The tops of two poles of height 20 m and 14 m are connected
Question: The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30 with horizontal, then the length of the wire is(a) 12 m(b) 10 m(c) 8 m(d) 6 m Solution: Letbe the length of wire. Given that wire makes an angle Now, $A C=(20-14) \mathrm{cm}=6 \mathrm{~cm}$, Here, we have to find length of wire. So we use trigonometric ratios. In a triangle, $\Rightarrow \sin B=\frac{A C}{A B}$' $\Rightarrow \sin 30^{\circ}=\frac{6}{h}$ $\Rightarrow \frac{1}{2}=\fra...
Read More →Find matrices X and Y, if X + Y
Question: Find matrices $X$ and $Y$, if $X+Y=\left[\begin{array}{ll}5 2 \\ 0 9\end{array}\right]$ and $X-Y=\left[\begin{array}{rr}3 6 \\ 0 -1\end{array}\right]$ Solution: Given : $(X+Y)+(X-Y)=\left[\begin{array}{ll}5 2 \\ 0 9\end{array}\right]+\left[\begin{array}{cc}3 6 \\ 0 -1\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{array}{ll}5+3 2+6 \\ 0+0 9-1\end{array}\right]$ $\Rightarrow 2 X=\left[\begin{array}{ll}8 8 \\ 0 8\end{array}\right]$ $\Rightarrow X=\frac{1}{2}\left[\begin{array}{ll}8 8 \...
Read More →If the sums of n terms of two arithmetic progressions are the ratio (2n + 3) :
Question: If the sums ofnterms of two arithmetic progressions are the ratio (2n+ 3) : (6n+ 5), then the ratio of their 13thterms is ___________. Solution: Leta1andd2represent the first term and common difference of first arithmetic progression. Also,a2andd2represent the first term and common difference of second arithmetic progression then sum ofnterms of first A.P $\left(S_{n}\right)_{1}=\frac{n}{2}\left\{2 a_{1}+(n-1) d_{1}\right\}$ and sum ofnterms of second A.P $\left(S_{n}\right)_{2}=\frac{...
Read More →In a trapezium ABCD, AB || DC and M is the midpoint of BC.
Question: In a trapeziumABCD,AB||DCandMis the midpoint ofBC. ThroughM, a linePQ||ADhas been drawn which meetsABinPandDCproduced inQ, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD). Solution: In△MQC and△MPB,MC = MB (M is the midpoint of BC)CMQ =BMP (Vertically opposite angles)MCQ =MBP (Alternate interior angles on the parallel lines AB and DQ)Thus,△MQC△MPB (ASA congruency)⇒⇒ar(△MQC) = ar(△MPB)⇒⇒ar(△MQC) + ar(APMCD) = ar(△MPB)+ ar(APMCD)⇒⇒ar(APQD) = ar(ABCD)...
Read More →Given the matrices
Question: Given the matrices $A=\left[\begin{array}{rrr}2 1 1 \\ 3 -1 0 \\ 0 2 4\end{array}\right], B=\left[\begin{array}{rrr}9 7 -1 \\ 3 5 4 \\ 2 1 6\end{array}\right]$ and $C=\left[\begin{array}{rrr}2 -4 3 \\ 1 -1 0 \\ 9 4 5\end{array}\right]$ Verify that (A+B) +C=A+ (B+C). Solution: Here, LHS $=(A+B)+C$ $=\left(\left[\begin{array}{ccc}2 1 1 \\ 3 -1 0 \\ 0 2 4\end{array}\right]+\left[\begin{array}{ccc}9 7 -1 \\ 3 5 4 \\ 2 1 6\end{array}\right]\right)+\left[\begin{array}{ccc}2 -4 3 \\ 1 -1 0 \\...
Read More →The angle of elevation of the top of a tower standing on a horizontal
Question: The angle of elevation of the top of a tower standing on a horizontal plane from a point A is . After walking a distance d towards the foot of the tower the angle of elevation is found to be . The height of the tower is (a) $\frac{d}{\cot \alpha+\cot \beta}$ (b) $\frac{d}{\cot \alpha-\cot \beta}$ (c) $\frac{d}{\tan \beta-\tan \alpha}$ (d) $\frac{d}{\tan \beta+\tan \alpha}$ Solution: The given information can be represented with the help of a diagram as below. Here,CD=his the height of ...
