M is the midpoint of the side AB of a parallelogram ABCD.
Question: Mis the midpoint of the sideABof a parallelogramABCD. If ar(AMCD) = 24 cm2, find ar(∆ABC). Solution: \ Join AC.AC divides parallelogram ABCD into two congruent triangles of equal areas. $\operatorname{ar}(\triangle \mathrm{ABC})=\operatorname{ar}(\triangle \mathrm{ACD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD})$ M is the midpoint of AB. So, CM is the median. $\mathrm{CM}$ divides $\triangle \mathrm{ABC}$ in two triangles with equal area. $\operatorname{ar}(\triangle \mathrm{AMC})=\o...
Read More →Solve the following
Question: If in an A.P.,Sn=n2pandSm=m2p, whereSrdenotes the sum ofrterms of the A.P., thenSpis equal to (a) $\frac{1}{2} p^{3}$ (b)mn p(c)P3(d) (m+n)p2 Solution: (c) $P^{3}$ Given: $S_{n}=n^{2} p$ $\Rightarrow \frac{n}{2}\{2 a+(n-1) d\}=n^{2} p$ $\Rightarrow 2 a+(n-1) d=2 n p$ $\Rightarrow 2 a=2 n p-(n-1) d$ ....(1) $S_{m}=m^{2} p$ $\Rightarrow \frac{m}{2}\{2 a+(m-1) d\}=m^{2} p$ $\Rightarrow 2 a+(m-1) d=2 m p$ $\Rightarrow 2 a=2 m p-(m-1) d$ ...(2) From $(1)$ and $(2)$, we have: $2 n p-(n-1) d=...
Read More →The angles of elevation of the top of a tower from two points
Question: 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 and in the same straight line with it are complementary. Prove that the height of the tower is 6m. Solution: Letbe tower of heightm and angle of elevation of the top of tower from two points areand Let,m andm and The corresponding figure is as follows So we use trigonometric ratios. In, $\Rightarrow \tan \theta=\frac{A B}{A C}$ $\Rightarrow \tan \theta=\frac{h}{4}$ Agai...
Read More →A flag-staff stands on the top of a 5 m high tower.
Question: A flag-staff stands on the top of a 5 m high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is 60 and from the same point, the angle of elevation of the top of the tower is 45. Find the height of the flag-staff. Solution: Letbe the tower height ofm. flag height ism and an angle of elevation of top of tower is 45 and an angle of elevation of the top of flag is 60. Let, $A C=h \mathrm{~m}$ and $B C=5 \mathrm{~m}$ and $\angle A D B=60^{\circ}, \angl...
Read More →BD is one of the diagonals of a quad.
Question: $B D$ is one of the diagonals of a quad. $A B C D$. If $A L \perp B D$ and $C M \perp B D$, show that ar (quad. $A B C D$ ) $=\frac{1}{2} \times B D \times(A L+C M)$. Solution: ar(quad.ABCD)=ar(∆ABD) +ar (∆DBC) $\operatorname{ar}(\Delta A B D)=\frac{1}{2} \times$ base $\times$ height $=\frac{1}{2} \times B D \times A L$ ...(i) $\operatorname{ar}(\Delta D B C)=\frac{1}{2} \times B D \times C L$ ...(ii) From (i) and (ii), we get: $\operatorname{ar}($ quad $A B C D)=\frac{1}{2} \times B ...
Read More →Solve the following
Question: If, $S_{1}$ is the sum of an arithmetic progression of ' $n$ ' odd number of terms and $S_{2}$ the sum of the terms of the series in odd places, then $\frac{S_{1}}{S_{2}}=$ (a) $\frac{2 n}{n+1}$ (b) $\frac{n}{n+1}$ (c) $\frac{n+1}{2 n}$ (d) $\frac{n+1}{n}$ Solution: (a) $\frac{2 n}{n+1}$ Letnbe an odd number. Given: $S_{1}=$ Sum of odd number of terms $=\frac{n}{2}\{2 a+(n-1) d\} \quad \ldots .(1)$ Since $n$ is odd, the number of odd places $=\frac{n+1}{2}$ $S_{2}=$ Sum of the terms of...
