The length of the shadow of a tower standing on level ground is found
Question: The length of the shadow of a tower standing on level ground is found to be 2x metres longer when the sun's elevation is 30than when it was 45. The height of the tower in metres is (a) $(\sqrt{3}+1) x$ (b) $(\sqrt{3}-1) x$ (c) $2 \sqrt{3} x$ (d) $3 \sqrt{2} x$ Solution: Letbe the height of tower Given that: angle of elevation of sun are $\angle D=30^{\circ}$ and $\angle C=45^{\circ}$. Then Distance $C D=2 x$ and we assume $B C=X$ Here, we have to find the height of tower. So we use tri...
Read More →ABCD is a trapezium in which AB || CD, AB = 16 cm and DC = 24 cm.
Question: $A B C D$ is a trapezium in which $A B \| C D, A B=16 \mathrm{~cm}$ and $D C=24 \mathrm{~cm}$. If $E$ and $F$ are respectively the midpoints of $A D$ and $B C$, prove that $\operatorname{ar}(A B F E)=\frac{9}{11} \operatorname{ar}(E F C D)$. Solution: Construction: Draw a perpendicular from point D to the opposite side CD, meeting CD at Q and EF at P.Let length AQ =hGiven, E and F are the midpoints of AD and BC respectively. So, $\mathrm{EF}\|\mathrm{AB}\| \mathrm{DC}$ and $\mathrm{EF}...
Read More →It is found that on walking x meters towards a chimney in a horizontal
Question: It is found that on walking x meters towards a chimney in a horizontal line through its base, the elevation of its top changes from 30 to 60. The height of the chimney is (a) $3 \sqrt{2} x$ (b) $2 \sqrt{3} x$ (c) $\frac{\sqrt{3}}{2} x$ (d) $\frac{2}{\sqrt{3}} x$ Solution: Letbe the height of chimney Given that: angle of elevation changes from angle $\angle D=30^{\circ}$ to $\angle C=60^{\circ}$. Then Distance becomes $C D=x$ and we assume $B C=y$ Here, we have to find the height of chi...
Read More →(i) Find a matrix X such that 2A + B + X = O, where
Question: (i) Find a matrix $X$ such that $2 A+B+X=0$, where $A=\left[\begin{array}{rr}-1 2 \\ 3 4\end{array}\right], B=\left[\begin{array}{rr}3 -2 \\ 1 5\end{array}\right]$ (ii) If $A=\left[\begin{array}{rr}8 0 \\ 4 -2 \\ 3 6\end{array}\right]$ and $B=\left[\begin{array}{rr}2 -2 \\ 4 2 \\ -5 1\end{array}\right]$, then find the matrix $X$ of order $3 \times 2$ such that $2 A+3 X=5 B$. Solution: (i) $2 A+B+X=0$ $\Rightarrow 2\left[\begin{array}{cc}-1 2 \\ 3 4\end{array}\right]+\left[\begin{array}...
Read More →In a trapezium ABCD, AB || DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides,
Question: In a trapeziumABCD,AB||DC, AB=acm, andDC=bcm. IfMandNare the midpoints of the nonparallel sides,ADandBCrespectively then find the ratio of ar(DCNM) and ar(MNBA). Solution: Construction: Draw a perpendicular from point D to the opposite side AB, meeting AB at Q and MN at P.Let length DQ =hGiven, M and N are the midpoints of AD and BC respectively. So, $\mathrm{MN}\|\mathrm{AB}\| \mathrm{DC}$ and $\mathrm{MN}=\frac{1}{2}(\mathrm{AB}+\mathrm{DC})=\left(\frac{a+b}{2}\right)$ M is the mid p...
Read More →A ladder 15 m long just reaches the top of a vertical wall.
Question: A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60 with the wall, then the height of the wall is (a) $15 \sqrt{3} \mathrm{~m}$ (b) $\frac{15 \sqrt{3}}{2} \mathrm{~m}$ (c) $\frac{15}{2} \mathrm{~m}$ (d) $15 \mathrm{~m}$ Solution: Suppose AB is the wall and AC is the ladder.It is given that, AC = 15 m andCAB = 60.In right ∆ABC, $\cos 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{AC}}$ $\Rightarrow \frac{1}{2}=\frac{\mathrm{AB}}{15}$ $\Rightarrow \mathr...
