In the given figure, ∠A = ∠CED, prove that ∆CAB ∼ ∆CED. Also, find the value of x.
Question: In the given figure, A = CED, prove that ∆CAB ∆CED. Also, find the value ofx. Solution: ComparingΔCABandΔCED, CAB=CED[Given] ACB=ECD[Common] $\therefore \triangle C A B \sim \triangle C E D$ $\Rightarrow \frac{C A}{C E}=\frac{A B}{E D}$ [In similar triangles, corresponding sides are in the same proportion] $\Rightarrow \frac{15 \mathrm{~cm}}{10 \mathrm{~cm}}=\frac{9 \mathrm{~cm}}{x}$ $\Rightarrow x=\frac{9 \times 10}{15} \mathrm{~cm}=6 \mathrm{~cm}$...
Read More →Prove that:
Question: (i) tan 8x tan 6x tan 2x= tan 8xtan 6xtan 2x (ii) $\tan \frac{\pi}{12}+\tan \frac{\pi}{6}+\tan \frac{\pi}{12} \tan \frac{\pi}{6}=1$ (iii) tan 36 + tan 9 + tan 36 tan 9 = 1 (iv) tan 13x tan 9x tan 4x= tan 13xtan 9xtan 4x Solution: (i) We know that $8 x=6 x+2 x$ Therefore, $\tan (8 x)=\tan (6 x+2 x)$ $\Rightarrow \tan (8 x)=\frac{\tan 6 x+\tan 2 x}{1-\tan 6 x \tan 2 x}$ $\Rightarrow \tan 8 x-\tan 8 x \tan 6 x \tan 2 x=\tan 6 x+\tan 2 x$ $\Rightarrow \tan 8 x-\tan 6 x-\tan 2 x=\tan 8 x \t...
Read More →In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:
Question: In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that: (i) $\triangle \mathrm{OMA} \sim \triangle \mathrm{OLC}$ (ii) $\mathrm{OAOC}=\mathrm{OMOL}$ Solution: i. In $\triangle O M A$ and $\triangle O L C, \angle A O M=\angle C O L$ Vertically opposite angles $\angle O M A=\angle O L C \quad 90^{\circ}$ each $\Rightarrow \triangle O M A \sim \triangle O L C$ AA similarityii. Since $\triangle O M A \sim \trian...
Read More →Show that
Question: $x \frac{d y}{d x}+2 y=x^{2} \log x$ Solution: The given differential equation is: $x \frac{d y}{d x}+2 y=x^{2} \log x$ $\Rightarrow \frac{d y}{d x}+\frac{2}{x} y=x \log x$ This equation is in the form of a linear differential equation as: $\frac{d y}{d x}+p y=Q$ (where $p=\frac{2}{x}$ and $Q=x \log x$ ) Now, I.F $=e^{\int p d x}=e^{\int_{x}^{2} d x}=e^{2 \log x}=e^{\log x^{2}}=x^{2}$ The general solution of the given differential equation is given by the relation, $y($ I.F. $)=\int($ ...
Read More →D is the mid-point of side BC of a ∆ABC.
Question: D is the mid-point of side BC of a ∆ABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE = EX = 3 : 1 Solution: Given:ABC is a triangle in which D is the mid point of BC, E is the mid point of AD. BE produced meets AC at X.To Prove:BE : EX = 3:1.Construction:We draw a line DY parallel to BX. In $\triangle \mathrm{BCX}$ and $\triangle \mathrm{DCY}, \angle \mathrm{CBX}=\angle \mathrm{CDY}$ Corresponding angles $\angle \mathrm{CXB}=\angle \mathrm{CYD}$ ...