Read More →An observed from the top of a 150 m tall light house, the angles of depression
Question: An observed from the top of a 150 m tall light house, the angles of depression of two ships approaching it are 30 and 45. If one ship is directly behind the other, find the distance between the two ships. Solution: Letbe the light house ofm. and angle of depression of two ship C and D areandrespectively. Let, $B C=x, C D=y$ and $\angle A D B=30^{\circ}, \angle A C B=45^{\circ}$. We use trigonometric ratios. In a triangle, $\Rightarrow \tan 45^{\circ}=\frac{A B}{B C}$ $\Rightarrow 1=\fr...
Read More →In the adjoining figure, ABCD is a trapezium in which AB || DC;
Question: In the adjoining figure,ABCDis a trapezium in whichAB||DC;AB= 7 cm;AD=BC= 5 cm and the distance betweenABandDCis 4 cm. Find the length ofDCand hence, find the area of trap.ABCD. Solution: ∆ADL is aright angle triangle. So, $D L=\sqrt{\left(5^{2}-4^{2}\right)}=\sqrt{9}=3 \mathrm{~cm}$ Similarly, in ∆BMC, we have: $M C=\sqrt{\left(5^{2}-4^{2}\right)}=\sqrt{9}=3 \mathrm{~cm}$ DC= DL+LM+MC= 3 + 7 + 3 = 13 cm Now, ar(trapezium. $A B C D)=\frac{1}{2} \times($ sum of parallel sides $) \times(...
Read More →If the first, second and last term of an A.P are a, b and 2a respectively,
Question: If the first, second and last term of an A.P area,band 2arespectively, then its sum is (a) $\frac{a b}{2(b-a)}$ (b) $\frac{a b}{b-a}$ (c) $\frac{3 a b}{2(b-a)}$ (d) none of these Solution: (c) $\frac{3 a b}{2(b-a)}$ Let the A.P. bea, a+d, a+2d........a+nd. Here, letdbe the common difference andnbe the total number of terms. Given: $a_{1}=a$ $a_{2}=b$ $\Rightarrow a+d=b$ $\Rightarrow d=b-a$ ....(1) $a_{n}=2 a$ $\Rightarrow a+(n-1) d=2 a$ $\Rightarrow(n-1) d=a$ $\Rightarrow d=\frac{a}{n-...
Read More →The angles of elevation of the top of a rock from the top and foot
Question: The angles of elevation of the top of a rock from the top and foot of a 100 m high tower are respectively 30 and 45. Find the height of the rock. Solution: Letbe the height of Rock which ism. and makes an angle of elevations 45 and 30 respectively from the bottom and top of tower whose height ism. Let $A E=h \mathrm{~m}, B C=x \mathrm{~m}$ and $C D=100 \cdot \angle A C B=45^{\circ}, \angle A D E=30^{\circ}$ We have to find the height of the rock We have the corresponding figure as So w...
Read More →(i) Calculate the area of quad. ABCD, given in Fig.
Question: (i) Calculate the area of quad.ABCD, given in Fig. (i).(ii) Calculate the area of trap.PQRS, given in Fig. (ii). Solution: (i) In△△BCD, $\mathrm{DB}^{2}+\mathrm{BC}^{2}=\mathrm{DC}^{2}$ $\Rightarrow \mathrm{DB}^{2}=17^{2}-8^{2}=225$ $\Rightarrow \mathrm{DB}=15 \mathrm{~cm}$ $\operatorname{Ar}(\triangle B C D)=\frac{1}{2} \times b \times h=\frac{1}{2} \times 8 \times 15=60 \mathrm{~cm}^{2}$ In $\triangle B A D$ $\mathrm{DA}^{2}+\mathrm{AB}^{2}=\mathrm{DB}^{2}$ $\Rightarrow \mathrm{AB}^{...
Read More →If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers,
Question: If the sum of firstneven natural numbers is equal toktimes the sum of firstnodd natural numbers, thenk= (a) $\frac{1}{n}$ (b) $\frac{n-1}{n}$ (c) $\frac{n+1}{2 n}$ (d) $\frac{n+1}{n}$ Solution: (d) $\frac{n+1}{n}$ Given: Sum of the even natural numbers =kSum of the odd natural numbers $\frac{n}{2}\{2 a+(n-1) d\}=k \times \frac{n}{2}\{2 a+(n-1) d\}$ $\Rightarrow\{2 \times 2+(n-1) 2\}=k \times\{2 \times 1+(n-1) 2\}$ $\Rightarrow \frac{4+(n-1) 2}{2+(n-1) 2}=k$ $\Rightarrow \frac{n+1}{n}=k...
