAP) (1) Also, CP is the median of aΔCAB ar(ΔCAP) = ar(ΔCPB) (2) From 1 and 2, we get ar(ΔARC) = (1/2) ar(ΔCPB) (3) PQ is the median of aΔPBC a...
Read More →Find the magnitude, in radians and degrees, of the interior angle of a regular
Question: Find the magnitude, in radians and degrees, of the interior angle of a regular (i) pentagon (ii) octagon (iii) heptagon (iv) duodecagon. Solution: (i) Sum of the interior angles of the polygon $=(n-2) \pi$ Number of sides in the pentagon $=5$ $\therefore$ Sum of the interior angles of the pentagon $=(5-2) \pi=3 \pi$ Each angle of the pentagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{3 \pi}{5} \mathrm{rad}$ Each angle of the pentagon ...
Read More →ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC.
Question: ABCD is a parallelogram. E is a point on BA such that BE = 2EA and F is point on DC such that DF = 2FC. Prove that AECF is a parallelogram whose area is one third of the area of parallelogram ABCD. Solution: Draw FGAB We have, BE = 2 EA and DF = 2FC ⇒AB - AE = 2 AE and DC - FC = 2 FC ⇒AB = 3 AE and DC = 3 FC ⇒ AE = (1/3) AB and FC = (1/3)DC (1) But AB = DC Then, AE = FC [opposite sides of ∥gm] Thus, AE = FC and AE ∥ FC Then, AECF is a parallelogram Now, area of parallelogram AECF =AE F...
Read More →Ritu can row downstream 20 km in 2 hours,
Question: Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current. Solution: Let the speed of rowing in still water bexkm/hr and the speed of the current beykm/hr Speed upstream $=(x-y) k n / h r$ Speed downstream $=(x+y) \mathrm{km} / \mathrm{hr}$ Now, Time taken to cover $20 \mathrm{~km}$ down stream $=\frac{20}{x+y} h r s$ Time taken to cover $4 \mathrm{~km}$ upstream $=\frac{4}{x-y} h r s$ But, time taken to...
Read More →ABCD is a parallelogram whose diagonals AC and BD intersect at O.
Question: ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O intersects AB at P and DC at Q. Prove that ar(ΔPOA) = ar(ΔQOC). Solution: In triangles POA and QOC, we have AOP = COQ AO = OC PAC = QCA So, by ASA congruence criterion, we have ΔPOA ΔQOC ⇒ ar(ΔPOA) = ar(ΔQOC)....
Read More →One angle of a triangle
Question: One angle of a triangle $\frac{2}{3} x$ grades and another is $\frac{3}{2} x$ degrees while the third is $\frac{\pi x}{75}$ radians. Express all the angles in degrees. Solution: One angle of the triangle $=\frac{2}{3} x \mathrm{grad}$ $=\left(\frac{2}{3} x \times \frac{9}{10}\right)^{\circ} \quad\left[\because 1 \operatorname{grad}=\left(\frac{9}{10}\right)^{\circ}\right]$ $=\left(\frac{3}{5} x\right)^{\circ}$ Another angle $=\left(\frac{3}{2} x\right)^{\circ}$ $\because 1 \operatornam...
Read More →ABCD is a parallelogram in which BC is produced to E such that CE = BC.
Question: ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F. (i) Prove thatar(ΔADF) = ar(ΔECF). (ii) If the area ofΔDFB=3 cm2, find the area of∥gm ABCD. Solution: In triangles ADF and ECF, we have ADF = ECF [Alternate interior angles, Since AD ∥ BE] AD = EC [since AD = BC = CE] AndDFA = CFA [Vertically opposite angles] So, by AAS congruence criterion, we have ΔADF ΔECF ⇒ar(ΔADF) = ar(ΔECF)and DF = CF . Now, DF = CF ⇒BF is a median inΔBCD. ⇒ ar(ΔBCD) = ...
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Question: $\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}$ Solution: $\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}=\frac{1}{\sqrt{\sin ^{3} x(\sin x \cos \alpha+\cos x \sin \alpha)}}$ $=\frac{1}{\sqrt{\sin ^{4} x \cos \alpha+\sin ^{3} x \cos x \sin \alpha}}$ $=\frac{1}{\sin ^{2} x \sqrt{\cos \alpha+\cot x \sin \alpha}}$ $=\frac{\operatorname{cosec}^{2} x}{\sqrt{\cos \alpha+\cot x \sin \alpha}}$ Let $\cos \alpha+\cot x \sin \alpha=t \Rightarrow-\operatorname{cosec}^{2} x \sin \alpha d x=d t$ $\ther...