Read More →Prove that:
Question: Prove that: (i) $\frac{\sin (A+B)+\sin (A-B)}{\cos (A+B)+\cos (A-B)}=\tan A$ (ii) $\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}=0$ (iii) $\frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A}=0$ (iv) $\sin ^{2} B=\sin ^{2} A+\sin ^{2}(A-B)-2 \sin A \cos B \sin (A-B)$ (v) $\cos ^{2} A+\cos ^{2} B-2 \cos A \cos B \cos (A+B)=\sin ^{2}(A+B)$ (vi) $\frac{\tan (A+B)}{\cot (A-B)}=\frac{\tan ^{2}...
Read More →Prove that:
Question: Prove that: (i) $\frac{\sin (A+B)+\sin (A-B)}{\cos (A+B)+\cos (A-B)}=\tan A$ (ii) $\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin (C-A)}{\cos C \cos A}=0$ (iii) $\frac{\sin (A-B)}{\sin A \sin B}+\frac{\sin (B-C)}{\sin B \sin C}+\frac{\sin (C-A)}{\sin C \sin A}=0$ (iv) $\sin ^{2} B=\sin ^{2} A+\sin ^{2}(A-B)-2 \sin A \cos B \sin (A-B)$ (v) $\cos ^{2} A+\cos ^{2} B-2 \cos A \cos B \cos (A+B)=\sin ^{2}(A+B)$ (vi) $\frac{\tan (A+B)}{\cot (A-B)}=\frac{\tan ^{2}...
Read More →In the given figure, PA, QB and RC are each perpendicular to AC.
Question: In the given figure, PA, QB and RC are each perpendicular to AC. Prove that1x+1z=1y. Solution: It is given that $P A, Q B$ and $R C$ are each perpendicular to $A C$. We have to prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$ In $\triangle P A C$ we have $B Q \| A P$ $\Rightarrow \frac{B Q}{A P}=\frac{C B}{C A}$ $\Rightarrow y x=C B C A$.....(1) Now in $\triangle A C R$, we have $B Q \| C R$ $\Rightarrow \frac{B Q}{C R}=\frac{A B}{A C}$ $\Rightarrow \frac{y}{z}=\frac{A B}{A C}$....(2) ...
Read More →Monica has a piece of Canvas whose area is 551m2.
Question: Monica has a piece of Canvas whose area is $551 \mathrm{~m}^{2}$. She uses it to have a conical tent made, with a base radius of $7 \mathrm{~m}$. Assuming that all the stitching margins and wastage incurred while cutting amounts to approximately $1 \mathrm{~m}^{2}$. Find the volume of the tent that can be made with it. Solution: It is given that: Area of the canvas $=551 \mathrm{~m}^{2}$ Area that is wasted $=1 \mathrm{~m}^{2}$ Radius of tent = 7m, Volume of tent (v) = ? Therefore the ...
Read More →In the given figure, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD.
Question: In the given figure, DE || BC such that AE = (1/4) AC. If AB = 6 cm, find AD. Solution: It is given that $D E \| B C, A E=\frac{1}{4} A C$ and $A B=6 \mathrm{~cm}$. We have to findAD. Since $\triangle A D E \sim \triangle A B C$ $\Rightarrow \frac{A D}{A B}=\frac{A E}{A C}$ So $\Rightarrow \frac{A D}{6 \mathrm{~cm}}=\frac{1 \mathrm{~cm}}{4 \mathrm{~cm}}$ $\Rightarrow 4 \mathrm{~cm} \times A D=6$ $\Rightarrow A D=\frac{6 \mathrm{~cm}}{4 \mathrm{~cm}}$ $\Rightarrow A D=\frac{3 \mathrm{~c...
Read More →Show that
Question: $\cos ^{2} x \frac{d y}{d x}+y=\tan x\left(0 \leq x\frac{\pi}{2}\right)$ Solution: The given differential equation is: $\cos ^{2} x \frac{d y}{d x}+y=\tan x$ $\Rightarrow \frac{d y}{d x}+\sec ^{2} x \cdot y=\sec ^{2} x \tan x$ This equation is in the form of: $\frac{d y}{d x}+p y=Q$ (where $p=\sec ^{2} x$ and $Q=\sec ^{2} x \tan x$ ) Now, I.F $=e^{\int p d x}=e^{\int \sec ^{2} x d x}=e^{\tan x}$. The general solution of the given differential equation is given by the relation, $y(\math...