Read More →If twice the 11th term of an A.P is equal to 7 times of its 21st terms,
Question: If twice the 11thterm of an A.P is equal to 7 times of its 21stterms, then the value of 25thterm is ___________. Solution: Let us suppose a denote the first term of A.P anddrepresents the common difference Then 11thterm isa+ 10d 21stterm isa+ 20d According to given condition, 2(11thterm) = 7(21stterm) i.e 2(a+ 10d) = 7(a+ 20d) i.e 2a+ 20d= 7a+ 140d i. 5a+ 120d= 0 i.e 5(a+ 24d) = 0 Since 5 0 ⇒a+ 24d= 0 i.e 25thterm of A.P is 0....
Read More →From a light house the angles of depression of two ships on opposite sides
Question: From a light house the angles of depression of two ships on opposite sides of the light house are observed to be 30 and 45. If the height of the light house is h metres, the distance between the ships is (a) $(\sqrt{3}+1)$ h metres (b) $(\sqrt{3}-1)$ hetres (c) $\sqrt{3} h$ metres (d) $1+\left(1+\frac{1}{\sqrt{3}}\right)$ h metres Solution: Let the height of the light house AB bemeters Given that: angle of depression of ship areand. Distance of the ship C =and distance of the ship D = ...
Read More →If n is even, then the sum of first n terms of the series 1 – 2 + 3 – 4 + 5 – 6 +...,
Question: Ifnis even, then the sum of firstnterms of the series 1 2 + 3 4 + 5 6 +..., is ___________. Solution: For even numbern, Consider 1 2 + 3 4 + 5 6 + ................ for 1 2 = 1 3 4 = 1 5 6 = 1 i.e Combining two terms at a time given 1 So, if in all,neven terms arethere Sum of the expression $1-2+3-4+$ $n-1-n$ $=-\frac{n}{2}$...
Read More →D is the midpoint of side BC of ∆ABC and E is the midpoint of BD. If O is the midpoint of AE,
Question: Dis the midpoint of side $B C$ of $\triangle A B C$ and $E$ is the midpoint of $B D$. If $O$ is the midpoint of $A E$, prove that ar $(\triangle B O E)=\frac{1}{8}$ ar( $\triangle A B C$ ). Solution: $D$ is the midpoint of side $B C$ of $\triangle A B C$. $\Rightarrow \mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$. $\Rightarrow \operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ACD})=\frac{1}{2} \operatorname{ar}(\triangle \mathrm{ABC})$ $E$ is the mi...
Read More →if
Question: IfA= diag (2 59),B= diag (11 4) andC= diag (6 3 4), find(i)A 2B(ii)B+C 2A(iii) 2A+ 3B 5C Solution: Here, $A=\left[\begin{array}{ccc}2 0 0 \\ 0 -5 0 \\ 0 0 9\end{array}\right], B=\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 -4\end{array}\right]$ and $C=\left[\begin{array}{ccc}-6 0 0 \\ 0 3 0 \\ 0 0 4\end{array}\right]$ (i) $A-2 B$ $\Rightarrow A-2 B=\left[\begin{array}{ccc}2 0 0 \\ 0 -5 0 \\ 0 0 9\end{array}\right]-2\left[\begin{array}{ccc}1 0 0 \\ 0 1 0 \\ 0 0 -4\end{array}\right]$ $\...
Read More →The sum of first n odd natural numbers
Question: The sum of firstnodd natural numbers is ___________. Solution: Firstnodd natural numbers are 1, 3, 5, 7, .............n Here first terma= 1 Common differenced= 2 Sum of A.P is $\frac{n}{2}\{2 a+(n-1) d\}$ $=\frac{n}{2}\{2(2)+(n-1) 2\}$ $=\frac{n}{2}\{4+2 n-2\}$ $=\frac{n}{2}\{2+2 n\}$ i.e Sum of firstneven numbers =n(n+ 1)....
Read More →The arithmetic mean of first n natural numbers
Question: The arithmetic mean of firstnnatural numbers is ___________. Solution: Firstnnatural numbers are 1, 2, 3, ...........n Sum of these numbers is $1+2+3+\ldots \ldots \ldots+n=\frac{n(n+1)}{2}$ Hence arithmetic mean of first $n$ natural numbers $=\frac{\text { Sum of numbers }}{\text { Total numbers }}$ $=\frac{n(n+1) / 2}{n}$ i.e Arithmetic mean of first $n$ natural number $=\frac{n+1}{2}$...
