If the altitude of the sum is at 60°,
Question: If the altitude of the sum is at 60, then the height of the vertical tower that will cast a shadow of length 30 m is (a) $10 \sqrt{3} \mathrm{~m}$ (b) $15 \mathrm{~m}$ (c) $\frac{30}{\sqrt{3}} m$ (d) $15 \sqrt{2} \mathrm{~m}$ Solution: Letbe the height of vertical tower Given that: altitude of sun isand shadow of lengthmeters. 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^{\circ}=\frac...
Read More →If the sum of n terms of an A.P. is given by
Question: If the sum ofnterms of an A.P. is given bySn= 3n+ 2n2, then the common difference of the A.P. is (a) 3 (b) 2 (c) 6 (d) 4 Solution: LetSndenote sum ofnterms of an A.P Such thatSn= 3n+ 2n2 i.eS1= 3(1) + 2(1)2 S1= 3 + 2 = 5 whereS1=a(first term only) S2=a+ (a+d) wherea + drepresents second term anddis common difference. $S_{2}=3(2)+2(2)^{2}$ $=6+2(4)$ $=6+8$ $S_{2}=14$ i.e2a+d= 14 i.e 10 +d= 14 d= 4 i.e common difference of A.P is 4 Hence, the correct answer is option D....
Read More →D and E are points on sides AB and AC respectively of ∆ABC such that ar(∆BCD)
Question: DandEare points on sidesABandACrespectively of ∆ABCsuch that ar(∆BCD) = ar(∆BCE). Prove thatDE||BC. Solution: ar(∆BCD) = ar(∆BCE) (Given)We know, triangles on the same base and having equal areas lie between the same parallels.Thus, DE || BC....
Read More →In the adjoining figure, ABC and ABD are two triangles on the same base AB.
Question: In the adjoining figure,ABCandABDare two triangles on the same baseAB. If line segmentCDis bisected byABatO, show that ar(ΔABC) = ar(ΔABD) Solution: Line segmentCDis bisected byABatO (Given)CO = OD .....(1)InΔCAO,AO is the median. (From (1))So, arΔCAO =arΔDAO .....(2)Similarly,InΔCBD,BO is the median (From (1))So,arΔCBO =arΔDBO .....(3)From (2) and (3) we havearΔCAO + arΔCBO =arΔDBO +arΔDAO⇒⇒ar(ΔABC) = ar(ΔABD)...
Read More →If the angle of elevation of a tower from a distance of 100 metres from its foot is 60°
Question: If the angle of elevation of a tower from a distance of 100 metres from its foot is 60, then the height of the tower is (a) $100 \sqrt{3} \mathrm{~m}$ (b) $\frac{100}{\sqrt{3}} m$ (c) $50 \sqrt{3}$ (d) $\frac{200}{\sqrt{3}} m$ Solution: Letbe the height of tower ismeters Given that: angle of elevation isfrom tower of foot and distancemeters. 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 6...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: Let Sndenote the sum of firstnterms of an A.P. IfS2n= 3Sn, thenS3n:Snis equal to (a) 4 (b) 6 (c) 8 (d) 10 Solution: As, $S_{2 n}=3 S_{n}$ $\Rightarrow \frac{2 n}{2}[2 a+(2 n-1) d]=\frac{3 n}{2}[2 a+(n-1) d]$ $\Rightarrow 2[2 a+(2 n-1) d]=3[2 a+(n-1) d]$ $\Rightarrow 4 a+2(2 n-1) d=6 a+3(n-1) d$ $\Rightarrow 4 a+4 n d-2 d=6 a+3 n d-3 d$ $\Rightarrow 6 a-4 a+3 n d-3 d-4 n d+2 d=0$ $\Rightarrow 2 a-n d-d=0$ $\Rightarrow 2 a-(n+1) d=0...
Read More →Show that a diagonal divides a parallelogram into two triangles of equal area.