Read More →One angle of a triangle
Question: One angle of a triangle $\frac{2}{3} x$ grades and another is $\frac{3}{2} x$ degrees while the third is $\frac{\pi x}{75}$ radians. Express all the angles in degrees. Solution: One angle of the triangle $=\frac{2}{3} x \mathrm{grad}$ $=\left(\frac{2}{3} x \times \frac{9}{10}\right)^{\circ} \quad\left[\because 1 \operatorname{grad}=\left(\frac{9}{10}\right)^{\circ}\right]$ $=\left(\frac{3}{5} x\right)^{\circ}$ Another angle $=\left(\frac{3}{2} x\right)^{\circ}$ $\because 1 \operatornam...
Read More →Roohi travels 300 km to her home partly by train and partly by bus.
Question: Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately. Solution: Let the speed of the train be x km/hour that of the bus be y km/hr, we have the following cases Case I: When Roohi travels 300 Km by train and the rest by bus Time taken by Roohi to travel $60 \mathrm...
Read More →ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that:
Question: ABCD is a parallelogram whose diagonals intersect at O .If P is any point on BO, prove that: (i) ar(ΔADO) = ar(ΔCDO). (ii) ar(ΔABP) = 2ar(ΔCBP). Solution: Given that ABCD is the parallelogram To Prove: (i)ar(ΔADO) = ar(ΔCDO). (ii)ar(ΔABP) = 2ar(ΔCBP). Proof: we know that diagonals of parallelogram bisect each other AO = OC and BO = OD (i) InΔDAC, since DO is a median. Thenar(ΔADO) = ar(ΔCDO). (ii) InΔBAC, since BO is a median. Thenar(ΔBAO) = ar(ΔBCO) (1) InΔPAC, since PO is a median. T...
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Question: $f^{\prime}(a x+b)[f(a x+b)]^{n}$ Solution: $f^{\prime}(a x+b)[f(a x+b)]^{n}$ Let $f(a x+b)=t \Rightarrow a f^{\prime}(a x+b) d x=d t$ $\Rightarrow \int f^{\prime}(a x+b)[f(a x+b)]^{n} d x=\frac{1}{a} \int t^{n} d t$ $=\frac{1}{a}\left[\frac{t^{n+1}}{n+1}\right]$ $=\frac{1}{a(n+1)}(f(a x+b))^{n+1}+\mathrm{C}$...
Read More →The difference between the two acute angles of a right-angled triangle is
Question: The difference between the two acute angles of a right-angled triangle is $\frac{2 \pi}{5}$ radians. Express the angles in degrees. Solution: Given: Difference between two acute angles of a right-angled triangle $=\frac{2 \pi}{5} \mathrm{rad}$ $\because 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ $\therefore \frac{2 \pi}{5} \mathrm{rad}=\left(\frac{180}{\pi} \times \frac{2 \pi}{5}\right)^{\circ}$ $=(36 \times 2)^{\circ}$ $=72^{\circ}$ Now, let one acute angle of the triangle b...
Read More →A point D is taken on the side BC of a ΔABC, such that BD = 2DC.
Question: A point D is taken on the side BC of aΔABC, such that BD = 2DC. Prove thatar(ΔABD) = 2ar(ΔADC). Solution: Given that, InΔABC, BD = 2DC To prove:ar(ΔABD) = 2ar(ΔADC). Construction: Take a point E on BD such that BE = ED Proof: Since, BE = ED and BD = 2 DC Then, BE = ED = DC We know that median of triangle divides it into two equal triangles. InΔABD, AE is the median. Then,ar(ΔABD) = 2ar(ΔAED) (1) InΔAEC, AD is the median. Then,ar(ΔAED) = 2ar(ΔADC) (2) Compare equation 1 and 2 ar(ΔABD) =...
Read More →The difference between the two acute angles of a right-angled triangle is
Question: The difference between the two acute angles of a right-angled triangle is $\frac{2 \pi}{5}$ radians. Express the angles in degrees. Solution: Given: Difference between two acute angles of a right-angled triangle $=\frac{2 \pi}{5} \mathrm{rad}$ $\because 1 \mathrm{rad}=\left(\frac{180}{\pi}\right)^{\circ}$ $\therefore \frac{2 \pi}{5} \mathrm{rad}=\left(\frac{180}{\pi} \times \frac{2 \pi}{5}\right)^{\circ}$ $=(36 \times 2)^{\circ}$ $=72^{\circ}$ Now, let one acute angle of the triangle b...
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