Read More →Find the value of λ, a non-zero scalar,
Question: Find the value of $\lambda$, a non-zero scalar, if $\lambda\left[\begin{array}{lll}1 0 2 \\ 3 4 5\end{array}\right]+2\left[\begin{array}{rrr}1 2 3 \\ -1 -3 2\end{array}\right]=\left[\begin{array}{lll}4 4 10 \\ 4 2 14\end{array}\right]$ Solution: Given : $\lambda\left[\begin{array}{lll}1 0 2 \\ 3 4 5\end{array}\right]+2\left[\begin{array}{ccc}1 2 3 \\ -1 -3 2\end{array}\right]=\left[\begin{array}{lll}4 4 10 \\ 4 2 14\end{array}\right]$ $\Rightarrow\left[\begin{array}{ccc}\lambda 0 2 \la...
Read More →if
Question: If $2\left[\begin{array}{ll}3 4 \\ 5 x\end{array}\right]+\left[\begin{array}{ll}1 y \\ 0 1\end{array}\right]=\left[\begin{array}{rr}7 0 \\ 10 5\end{array}\right]$, find $x$ and $y$ Solution: Given : $2\left[\begin{array}{ll}3 4 \\ 5 x\end{array}\right]+\left[\begin{array}{cc}1 y \\ 0 1\end{array}\right]=\left[\begin{array}{cc}7 0 \\ 10 5\end{array}\right]$ $\Rightarrow\left[\begin{array}{cc}6 8 \\ 10 2 x\end{array}\right]+\left[\begin{array}{ll}1 y \\ 0 1\end{array}\right]=\left[\begin...
Read More →In the adjoining figure, the point D divides the side BC of ∆ABC in the
Question: In the adjoining figure, the pointDdivides the sideBCof∆ABCin the ratiom:n. Prove that ar(∆ABD) : ar(∆ADC) =m:n. Solution: Given:D is a point on BC of∆ABC,such that BD : DC = m:nTo prove:ar(∆ABD) : ar(∆ADC) =m:nConstruction:DrawALBC.Proof: $\operatorname{ar}(\triangle A B D)=\frac{1}{2} \times B D \times A L$ .........(i) $\operatorname{ar}(\triangle A D C)=\frac{1}{2} \times D C \times A L$ ..........(ii) Dividing (i) by (ii), we get: $\frac{\operatorname{ar}(\triangle A B D)}{\operat...
Read More →The angle of depression of a car, standing on the ground,
Question: The angle of depression of a car, standing on the ground, from the top of a 75 m tower, is 30. The distance of the car from the base of the tower (in metres) is (a) $25 \sqrt{3}$ (b) $50 \sqrt{3}$ (c) $75 \sqrt{3}$ (d) 150 [CBSE 2013] Solution: Suppose AB is the tower and C is the position of the car from the base of the tower.It is given that, AB = 75 m Now, $\angle A C B=\angle C A D=30^{\circ}$ (Alternate angles) In right ∆ABC, $\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$ $\Rig...
Read More →If the sums of n terms of two AP.'s are in the ratio (3n + 2) : (2n + 3),
Question: If the sums ofnterms of two AP.'s are in the ratio (3n+ 2) : (2n + 3), then find the ratio of their12th terms. Solution: Let the first terms of the two A.P.'s beaanda'; and their common difference bedandd'. Now, $\frac{S_{n}}{S_{n}{ }^{\prime}}=\frac{(3 n+2)}{(2 n+3)}$ $\Rightarrow \frac{\frac{n}{2}[2 a+(n-1) d]}{\frac{n}{2}\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{(3 n+2)}{(2 n+3)}$ $\Rightarrow \frac{[2 a+(n-1) d]}{\left[2 a^{\prime}+(n-1) d^{\prime}\right]}=\frac{(3 n+2)}{(2...
Read More →If mth term of an A.P. is n and nth term is m,
Question: Ifmth term of an A.P. isnandnth term ism, then write itspth term. Solution: Given: $a_{m}=n$ $\Rightarrow a+(m-1) d=n \quad \ldots(1)$ $a_{n}=m$ $\Rightarrow a+(n-1) d=m \quad \ldots .(2)$ Solving equations $(1)$ and $(2)$, we get: $d=-1$ $a=n+m-1$ pth term: $a_{p}=a+(p-1) d$ $=n+m-1+(p-1)(-1)$ $=n+m-p$ Hence, the pth term is $n+m-p$....
