The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD,
Question: The quadrilateral formed by joining the midpoints of the sides of a quadrilateralABCD, taken in order, is a rhombus, if(a)ABCDis a Parallelogram(b)ABCDis rhombus(c) diagonals ofABCDare equal(4) diagonals ofABCDare perpendicular to each other. Solution: Given: The quadrilateralABCDis a rhombus.So, the sidesAB,BC,CDandADare equal. Now, in $\triangle P Q S$, we have $D C=\frac{1}{2} Q S$ (Using mid-point theorem) ...(1) Similarly, in $\triangle P S R$, $B C=\frac{1}{2} P R$ ..(2) As,BC = ...
Read More →A statue 1.6 m tall stands on the top of pedestal.
Question: A statue 1.6 m tall stands on the top of pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60 and from the same point the angle of elevation of the top of the pedestal is 45. Find the height of the pedestal. Solution: Letbe the pedestal of heightm andthe statue of heightmeter and angle of elevation at top of statue is 60 and angle of elevation of pedestal at the same point is 45. Here we have to find height of pedestal. The corresponding figure is...
Read More →Insert 4 A.M.s between 4 and 19.
Question: Insert 4 A.M.s between 4 and 19. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the four A.M.s between 4 and 19. Then, 4, $A_{1}, A_{2}, A_{3}, A_{4}$ and 19 are in A.P. whose common difference is as follows: $d=\frac{19-4}{4+1}$ = 3 $A_{1}=4+d=4+3=7$ $A_{2}=4+2 d=4+6=10$ $A_{3}=4+3 d=4+9=13$ $A_{4}=4+4 d=4+12=16$ Hence, the required A.M.s are 7, 10, 13, 16....
Read More →Insert 4 A.M.s between 4 and 19.
Question: Insert 4 A.M.s between 4 and 19. Solution: Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the four A.M.s between 4 and 19. Then, 4, $A_{1}, A_{2}, A_{3}, A_{4}$ and 19 are in A.P. whose common difference is as follows: $d=\frac{19-4}{4+1}$ = 3 $A_{1}=4+d=4+3=7$ $A_{2}=4+2 d=4+6=10$ $A_{3}=4+3 d=4+9=13$ $A_{4}=4+4 d=4+12=16$ Hence, the required A.M.s are 7, 10, 13, 16....
Read More →Find the A.M. between:
Question: Find the A.M. between: (i) 7 and 13 (ii) 12 and 8 (iii) (xy) and (x+y). Solution: (i) Let $A_{1}$ be the A.M. between 7 and $13 .$ $A_{1}=\frac{a+b}{2}$ $=\frac{7+13}{2}$ = 10 (ii) Let $A_{1}$ be the A.M. between 12 and $-8$. $A_{1}=\frac{a+b}{2}$ $=\frac{12+(-8)}{2}$ =2 (iii) Let $A_{1}$ be the A.M. between $(x-y)$ and $(x+y)$. $A_{1}=\frac{a+b}{2}$ $=\frac{(x-y)+(x+y)}{2}$ $=x$...
Read More →The angles of depression of the top and bottom of 8 m tall building from the
Question: The angles of depression of the top and bottom of 8 m tall building from the top of a multistoried building are 30 and 45 respectively. Find the height of the multistoried building and the distance between the two buildings. Solution: LetADbe the multistoried building of heighthm. And angle of depression of the top and bottom are 30 and 45. We assume thatBE= 8,CD= 8 andBC=x,ED = xandAC=h 8. Here we have to find height and distance of building. We use trignometrical ratio. $\Rightarrow ...
Read More →The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if
Question: The quadrilateral formed by joining the midpoints of the sides of a quadrilateralABCD, taken in order, is a rectangle, if(a)ABCDis a Parallelogram(b)ABCDis rectangle(c) diagonals ofABCDare equal(d) diagonals ofABCDare perpendicular to each other. Solution: Since,The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram,The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,The quadrilateral formed by joining ...
Read More →The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
Question: The figure formed by joining the mid-points of the adjacent sides of a rhombus is a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (c) RectangleThe figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle....
Read More →are consecutive terms of an A.P., if x, y and z are in A.P.