Read More →The vertex A of ∆ABC is joined to a point D on the side BC.
Question: The vertexAof∆ABCis joined to a pointDon the sideBC. The midpoint ofADisE. Prove that $\operatorname{ar}(\Delta B E C)=\frac{1}{2} \operatorname{ar}(\Delta A B C)$. Solution: Given: Dis the midpoint ofBCand E is the midpoint ofAD. To prove: $\operatorname{ar}(\Delta B E C)=\frac{1}{2} \operatorname{ar}(\Delta A B C)$ Proof:SinceEis the midpoint ofAD,BEis the median of∆ABD.We know that a median of a triangle divides it into two triangles of equal areas. i.e., $\operatorname{ar}(\Delta B...
Read More →The number of terms in an A.P. whose first term is 10,
Question: The number of terms in an A.P. whose first term is 10, last term is 50 and the sum of all terms is 300, is ___________. Solution: First term of A.P is given = 10 Last term of A.P is given = 50 Sum of all termsSn= 300 (given) Since sum of all terms $S_{n}=\frac{n \text { (First term+Last term) }}{2}$ i. e $300=n\left(\frac{10+50}{2}\right)^{2}$ i. e $300=n\left(\frac{60}{2}\right)$ i. e $300=n(30)$ i. e $n=10$ Hence, number of terms of an A.P is 10....
Read More →If the angles of elevation of the top of a tower from two points distant a and b from
Question: If the angles of elevation of the top of a tower from two points distant a and b from the base and in the same straight line with it are complementary, then the height of the tower is (a) $a b$ (b) $\sqrt{a b}$ (c) $\frac{a}{b}$ (d) $\sqrt{\frac{a}{b}}$ Solution: Letbe the height of tower. Given that: angle of elevation of top of the tower areand. Distanceand Here, we have to find the height of tower. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $...
Read More →If the first, second and last terms of an A.P. are a, b and 2a respectively,
Question: If the first, second and last terms of an A.P. area,band 2arespectively, then the sum of its terms is ___________. Solution: Let first term of A.P bea Second term of A.P beband last term of A.P be given by 2a Sinced=a2a1=ba $n^{\text {th }}$ term i. e $a_{n}=a+(n-1) d$ $\Rightarrow a+(n-1)(b-a)=2 a$ $\Rightarrow(n-1)(b-a)=a$ i. e $n-1=\frac{a}{b-a}$ i. e $n=\frac{a}{b-a}+1=\frac{b}{b-a}$ Sni.e sum ofnterms of A.P is $\frac{n}{2}\left(a_{1}+a_{n}\right)$ $=\frac{n}{2}(a+2 a)$ $=\frac{b}...
Read More →The minimum value of
Question: The minimum value of 4x+ 41 x,xR, is ___________. Solution: Since Arithmetic mean geometric mean of 4xand 41 x i. e $\frac{4^{x}+4^{-x}}{2} \geq \sqrt{4^{x}} \cdot 4^{1-x}$ i. e $\frac{4^{x}+4^{1-x}}{2} \geq \sqrt{4^{x} \cdot 4 \cdot 4^{-x}}$ i. e $\frac{4^{x}+4^{1-x}}{2} \geq 2$ i. e $4^{x}+4^{1-x} \geq 2 \times 2=4$ i.e minimum value of4x+ 41 xis 4....
Read More →In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O.
Question: In the adjoining figure, the diagonalsACandBDof a quadrilateralABCDintersect atO.IfBO=OD, prove thatar(∆ABC) = ar(∆ADC), Solution: Given: BO=ODTo prove:ar(∆ABC) = ar(∆ADC)Proof:SinceBO=OD,Ois the mid point ofBD.We know that a median of a triangle divides it into two triangles of equal areas.COis a median of∆BCD.i.e., ar(∆COD)=ar(∆COB) ...(i)AOis amedian of∆ABD.i.e., ar(∆AOD)=ar(∆AOB) ...(ii)From (i) and (ii), we have:ar(∆COD)+ ar(∆AOD)=ar(∆COB) +ar(∆AOB) ar(∆ADC) =ar(∆ABC)...