Read More →The length of shadow of a tower on the plane ground is
Question: The length of shadow of a tower on the plane ground is $\sqrt{3}$ times the height of the tower. The angle of elevation of sun is (a) 45 (b) 30 (c) 60 (d) 90 [CBSE 2012] Solution: Let the angle of elevation of the sun be.Suppose AB is the height of the tower and BC is the length of its shadow. It is given that, $\mathrm{BC}=\sqrt{3} \mathrm{AB}$ In right ∆ABC, $\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}$ $\Rightarrow \tan \theta=\frac{\mathrm{AB}}{\sqrt{3} \mathrm{AB}}=\frac{1}{\sqrt{...
Read More →In the adjoining figure, CE || AD and CF || BA.
Question: In the adjoining figure,CE||ADandCF||BA. Prove that ar(∆CBG) = ar(∆AFG). Solution: $\operatorname{ar}(\triangle \mathrm{CFA})=\operatorname{ar}(\triangle \mathrm{CFB})$(Triangles on the same base CF and between the same parallels CF || BA will be equal in area) $\Rightarrow \operatorname{ar}(\triangle$ CFA $)-\operatorname{ar}(\triangle$ CFG $)=\operatorname{ar}(\triangle$ CFB $)-\operatorname{ar}(\triangle$ CFG $)$ $\Rightarrow \operatorname{ar}(\triangle \mathrm{AFG})=\operatorname{a...
Read More →Find x, y satisfying the matrix equations
Question: Findx,ysatisfying the matrix equations (i) $\left[\begin{array}{crr}x-y 2 -2 \\ 4 x 6\end{array}\right]+\left[\begin{array}{rrr}3 -2 2 \\ 1 0 -1\end{array}\right]=\left[\begin{array}{ccc}6 0 0 \\ 5 2 x+y 5\end{array}\right]$ (ii) $\left[\begin{array}{lll}x y+2 z-3\end{array}\right]+\left[\begin{array}{lll}y 4 5\end{array}\right]=\left[\begin{array}{lll}4 9 12\end{array}\right]$ (iii) $x\left[\begin{array}{l}2 \\ 1\end{array}\right]+y\left[\begin{array}{l}3 \\ 5\end{array}\right]+\left[...
Read More →A tower subtends an angle of 30° at a point on the same level as its foot.
Question: A tower subtends an angle of 30 at a point on the same level as its foot. At a second pointhmetres above the first, the depression of the foot of the tower is 60. The height of the tower is (a) $\frac{h}{2} m$ (b) $\sqrt{3 h} m$ (c) $\frac{h}{3} m$ (d) $\frac{h}{\sqrt{3}} m$ Solution: LetABbe the tower andCis a point on the same level as its foot such thatACB= 30 The given situation can be represented as, HereDis a pointhm above the pointC. InΔBCD, $\Rightarrow \tan B=\frac{C D}{C B}$ ...
Read More →then find the value of n
Question: If $\frac{3+5+7+\ldots+\text { upto } n \text { terms }}{5+8+11+\ldots . \text { upto } 10 \text { terms }} 7$, then find the value of $n$. Solution: $\frac{3+5+7+\ldots+\text { upto } n \text { terms }}{5+8+11+\ldots . \text { upto } 10 \text { terms }}=7$ $\Rightarrow \frac{\frac{n}{2}\{2 \times 3+(n-1) 2\}}{\frac{10}{2}\{2 \times 5+(10-1) 3\}}=7$ $\Rightarrow \frac{n(4+2 n)}{370}=7$ $\Rightarrow n^{2}+2 n-1295=0$ $\Rightarrow n^{2}+37 n-35 n-1295=0$ $\Rightarrow(n+37)(n-35)$ $\Right...
Read More →The given figure shows a pentagon ABCDE. EG, drawn parallel to DA, meets BA produced at G, and CF, drawn parallel to DB, meets AB produced at F.
Question: The given figure shows a pentagonABCDE.EG, drawn parallel toDA, meetsBAproduced atG, andCF, drawn parallel toDB, meetsABproduced atF. Show that ar(pentagonABCDE) = ar(∆DGF). Solution: Given: ABCDEis a pentagon.EG|| DAandCF|| DB.To prove:ar(pentagonABCDE) = ar(∆DGF)Proof:ar(pentagonABCDE) =ar(∆DBC) + ar(∆ADE) +ar(∆ABD) ...(i)Also, ar(∆DGF) =ar(∆DBF) +ar(∆ADG) +ar(∆ABD) ...(ii)Now, ∆DBCand ∆DBFlie on the same base and between the same parallel lines.ar(∆DBC) =ar(∆DBF) ...(iii) Similarly...