Question: Show thatx2+xy+y2,z2+zx+x2andy2+yz+z2are consecutive terms of an A.P., ifx,yandzare in A.P. [NCERT EXEMPLAR] Solution: As, $x, y$ and $z$ are in A.P. So, $y=\frac{x+z}{2} \quad \ldots \ldots($ i) Now, $\left(x^{2}+x y+y^{2}\right)+\left(y^{2}+y z+z^{2}\right)$ $=x^{2}+z^{2}+2 y^{2}+x y+y z$ $=x^{2}+z^{2}+2 y^{2}+y(x+z)$ $=x^{2}+z^{2}+2\left(\frac{x+z}{2}\right)^{2}+\left(\frac{x+z}{2}\right)(x+z) \quad[$ Using $(\mathrm{i})]$ $=x^{2}+z^{2}+2\left(\frac{(x+z)^{2}}{4}\right)+\frac{(x+z)^...
Read More →Evaluate each of the following:
Question: Evaluate each of the following: (i) $\sin \left(\sin ^{-1} \frac{7}{25}\right)$ (ii) $\sin \left(\cos ^{-1} \frac{5}{13}\right)$ (iii) $\sin \left(\tan ^{-1} \frac{24}{7}\right)$ (iv) $\sin \left(\sec ^{-1} \frac{17}{8}\right)$ (v) $\operatorname{cosec}\left(\cos ^{-1} \frac{3}{5}\right)$ (vi) $\sec \left(\sin ^{-1} \frac{12}{13}\right)$ (vii) $\tan \left(\cos ^{-1} \frac{8}{17}\right)$ (viii) $\cot \left(\cos ^{-1} \frac{3}{5}\right)$ (ix) $\cos \left(\tan ^{-1} \frac{24}{7}\right)$ S...
Read More →From a point on the ground the angles of elevation of the bottom and top of a
Question: From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are 45 and 60 respectively. Find the height of the transmission tower. Solution: Letbe the building of heightm andthe transmission tower of heightmeter. Again let the angle of elevation of the bottom and top of tower at the pointis 45 and 60 respectively. $\Rightarrow \quad \tan 45^{\circ}=\frac{A B}{O A}$ $\Rightarrow \quad 1=\frac{20}{x}$ $\Rightarro...
Read More →The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
Question: The figure formed by joining the mid-points of the adjacent sides of a rectangle is a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (a) rhombusThe figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus....
Read More →The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
Question: The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (d)parallelogram.The figure made by joining the mid points of the adjacent sides of a parallelogram is a parallelogram....
Read More →The shadow of a tower standing on a level ground is found to be 40 m
Question: The shadow of a tower standing on a level ground is found to be 40 m longer when Sun's altitude is 30 than when it was 60. Find the height of the tower. Solution: Letbe the tower of height.given the shadow of towerm. attitude of sun areand. Here we have to find height of tower. Letand. In $\triangle A C B$ $\Rightarrow \quad \tan C=\frac{A B}{B C}$ $\Rightarrow \quad \tan 60^{\circ}=\frac{h}{x}$ $\Rightarrow \quad \sqrt{3}=\frac{h}{x}$ $\Rightarrow \quad x=\frac{h}{\sqrt{3}}$ Again in ...
Read More →The figure formed by joining the mid-points of the adjacent sides of a square is a
Question: The figure formed by joining the mid-points of the adjacent sides of a square is a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (b) SquareThe figure formed by joining the mid points of the adjacent sides of a square is a square....
Read More →are in A.P., prove that a, b, c are in A.P.
Question: If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P., prove that $a, b, c$ are in A.P. Solution: Given; $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P. By adding 1 to each term, we get: $a\left(\frac{1}{b}+\frac{1}{c}\right)+1, b\left(\frac{1}{c}+\frac{1}{a}\right)+1, c\left(\frac{1}{a}+\frac{1}{b}\right)+1$ are in A.P. $\Righ...
Read More →The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
Question: The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (d)parallelogramThe figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram....
Read More →A 1.5 m tall boy is standing at some distance from a 30 m tall building.