Read More →Let
Question: Let $A=\left[\begin{array}{rrr}-1 0 2 \\ 3 1 4\end{array}\right], B=\left[\begin{array}{lll}0 -2 5 \\ 1 -3 1\end{array}\right]$ and $C=\left[\begin{array}{rrr}1 -5 2 \\ 6 0 -4\end{array}\right]$. Compute $2 A-3 B+4 C$. Solution: Here, $2 A-3 B+4 C=2\left[\begin{array}{ccc}-1 0 2 \\ 3 1 4\end{array}\right]-3\left[\begin{array}{ccc}0 -2 5 \\ 1 -3 1\end{array}\right]+4\left[\begin{array}{ccc}1 -5 2 \\ 6 0 -4\end{array}\right]$ $\Rightarrow 2 A-3 B+4 C=\left[\begin{array}{ccc}-2 0 4 \\ 6 2...
Read More →If the angles of elevation of a tower from two points distant a and b (a>b)
Question: If the angles of elevation of a tower from two points distant a and b (ab) from its foot and in the same straight line from it are 30 and 60, then the height of the tower is (a) $\sqrt{a+b}$ (b) $\sqrt{a b}$ (c) $\sqrt{a-b}$ (d) $\sqrt{\frac{a}{b}}$ Solution: Letbe the height of tower Given that: angle of elevation areand. Distanceand Here, we have to find the height of tower. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $\Rightarrow \tan 60^{\cir...
Read More →The sum of the terms equidistant from the beginning and end in an A.P.
Question: The sum of the terms equidistant from the beginning and end in an A.P. is always same and is equal to the sum of ________ and ________ terms. Solution: Leta1,a2, ............anbe an A.P with common differenced. Thenkthterm isak=a1+ (k 1)dandkthterm from last is the (n k+ 1)thterm from beginning i. e $a_{n-k+1}=a_{1}+(n-k+1-1) d$ $a_{n-k+1}=a_{1}+(n-k) d$ kthterm from beginning +kthterm from the end is $a_{1}+(k-1) d+a_{1}+(n-k) d$ $=2 a_{1}+d(k-1+n-k) d$ $=2 a_{1}+(n-1) d$ $=a_{1}+\lef...
Read More →P is any point on the diagonal AC of a parallelogram ABCD.
Question: Pis any point on the diagonalACof a parallelogramABCD. Prove that ar(∆ADP) = ar(∆ABP). Solution: Join BD.Let BD and AC intersect at point O.O is thus the midpoint of DB and AC.PO is the median of△DPB,So, $\operatorname{ar}(\triangle \mathrm{DPO})=\operatorname{ar}(\triangle \mathrm{BPO}) \quad \ldots(1)$ $\operatorname{ar}(\triangle \mathrm{ADO})=\operatorname{ar}(\triangle \mathrm{ABO}) \quad \ldots . .(2)$ Case 1 . $(2)-(1)$ $\Rightarrow \operatorname{ar}(\triangle \mathrm{ADO})-\ope...
Read More →if
Question: If $A=\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right], B=\left[\begin{array}{rrr}-1 0 2 \\ 3 4 1\end{array}\right], C=\left[\begin{array}{rrr}-1 2 3 \\ 2 1 0\end{array}\right]$, find (i) $A+B$ and $B+C$ (ii) $2 B+3 A$ and $3 C-4 B$. Solution: (i) $A+B=\left[\begin{array}{ll}2 3 \\ 5 7\end{array}\right]+\left[\begin{array}{ccc}-1 0 2 \\ 3 4 1\end{array}\right]$ It is not possible to add these matrices because the number of elements inAare not equal to thenumber of elements inB. So,A...
Read More →If 9 times the 9th term of an A.P. is equal to 13 times the 13th term,
Question: If 9 times the 9thterm of an A.P. is equal to 13 times the 13th term, then the 22ndterm of the A.P. is (a) 0 (b) 22 (c) 220 (d) 198 Solution: Let a denote the first term of A.P andddenote the common difference of the A.P Then ninth term isa +8dand 13thterm isa+ 12d According to given condition, 9(a+ 8d) = 13(a+ 12d) i.e 9a+ 72d= 13a+ 156d i.e 156d+ 72d =4a i.e 84d= 4a i.ea= 21d i.ea+ 21d= 0 which represents 22ndterm of A.P. The 22ndterm of the A.P is 0 Hence, the correct answer is opti...
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