Read More →Prove that
Question: Prove that $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$ Solution: $\cos ^{-1} \frac{12}{13}$ $=\sin ^{-1} \sqrt{1-\left(\frac{12}{13}\right)^{2}}$ $=\sin ^{-1} \sqrt{1-\frac{144}{169}}$ $\left(\cos ^{-1} x=\sin ^{-1} \sqrt{1-x^{2}}\right)$ $=\sin ^{-1} \sqrt{\frac{25}{169}}$ $=\sin ^{-1} \frac{5}{13}$ $\therefore \cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$ $=\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{3}{5}$ $=\sin ^{-1}\left(\frac{5}{13} \times \sqrt{1-...
Read More →The horizontal distance between two poles is 15 m.
Question: The horizontal distance between two poles is 15 m. The angle of depression of the top of the first pole as seen from the top of the second pole is 30. If the height of the second pole is 24 m, find the height of the first pole.$(\sqrt{3}=1.732)$ Solution: Let AB be the first pole and CD be the second pole.Distance between the two poles, BD = 15 mHeight of the second pole, CD = 24 mSuppose the height of the first pole behm.Draw AE CD. CE = CD ED = (24 h) m [AB = ED =hm]AE = BD = 15 m No...
Read More →Find the area of a trapezium whose parallel sides are 9 cm
Question: Find the area of a trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm. Solution: $\operatorname{ar}($ trapezium $)=\frac{1}{2} \times($ sum of parallel sides $) \times($ distance between them $)$ $=\frac{1}{2} \times(9+6) \times 8$ = 60 cm2Hence, the area of the trapezium is60cm2....
Read More →Prove that
Question: Prove that $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$ Solution: L. H. S $=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$ $=\cos ^{-1}\left[\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \sqrt{1-\left(\frac{12}{13}\right)^{2}}\right]$ $\left[\because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\right]$ $=\cos ^{-1}\left[\frac{48}{65}-\frac{3}{5} \times \frac{5}{13}\right]$ $=\cos ^{-1}\left(\...
Read More →Find the area of a figure formed by joining the midpoints
Question: Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm. Solution: LetABCDbe a rhombus andP,Q, RandSbe the midpoints ofAB, BC, CDandDA, respectively.Join the diagonals,ACandBD.In∆ABC,we have: $P Q \| A C$ and $P Q=\frac{1}{2} A C$ [By midpoint theorem] $\mathrm{PQ}=\frac{1}{2} \times 16=8 \mathrm{~cm}$ Again, in∆DAC,the pointsSandRare the midpoints ofADandDC,respectively. $\therefore S R \| A C$ and $S R=\frac{1}{2} A ...
Read More →The angles of depression of two ships from the top of a light house and on the same side
Question: The angles of depression of two ships from the top of a light house and on the same side of it are found to be 45 and 30 respectively. If the ships are 200 m apart, find the height of the light house. [CBSE 2012] Solution: Let CD be the the light house and A and B be the positions of the two ships.AB = 200 m (Given)Suppose CD =hm and BC =xmNow,DAC =ADE = 30 (Alternate angles)DBC =EDB = 45 (Alternate angles)In right ∆BCD, $\tan 45^{\circ}=\frac{\mathrm{CD}}{\mathrm{BC}}$ $\Rightarrow 1=...
Read More →The first and last term of an A.P. are a and l respectively.
Question: The first and last term of an A.P. are a and $/$ respectively. If $S$ is the sum of all the terms of the A.P. and the common difference is given by $\frac{l^{2}-a^{2}}{k-(l+a)}$, then $k=$ (a)S (b) 2S (c) 3S (d) none of these Solution: (b) 2S Given: $S=\frac{n}{2}(l+a)$ $\Rightarrow(l+a)=\frac{2 S}{n}$ Also, $d=\frac{l^{2}-a^{2}}{k-(l+a)}$ $\Rightarrow d=\frac{(l+a)(l-a)}{k-(l+a)}$ $\Rightarrow d=\frac{[(n-1) d] \times \frac{2 S}{n}}{k-\frac{25}{n}}$ $\Rightarrow k-\frac{2 S}{n}=(n-1) ...