Question: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increase from 30 to 60 as he walks towards the building. Find the distance he walked towards the building. Solution: LetBGbe the distance of tall Boyxand he walks towards the building, makes an angle of elevation at top of building increase from 30 to 60. ThereforeA= 30 andF= 60 givenCE= 30 m,AB= 15 m,FG= 1.5 and DE = 28.5,GC=X xandFD=X x We have to f...
Read More →If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q,
Question: If bisectors of Aand Bof a quadrilateralABCDintersect each other atP, of Band C atQ, of Cand DatRand of Dand AatSthenPQRSis a(a) rectangle(b) parallelogram(c) rhombus(d) quadrilateral whose opposite angles are supplementary Solution: Given:In quadrilateralABCD,AS,BQ,CQandDSareanglebisectorsofanglesA,B,CandD,respectively.QPS=APB (Verticallyoppositeangles) ...(1)InΔ∆APB,APB +PAB +ABP=180 (Angle sum property of triangle.) $\Rightarrow \angle A P B+\frac{1}{2} \angle A+\frac{1}{2} \angle B...
Read More →A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts
Question: A 1.6 m tall girl stands at a distance of 3.2 m from a lamp-post and casts a shadow of 4.8 m on the ground. Find the height of the lamp-post by using (i) trignometric ratios (ii) property of similar triangles. Solution: LetACbe the lamp post of height. We assume thatED=1.6m,BE = 4.8m andEC =3.2 m We have to find the height of the lamp post Now we have to find height of lamp post using similar triangles. $\frac{A C}{B C}=\frac{E D}{B E}$ $\frac{h}{4.8+3.2}=\frac{1.6}{4.8}$ $h=\frac{8}{3...
Read More →Write each of the following in the simplest form:
Question: Write each of the following in the simplest form: (i) $\cot ^{-1} \frac{a}{\sqrt{x^{2}-a^{2}}},|x|a$ (ii) $\tan ^{-1}\left\{x+\sqrt{1+x^{2}}\right\}, x \in R$ (iii) $\tan ^{-1}\left\{\sqrt{1+x^{2}}-x\right\}, x \in R$ (iv) $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-1}{x}\right\}, x \neq 0$ (v) $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}+1}{x}\right\}, x \neq 0$ (vi) $\tan ^{-1} \sqrt{\frac{a-x}{a+x}},-axa$ (vii) $\tan ^{-1}\left\{\frac{x}{a+\sqrt{a^{2}-x^{2}}}\right\},-axa$ (viii) $\sin ^{-1}\l...
Read More →The bisectors of the angles of a parallelogram enclose a
Question: The bisectors of the angles of a parallelogram enclose a(a) rhombus(b) square(c) rectangle(d) parallelogram Solution: (c) RectangleThe bisectors of the angles of a parallelogram encloses a rectangle....
Read More →From a point P on the ground the angle of elevation of a 10 m tall building is 30°.
Question: From a pointPon the ground the angle of elevation of a 10 m tall building is 30. A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff fromPis 45. Find the length of the flag-staff and the distance of the building from the point P.(Take $\sqrt{3}=1.732$ ). Solution: Letbe the flag of lengthm on the building. We assume that $B C=10, C P=y$ and $\angle A P C=45^{\circ}, \angle B P C=30^{\circ}$ Now we have to find height of flag-staff and di...
Read More →The bisectors of any two adjacent angles of a parallelogram intersect at
Question: The bisectors of any two adjacent angles of a parallelogram intersect at(a) 40(b) 45(c) 60(d) 90 Solution: (d)90Explanation:Sum of two adjacent angles = 180o Now, sum of angle bisectors of two adjacent angles $=\frac{1}{2} \times\left(180^{\circ}\right)=90^{\circ}$ $\therefore$ Intersection angle of bisectors of two adjacent angles $=180^{\circ}-90^{\circ}=90^{\circ}$...
Read More →A tree breaks due to storm and the broken part bends so that the top of the tree touches
Question: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 30 with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. Solution: LetABbe the tree of heighth. And the top of tree makes an angle 30 with ground. The distance between foot of tree to the point where the top touches is 8m. Let. And. Here we have to find height of tree. So we trigono...
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