Question: Show that a diagonal divides a parallelogram into two triangles of equal area. Solution: LetABCDbe a parallelogram and BD be its diagonal.To prove:ar(∆ABD) =ar(∆CDB)Proof:In ∆ABDand∆CDB,we have:AB = CD [Opposite sides of a parallelogram]AD = CB [Opposite sides of a parallelogram] BD = DB [Common] i.e., ∆ABD∆CDB [ SSS criteria] ar(∆ABD) =ar(∆CDB)...
Read More →Prove that a median divides a triangle into two triangles of equal area.
Question: Prove that a median divides a triangle into two triangles of equal area. Solution: LetADis a median of∆ABCandDis the midpoint ofBC. AD divides∆ABCin two triangles: ∆ABDand∆ADC.To prove:ar(∆ABD) =ar(∆ADC)Construction:DrawALBC.Proof:SinceDis the midpoint ofBC, we have:BD = DC Multiplying with $\frac{1}{2} A L$ on both sides, we get: $\frac{1}{2} \times B D \times A L=\frac{1}{2} \times D C \times A L$ ⇒ ar(∆ABD) =ar(∆ADC)...
Read More →Mark the correct alternative in each of the following:
Question: Mark the correct alternative in each of the following: The ratio of the length of a rod and its shadow is $1: \sqrt{3}$. The angle of elevation of the sum is The angle of elevation of the sum is(a) 30(b) 45(c) 60(d)90 Solution: Letbe angle of elevation of sun. Given that: length of roadand its shadow Here, we have to find angle of elevation of sun. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $\Rightarrow \tan \theta=\frac{1}{\sqrt{3}}$ $\left[\be...
Read More →In the adjoining figure, DE || BC. Prove that
Question: In the adjoining figure,DE||BC. Prove that(i) ar(∆ACD) = ar(∆ABE),(ii) ar(∆OCE) = ar(∆OBD), Solution: ∆DECand∆DEBlies on the same base and between the same parallel lines.So, ar(∆DEC) = ar(∆DEB) ...(1)(i) On addingar(∆ADE) in both sides of equation (1), we get:ar(∆DEC) + ar(∆ADE) = ar(∆DEB) +ar(∆ADE) ⇒ar(∆ACD) = ar(∆ABE)(ii) On subtracting ar(ODE) from both sides of equation (1), we get:ar(∆DEC)-ar(∆ODE) = ar(∆DEB)-ar(∆ODE) ⇒ ar(∆OCE) = ar(∆OBD)...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: If in an A.P. $S_{n}=n^{2} q$ and $S_{m}=m^{2} q$, where $S_{r}$ denotes the sum of $r$ terms of the A.P., then $S_{q}$ equals (a) $\frac{q^{3}}{2}$ (b) mnq (c) $q^{3}$ (d) $\left(m^{2}+n^{2}\right) q$ Solution: As, $S_{n}=n^{2} q$ $\Rightarrow \frac{n}{2}[2 a+(n-1) d]=n^{2} q$ $\Rightarrow 2 a+(n-1) d=\frac{n^{2} q \times 2}{n}$ $\Rightarrow 2 a+(n-1) d=2 n q$ ....(1) Also, $S_{m}=m^{2} q$ $\Rightarrow \frac{m}{2}[2 a+(m-1) d]=m^...
Read More →In Fig. 12.58, what are the angles of depression from
Question: In Fig. 12.58, what are the angles of depression from the observing position O1and O2of the object at A? Solution: In a triangle, $\Rightarrow \angle A=180^{\circ}-\left(90^{\circ}+60^{\circ}\right)$ $\Rightarrow \angle A=180^{\circ}-150^{\circ} \quad$ We know that $\left[\angle A+\angle B+\angle C=180^{\circ}\right]$ $\Rightarrow \angle A=30^{\circ}$ Again, In a triangle, $\Rightarrow \angle O_{2}=180^{\circ}-\left(90^{\circ}+45^{\circ}\right)$ $\Rightarrow \angle O_{2}=180^{\circ}-13...
Read More →In the adjoining figure, ABCD is a trapezium in which AB || DC and its diagonals AC and BD intersect at O.
