What special name can be given to a quadrilateral PQRS if
Question: What special name can be given to a quadrilateralPQRSif P+ S= 180? Solution: Since, in quadrilateralPQRS,P+ S= 180.i.e. the sum of adjacent angles is 180.So,PQRSis a parallelogram....
Read More →If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake be b,
Question: If the angle of elevation of a cloud from a point h metres above a lake is a and the angle of depression of its reflection in the lake be b, prove that the distance of the cloud from the point of observation is $\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}$ Solution: Let $C^{\prime}$ be the image of cloud $C$. We have $\angle C A B=\alpha$ and $\angle B A C^{\prime}=\beta$. Again let.andbe the distance of cloud from point of observation. We have to prove that $A C=\frac{2 h \sec \alp...
Read More →If the sum of n terms of an A.P.
Question: If the sum ofnterms of an A.P. be 3n2nand its common difference is 6, then its first term is (a) 2 (b) 3 (c) 1 (d) 4 Solution: (a) 2 $S_{n}=3 n^{2}-n$ $\Rightarrow S_{1}=3(1)^{2}-1$ $\Rightarrow S_{1}=2$ $\therefore a_{1}=2$...
Read More →Diagonals of a parallelogram are perpendicular to each other.
Question: Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer. Solution: No, the statement is incorrect because the diagonals of a parallelogram bisect each other....
Read More →In a quadrilateral PQRS, opposite angles are equal.
Question: In a quadrilateralPQRS, opposite angles are equal. IfSR= 2 cm andPR= 5 cm then determinePQ. Solution: Since, the opposite angles of thequadrilateralPQRSare equal.⇒QuadrilateralPQRSis a parallelogram. $\Rightarrow S R=P Q$ (Opposite sides of a parallelogram are equal.) $\therefore S R=P Q=2 \mathrm{~cm}$...
Read More →If the sum of p terms of an A.P. is q and the sum of q terms is p,
Question: If the sum ofpterms of an A.P. is q and the sum ofqterms isp, then the sum ofp+qterms will be (a) 0 (b)pq (c)p+q (d) (p+q) Solution: (d) $-(p+q)$ $S_{p}=q$ $\Rightarrow \frac{p}{2}\{2 a+(p-1) d\}=q$ $\Rightarrow 2 a p+(p-1) p d=2 q$ .....(1) $S_{q}=p$ $\Rightarrow \frac{q}{2}\{2 a+(q-1) d\}=p$ $\Rightarrow 2 a q+(q-1) q d=2 p$ ....(2) Multiplying equation (1) by $q$ and equation (2) by $p$ and then solving, we get: $d=\frac{-2(p+q)}{p q}$ Now, $S_{p+q}=\frac{(p+q)}{2}[2 a+(p+q-1) d]$ $...
Read More →Three statements are given below:
Question: Three statements are given below:I. In a rectangleABCD, the diagonalACbisectsAas well as C.II. In a squareABCD, the diagonalACbisects Aas well as C.III. In a rhombusABCD, the diagonalACbisects Aas well as C.(a) I only(b) II and III(c) I and III(d) I and II Solution: (b) II and III Clearly, statements II and III are true. Statement I is false, as diagonal of a rectangle does not bisect AandC( adjacent sides are not equal)....
Read More →The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15°
Question: The angle of elevation of a stationery cloud from a point 2500 m above a lake is 15 and the angle of depression of its reflection in the lake is 45. What is the height of the cloud above the lake level? (Use tan 15 = 0.268) Solution: Let AB be the surface of lake and P be the point of observation such thatm. Let C be the position of cloud andbe the reflection in the lake. Then Letbe the perpendicular from P on CB. Let $\mathrm{PQ}=x \mathrm{~m}, \mathrm{CQ}=h, \mathrm{QB}=2500 \mathrm{...
Read More →Three statements are given below:
Question: Three statements are given below:I. In a || gm, the angle bisectors of two adjacent angles enclose a right angle.II. The angle bisectors of a || gm form a rectangle.III. The triangle formed by joining the mid-points of the sides of an isosceles triangle is not necessarily an isosceles triangle.Which is true?(a) I only(b) II only(c) I and II(d) II and III Solution: (c) I and IIStatements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the si...
Read More →The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30° and ∠AOB = 70°.