Read More →In the adjoining figure, BD || CA, E is the midpoint of CA and BD
Question: In the adjoining figure, $B D \| C A, E$ is the midpoint of $C A$ and $B D=\frac{1}{2} C A$. Prove that $\operatorname{ar}(\triangle A B C)=2 \operatorname{ar}(\Delta D B C)$. Solution: E is the midpoint of CA.So, AE = EC .....(1) Also, $\mathrm{BD}=\frac{1}{2} \mathrm{CA}$ (Given) So, BD = AE .....(2)From (1) and (2) we haveBD = ECBD || CA and BD = EC so, BDEC is a parallelogram $B E$ acts as the median of $\triangle A B C$ so, $\operatorname{ar}(\triangle \mathrm{BCE})=\operatorname{...
Read More →Write the value of n for which nth terms of the A.P.s 3, 10, 17,
Question: Write the value ofnfor whichnth terms of the A.P.s 3, 10, 17, ... and 63, 65, 67, .... are equal. Solution: For the first series, $a=3, d_{1}=7$ For the second series, $b=63, d_{2}=2$ Given: $a_{n}=b_{n}$ $\Rightarrow a+(n-1) d_{1}=b+(n-1) d_{2}$ $\Rightarrow 3+(n-1) 7=63+(n-1) 2$ $\Rightarrow 3+7 n-7=63+2 n-2$ $\Rightarrow 5 n=65$ $\Rightarrow n=13$ Hence, the 13th terms of both the series are the same....
Read More →The base BC of ∆ABC is divided at D such that BD
Question: The base $B C$ of $\triangle A B C$ is divided at $D$ such that $B D=\frac{1}{2} D C$. Prove that $\operatorname{ar}(\Delta A B D)=\frac{1}{3} \times \operatorname{ar}(\Delta A B C)$. Solution: Given: $\mathrm{D}$ is a point on $\mathrm{BC}$ of $\triangle A B C$, such that $\mathrm{BD}=\frac{1}{2} D C$ To prove: $\operatorname{ar}(\Delta A B D)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$ Construction:Draw AL BC.Proof:In ∆ABC,we have:BC = BD + DC⇒BD + 2 BD = 3 BDNow, we have: $\opera...
Read More →Write the sum of first n even natural numbers.
Question: Write the sum of firstneven natural numbers. Solution: We need to find the sum of 2, 4, 6, 8...uptonterms. Here,a= 2,d = 2 We know: $S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$ $=\frac{n}{2}\{2 \times 2+(n-1) 2\}$ $=n(n+1)$ Therefore, the sum of the first $n$ odd numbers is $n(n+1)$....
Read More →if
Question: If $A=\left[\begin{array}{rrr}1 -3 2 \\ 2 0 2\end{array}\right]$ and, $B=\left[\begin{array}{rrr}2 -1 -1 \\ 1 0 -1\end{array}\right]$, find the matrix $C$ such that $A+B+C$ is zero matrix. Solution: Given : $A+B+C=\left[\begin{array}{lll}0 0 0 \\ 0 0 0\end{array}\right]$ $\Rightarrow\left[\begin{array}{ccc}1 -3 2 \\ 2 0 2\end{array}\right]+\left[\begin{array}{ccc}2 -1 -1 \\ 1 0 -1\end{array}\right]+C=\left[\begin{array}{lll}0 0 0 \\ 0 0 0\end{array}\right]$ $\Rightarrow\left[\begin{arr...
Read More →In a triangle ABC, the medians BE and CF intersect at G.
Question: In a triangleABC, the mediansBEandCFintersect atG. Prove that ar(∆BCG) = ar(AFGE). Solution: Figure $C F$ is median of $\triangle A B C$. $\Rightarrow \operatorname{ar}(\triangle \mathrm{BCF})=\frac{1}{2}(\triangle \mathrm{ABC})$ Similarly, $B E$ is the median of $\triangle A B C$, $\Rightarrow \operatorname{ar}(\triangle \mathrm{ABE})=\frac{1}{2}(\triangle \mathrm{ABC})$ From (1) and (2) we have $\operatorname{ar}(\triangle \mathrm{BCF})=\operatorname{ar}(\triangle \mathrm{ABE})$ $\Ri...
Read More →Write the sum of first n odd natural numbers.
Question: Write the sum of firstnodd natural numbers. Solution: We need to find the sum of 1, 3, 5, 7... uptonterms. Here,a =1,d= 2 We know: $S_{n}=\frac{n}{2}\{2 a+(n-1) d\}$ $=\frac{n}{2}\{2 \times 1+(n-1) 2\}$ $=n^{2}$ Therefore, the sum of the first $n$ odd numbers is $n^{2}$....
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