Question: In the adjoining figure,ABCDis a trapezium in whichAB||DCand its diagonalsACandBDintersect atO. Prove that ar(∆AOD) = ar(∆BOC). Solution: ∆CDAand∆CBDlies on the same base and between the same parallel lines.So, ar(∆CDA) = ar(CDB) ...(i)Subtracting ar(∆OCD) from both sides of equation (i), we get:ar(∆CDA)-ar(∆OCD) = ar(∆CDB)-ar (∆OCD)⇒ ar(∆AOD) = ar(∆BOC)...
Read More →Let
Question: Let $A=\left[\begin{array}{ll}2 4 \\ 3 2\end{array}\right], B=\left[\begin{array}{rr}1 3 \\ -2 5\end{array}\right]$ and $C=\left[\begin{array}{rr}-2 5 \\ 3 4\end{array}\right]$. Find each of the following: (i) $2 A-3 B$ (ii) $B-4 C$ (iii) $3 A-C$ (iv) $3 A-2 B+3 C$ Solution: (i) $2 A-3 B$ $\Rightarrow 2 A-3 B=2\left[\begin{array}{ll}2 4 \\ 3 2\end{array}\right]-3\left[\begin{array}{cc}1 3 \\ -2 5\end{array}\right]$ $\Rightarrow 2 A-3 B=\left[\begin{array}{ll}4 8 \\ 6 4\end{array}\right...
Read More →If the angles of elevation of the top of a tower from two points at a distance of
Question: If the angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower in the same straight line with it are complementary, find the height of the tower. Solution: Letbe the height of tower ismeters. Given that angle of elevation are $\angle B=90^{\circ}-\theta$ and $\angle D=\theta$ and also $C D=4 \mathrm{~m}$ and $B C=9 \mathrm{~m}$. Here we have to find height of tower. So we use trigonometric ratios. In a triangle, $\tan \theta=\fr...
Read More →In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that
Question: In the adjoining figure,MNPQandABPQare parallelograms andTis any point on the side BP. Prove that(i) ar(MNPQ) = are(ABPQ) (ii) $\operatorname{ar}(\triangle A T Q)=\frac{1}{2} \operatorname{ar}(M N P Q)$. Solution: (i) We know that parallelograms on the same base and between the same parallels are equal in area.So,ar(MNPQ) = are(ABPQ) (Same base PQ and MB || PQ) .....(1)(ii) If a parallelogram and a triangle are on the same base and between the same parallels then the area of the triang...
Read More →If P and Q are any two points lying respectively on the sides DC and AD
Question: IfPandQare any two points lying respectively on the sidesDCandADof a parallelogramABCDthen show that ar(∆APB) = ar(∆BQC). Solution: We know $\operatorname{ar}(\Delta \mathrm{APB})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$ .....(1) [If a triangle and a parallelogram are on the same base and between the same parallels then the area of the triangle is equal to half the area of the parallelogram] Similarly, $\operatorname{ar}(\Delta \mathrm{BQC})=\frac{1}{2} \operatorname{ar}(\mathrm{A...
Read More →From a point on the ground, 20 m away from the foot of a vertical tower
Question: From a point on the ground, 20 m away from the foot of a vertical tower, the angle elevation of the top of the tower is 60, What is the height of the tower? Solution: Letbe the height of tower ismeters. Given that: angle of elevation isandmeters. Here we have to find height of tower. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $\Rightarrow \tan 60^{\circ}=\frac{h}{20}\left[\because \tan 60^{\circ}=\sqrt{3}\right]$ $\Rightarrow \sqrt{3}=\frac{h}{2...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: The 10th common term between the A.P.s 3, 7, 11, 15, ... and 1, 6, 11, 16, ... is (a) 191 (b) 193 (c) 211 (d) none of these Solution: As, the common difference of the A.P. $3,7,11,15, \ldots=7-3=4$ and the common difference of the A.P. $1,6,11,16, \ldots=6-1=5$ And, the common terms of both the A.P.s will be in A.P. So, the common difference of the A.P. of the common terms, $d=\operatorname{LCM}(4,5)=4 \times 5=20$ and its first c...