Question: The diagonalsACandBDof a parallelogramABCDintersect each other at the pointOsuch thatDAC= 30 and AOB= 70. Then, DBC= ?(a) 40(b) 35(c) 45(d) 50 Solution: (a)40Explanation:OAD=OCB= 30o (Alternate interior angles)AOB+BOC= 180o (Linear pair of angles)BOC= 180o 70o= 110o (AOB= 70o)In∆BOC, we have:OBC= 180o (110o+ 30o) = 40o DBC= 40o...
Read More →If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is
Question: If 7th and 13th terms of an A.P. be 34 and 64 respectively, then its 18th term is (a) 87 (b) 88 (c) 89 (d) 90 Solution: (c) 89 $a_{7}=34$ $\Rightarrow a+6 d=34$ ....(1) Also, $a_{13}=64$ $\Rightarrow a+12 d=64$ ....(2) Solving equations $(1)$ and $(2)$, we get: a= 4 andd= 5 $\therefore a_{18}=a+17 d$ $=4+17(5)$ $=89$...
Read More →The angle of elevation of the top of a vertical tower PQ from a point X
Question: The angle of elevation of the top of a vertical towerPQfrom a pointXon the ground is 60. At a pointY, 40 m vertically aboveX, the angle of elevation of the top is 45. Calculate the height of the tower. Solution: LetPQbe the tower of heightHm and an angle of elevation of the top of towerPQfrom point X is 60. Angle of elevation at 40 m vertical from point X is 45. LetPQ = Hm andSX= 40m.OX = x,,. Here we have to find height of tower. The corresponding figure is as follows We use trigonome...
Read More →Find the value
Question: Find the value of $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ Solution: We know $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}, x y-1$Now, $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ $=\tan ^{-1}\left\{\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y}\left(\frac{x-y}{x+y}\right)}\right\}$ $=\tan ^{-1}\left\{\frac{\frac{x^{2}+x y-x y+y^{2}}{y(x+y)}}{\frac{x^{2}+y^{2}+x y-x y}{3(x+y)}}\right\}$ $=\tan ^{-1} 1$ $=\tan...
Read More →In the given figure, AD is a median of ∆ABC and E is the mid-point of AD.
Question: In the given figure,ADis a median of∆ABCandEis the mid-point ofAD. IfBEis joined and produced to meetACinF, thenAF= ? (a) $\frac{1}{2} A C$ (b) $\frac{\overline{1}}{3} A C$ (3) $\frac{2}{3} A C$ (4) $\frac{3}{4} A C$ Solution: (b) $1 / 3 A C$ Explanation: LetGbe the mid point ofFC. JoinDG.In∆BCF,Dis the mid point ofBCandGis the mid point ofFC.DG||BF⇒ DG||EFIn∆ADG,Eis the mid point ofADandEF||DG.i.e.,Fis the mid point ofAG.Now,AF = FG = GC [∵Gis the mid point ofFC] $\therefore A F=1 /...
Read More →A tree standing on a horizontal plane is leaning towards east.
Question: A tree standing on a horizontal plane is leaning towards east. At two points situated at distances a and b exactly due west on it, the angles of elevation of the top are respectively and . Prove that the height of the top from the ground is $\frac{(b-a) \tan \alpha \tan \beta}{\tan \alpha-\tan \beta}$ Solution: LetOPbe the tree andA,Bbe the two points suchOA = aandOB = band angle of elevation to the tops areandrespectively. LetOL = xandPL = h We have to prove the following $h=\frac{(b-...
Read More →In a cricket team tournament 16 teams participated.
Question: In a cricket team tournament 16 teams participated. A sum of₹8000 is to be awardedamong themselves as prize money. If the last place team is awarded₹275 in prize money and the award increases by the same amount for successive finishing places, then how much amount will the first place team receive? Solution: the total sum of prize money to be awarded among 16 teams,S16=₹8000 and the prize money awarded to the last place team i.e.a16=₹275 As, the award increases by the same amount for s...
Read More →A man saved ₹66000 in 20 years.
Question: A man saved₹66000 in 20 years. In each succeeding year after the first year he saved₹200 more than what he saved in the previous year. How much did he save in the first year? Solution: As, in each succeeding year after the first year he saved₹200more than what he saved in the previous year.So, the savings of each year are in A.P.We have, the total savings of the man in 20 years,S20=₹66000 and the difference of his savings in each succeeding year,d=₹200 Let his savings in the first year...
Read More →Two parallelograms stand on equal bases and between the same parallels.