Read More →Solve
Question: Solve $\cos ^{-1} \sqrt{3} x+\cos ^{-1} x=\frac{\pi}{2}$ Solution: $\cos ^{-1} \sqrt{3} x+\cos ^{-1} x=\frac{\pi}{2}$ $\Rightarrow \cos ^{-1}\left[\sqrt{3} x \times x-\sqrt{1-(\sqrt{3} x)^{2}} \sqrt{1-x^{2}}\right]=\frac{\pi}{2}$ $\left[\because \cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)\right]$ $\Rightarrow \cos ^{-1}\left[\sqrt{3} x^{2}-\sqrt{1-3 x^{2}} \sqrt{1-x^{2}}\right]=\frac{\pi}{2}$ $\Rightarrow \sqrt{3} x^{2}-\sqrt{1-3 x^{2}} \sqrt{1-x^...
Read More →In a parallelogram ABCD, it is being given that AB = 10 cm
Question: In a parallelogramABCD, it is being given thatAB= 10 cm and the altitudes corresponding to the sidesABandADareDL= 6 cm andBM= 8 cm, respectively. FindAD. Solution: ar(parallelogramABCD) = base⨯ height⇒AB⨯DL=AD⨯BM⇒ 10⨯ 6 =AD⨯BM⇒AD⨯ 8 = 60 cm2⇒AD= 7.5 cmAD= 7.5 cm...
Read More →Solve the equation
Question: Solve the equation $\cos ^{-1} \frac{a}{x}-\cos ^{-1} \frac{b}{x}=\cos ^{-1} \frac{1}{b}-\cos ^{-1} \frac{1}{a}$ Solution: $\cos ^{-1} \frac{a}{x}-\cos ^{-1} \frac{b}{x}=\cos ^{-1} \frac{1}{b}-\cos ^{-1} \frac{1}{a}$ $\Rightarrow \cos ^{-1} \frac{a}{x}+\cos ^{-1} \frac{1}{a}=\cos ^{-1} \frac{1}{b}+\cos ^{-1} \frac{b}{x}$ $\Rightarrow \cos ^{-1}\left[\frac{a}{x} \times \frac{1}{a}-\sqrt{1-\left(\frac{a}{x}\right)^{2}} \sqrt{1-\left(\frac{1}{a}\right)^{2}}\right]=\cos ^{-1}\left[\frac{b}...
Read More →A straight highway leads to the foot of a tower of height 50 m.
Question: A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, the angles of depression of two cars standing on the highway are 30 and 60 respectively. What is the distance the two cars and how far is each car from the tower? Solution: Letbe the height of towerm and angle of depression from the top of tower areandrespectively at two observing Car C and D. Letm,m and, We have the corresponding figure as follows So we use trigonometric ratios. In a triangle, $...
Read More →Solve the following
Question: LetSndenote the sum ofnterms of an A.P. whose first term isa. If the common differencedis given byd=SnkSn1+Sn 2, thenk= (a) 1 (b) 2 (c) 3 (d) none of these Solution: (b) 2 Let the A.P. bea, a+d, a+2d, a+3d... Given: $d=S_{n}-k S_{n-1}+S_{n-2}$ Forn= 3, we have: $d=(3 a+3 d)-k(2 a+d)+a$ $\Rightarrow 4 a+2 d-k(2 a+d)=0$ $\Rightarrow 2(2 a+d)=k(2 a+d)$ $\Rightarrow 2=k$...
Read More →In the adjoining figure, show that ABCD is a parallelogram.
Question: In the adjoining figure, show thatABCDis a parallelogram.Calculate the area of || gmABCD. Solution: Given: A quadrilateralABCDandBDis a diagonal.To prove:ABCDis a parallelogram.Construction: Draw AM DC and CL AB (extend DC and AB). Join AC, the other diagonal of ABCD.Proof: ar(quad.ABCD) = ar(∆ABD) + ar(∆DCB)= 2 ar(∆ABD) [∵ ar(∆ABD) = ar(∆DCB)] $\therefore \operatorname{ar}(\Delta A B D)=\frac{1}{2} \operatorname{ar}($ quad. $A B C D)$ ...(i) Again, ar(quad.ABCD) =ar(∆ABC) +ar(∆CD...
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