Read More →What is the angle of elevation of the Sun when the
Question: What is the angle of elevation of the Sun when the length of the shadow of a vetical pole is equal to its height? Solution: Letbe the angle of elevation of sun is. Given that: Height of pole ismeters and length of shadow ismeters. Because length of shadow is equal to the height of pole. Here we have to find angle of elevation of sun. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan \theta=\frac{A B}{B C}$ $\Rightarrow \tan \theta=\frac{h}{h} \quad\left[\because \tan 45^...
Read More →In the adjoining figure, ABCD is a quadrilateral in which diag.
Question: In the adjoining figure,ABCDis a quadrilateral in which diag.BD= 14 cm. IfALBDandCMBDsuch thatAL= 8 cm andCM= 6 cm, find the area of quad.ABCD. Solution: $\operatorname{ar}($ quad $A B C D)=\operatorname{ar}(\triangle A B D)+\operatorname{ar}(\triangle B D C)$ $=\frac{1}{2} \times B D \times A L+\frac{1}{2} \times B D \times C M$ $=\frac{1}{2} \times B D \times(A L+C M)$ By substituting the values, we have; $\operatorname{ar}($ quad $A B C D)=\frac{1}{2} \times 14 \times(8+6)$ = 7 ⨯14...
Read More →Mark the correct alternative in the following question:
Question: Mark the correct alternative in the following question: If in an A.P., the pth term isqand (p+q)th term is zero, then theqth term is (a)-p (b)p (c)p+q (d)p-q Solution: As, $a_{p}=q$ $\Rightarrow a+(p-1) d=q \quad \ldots \ldots(\mathrm{i})$ Also, $a_{(p+q)}=0$ $\Rightarrow a+(p+q-1) d=0 \quad \ldots$. (ii) Subtracting (i) from (ii), we get $a+(p+q-1) d-a-(p-1) d=0-q$ $\Rightarrow(p+q-1-p+1) d=-q$ $\Rightarrow q d=-q$ $\Rightarrow d=\frac{-q}{q}$ $\Rightarrow d=-1$ Substituting $d=-1$ in...
Read More →If the ratio of the height of a tower and the length of its shadow is
Question: If the ratio of the height of a tower and the length of its shadow is $\sqrt{3}: 1$, what is the angle of elevation of the Sun? Solution: Letbe the angle of elevation of sun is. Given that: Height of tower ismeters and length of shadow is 1. Here we have to find angle of elevation of sun. In a triangle, $\Rightarrow \tan \theta=\frac{A B}{B C}$ $\Rightarrow \tan \theta=\frac{\sqrt{3}}{1}\left[\because \tan 60^{\circ}=\sqrt{3}\right]$ $\Rightarrow \tan \theta=\sqrt{3}$ $\Rightarrow \the...
Read More →Compute the following sums:
Question: Compute the following sums: (i) $\left[\begin{array}{rr}3 -2 \\ 1 4\end{array}\right]+\left[\begin{array}{rr}-2 4 \\ 1 3\end{array}\right]$ (ii) $\left[\begin{array}{rrr}2 1 3 \\ 0 3 5 \\ -1 2 5\end{array}\right]+\left[\begin{array}{rrr}1 -2 3 \\ 2 6 1 \\ 0 -3 1\end{array}\right]$ Solution: (i) $\left[\begin{array}{cc}3 -2 \\ 1 4\end{array}\right]+\left[\begin{array}{cc}-2 4 \\ 1 3\end{array}\right] \Rightarrow\left[\begin{array}{cc}3-2 -2+4 \\ 1+1 4+3\end{array}\right] \Rightarrow\lef...
Read More →The height of a tower is 10 m.
Question: The height of a tower is 10 m. What is the length of its shadow when Sun's altitude is 45? Solution: Letbe the length of shadow ism Given that: Height of tower ismeters and altitude of sun is Here we have to find length of shadow. So we use trigonometric ratios. In a triangle, $\Rightarrow \tan C=\frac{A B}{B C}$ $\Rightarrow \tan 45^{\circ}=\frac{A B}{A C}$ $\Rightarrow 1=\frac{10}{x}$ $\Rightarrow x=10$ Hence the length of shadow is $10 \mathrm{~m}$....
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