Question: Two parallelograms stand on equal bases and between the same parallels. The ratio of their areas is(a) 1 : 2(b) 2 : 1(c) 1 : 3(d) 1 : 1 Solution: (d) 1:1Area of a parallelogram = base ⨯heightIf both parallelograms stands on the same base and between the same parallels, then their heights are the same.So, their areas will also be the same....
Read More →The horizontal distance between two trees of different heights is 60 m.
Question: The horizontal distance between two trees of different heights is 60 m. The angle of depression of the top of the first tree when seen from the top of the second tree is 45. If the height of the second tree is 80 m, find the height of the first tree. Solution: Let the difference between two trees beDE= 60 m and angle of depression of the first tree from the top to the top of the second tree is. LetBE = Hm,AC = hm,AD= 80m. We have to find the height of the first tree The corresponding f...
Read More →In a trapezium ABCD, if AB || CD, then
Question: In a trapezium $A B C D$, if $A B \| C D$, then $\left(A C^{2}+B D^{2}\right)=?$ (a) $B C^{2}+A D^{2}+2 B C \cdot A D$ (b) $A B^{2}+C D^{2}+2 A B \cdot C D$ (c) $A B^{2}+C D^{2}+2 A D \cdot B C$ (d) $B C^{2}+A D^{2}+2 A B \cdot C D$ Solution: (c) $B C^{2}+A D^{2}+2 A B \cdot C D$ Explanation: Draw perpendicular fromDandConABwhich meetsABatEandF, respectively.DEFCis a parallelogram andEF = CD. In∆ABC,Bis acute. $\therefore A C^{2}=B C^{2}+A B^{2}-2 A B \cdot A E$ In∆ABD,Ais acute. $\t...
Read More →From the top of a 50 m high tower, the angles of depression of the top and bottom
Question: From the top of a 50 m high tower, the angles of depression of the top and bottom of a pole are observed to be 45 and 60 respectively. Find the height of the pole. Solution: LetHbe the height of pole, makes an angle of depression from top of tower to top and bottom of poles are 45 and 60 respectively. Let $A B=H, C E=h, A D=x$ and $D E=50 \mathrm{~m} . \angle C B E=45^{\circ}$ and $\angle D A E=60^{\circ}$. Here we have to find height of pole. The corresponding figure is as follows $\l...
Read More →Prove the following results:
Question: Prove the following results: (i) $\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}=\tan ^{-1} \frac{2}{9}$ (ii) $\sin ^{-1} \frac{12}{13}+\cos ^{-1} \frac{13}{5}+\tan ^{-1} \frac{63}{16}=\pi$ (iii) $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{5}{9}=\sin ^{-1} \frac{1}{\sqrt{5}}$ Solution: (i) LHS $=\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{13}$ $=\tan ^{-1}\left(\frac{\frac{1}{\tau}+\frac{1}{13}}{1-\frac{1}{7} \times \frac{1}{13}}\right) \quad\left[\because \tan ^{-1} x+\tan ^{-1} y=\tan ^{...
Read More →A man accepts a position with an initial salary of ₹5200 per month.
Question: A man accepts a position with an initial salary of₹5200 per month. It is understood that hewill receive an automatic increase of₹320 in the very next month and each monththereafter. (i) Find his salary for the tenth month. (ii) What is his total earnings during the first year? Solution: We have, the initial salary, a1=₹5200, the salary of the second month,a2=₹5200 +₹320 =₹5520, the salary of the third month,a3=₹5520 +₹320 =₹5840, As, $a_{2}-a_{1}=5520-5200=320$ and $a_{3}-a_{2}=5840-55...
Read More →An aeroplane flying horizontally 1 km above the ground is observed
Question: An aeroplane flying horizontally 1 km above the ground is observed t an elevation of 60. After 10 seconds, its elevation is observed to be 30. Find the speed of the aeroplane in km/hr. Solution: An aero plane is flyingkm above the ground making an angle of elevation of aero plane 60. Afterseconds angle of elevation is changed to 30. Let,,,,km andkm. Here we have to find speed of aero plane. The corresponding figure is as follows So we use trigonometric ratios. In $\triangle D C E$ $\Ri...
Read More →In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point.
Question: In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes? Solution: We have: the distance travelled to bring the first potato, $a_{1}=2 \times 24=48 \mathrm{~m}$, the distance travelled to bring the second potato, $a_{2}=2 \times(24+4)=56 \mathrm{~m}$, the distance trave...
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