Read More →In the given figure, ∠ABC = 90° and BD ⊥ AC.
Question: In the given figure, $\angle \mathrm{ABC}=90^{\circ}$ and $\mathrm{BD} \perp \mathrm{AC}$. If $\mathrm{AB}=5.7 \mathrm{~cm}, \mathrm{BD}=3.8 \mathrm{~cm}$ and $\mathrm{CD}=5.4 \mathrm{~cm}$, find $\mathrm{BC}$. Solution: It is given that $B D \perp A C, A B=5.7 \mathrm{~cm}, D B=3.8 \mathrm{~cm}, C D=5.4 \mathrm{~cm}$ and $\angle A B C=90^{\circ}$ We have to find $B C$. Since $\triangle A B C \sim \triangle B D C$ $\Rightarrow \frac{A B}{B D}=\frac{B C}{C D}$ So $\Rightarrow \frac{5.7 ...
Read More →A conical pit of top diameter 3.5 m is 12 m deep.
Question: A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres? Solution: It is given that: Diameter of the conical pit (d) = 3.5 m Height of the conical pit (h) = 12 m Radius of the conical pit (r) = ? Volumeof the conical pit (v) = ? Radius of the conical pit(r) = d/2 = 3.5/2 = 1.75 m Volume of the cone $(\mathrm{v})=1 / 3 \pi \mathrm{r}^{2} \mathrm{~h}$ $=1 / 3 * 3.14 * 1.75^{2} * 12=38.5 \mathrm{~m}^{3}$ Capacity of the pit $=(38.5 * 1)$ kilolitres $=38.5$ kil...
Read More →Prove that:
Question: Prove that: (i) $\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}=\frac{\sqrt{3}}{4}$ (ii) sin2(n+ 1)A sin2nA= sin (2n+ 1)AsinA. Solution: (i) $\cos ^{2} 45^{\circ}-\sin ^{2} 15^{\circ}$ $=\cos \left(45^{\circ}+15^{\circ}\right) \cos \left(45^{\circ}-15^{\circ}\right) \quad\left[\cos ^{2} X-\sin ^{2} Y=\cos (X+Y) \cos (X-Y)\right]$ $=\cos 60^{\circ} \cos 30$ $=\frac{1}{2} \times \frac{\sqrt{3}}{2}$ $=\frac{\sqrt{3}}{4}$ Hence proved. (ii) LHS $=\sin ^{2}(n+1) A-\sin ^{2} n A$ $=\sin [(n+1) A+...
Read More →In the given figure, ∠ABC = 90° and BD ⊥ AC. If BD = 8 cm and AD = 4 cm, find CD.
Question: In the given figure, $\angle \mathrm{ABC}=90^{\circ}$ and $\mathrm{BD} \perp \mathrm{AC}$. If $\mathrm{BD}=8 \mathrm{~cm}$ and $\mathrm{AD}=4 \mathrm{~cm}$, find $\mathrm{CD}$. Solution: It is given that $\angle A B C=90^{\circ}$ and $B D \perp A C$. When $B D=8 \mathrm{~cm}, A D=4 \mathrm{~cm}$ we have to find the $C D$. Since $A B C$ is right angle triangle and $B D$ is perpendicular on $A C$, so $\triangle D B A \sim \triangle D C B$ (AA similarity) $\frac{B D}{C D}=\frac{A D}{B D}$...
Read More →The volume of a right circular cone is 9856 cm3
Question: The volume of a right circular cone is $9856 \mathrm{~cm}^{3}$. If the diameter of the base is $28 \mathrm{~cm}$. Find: (a) Height of the cone (b) Slant height of the cone (c) Curved surface area of the cone Solution: (a)It is given that diameter of the cone (d) = 28 cm Radius of the cone(r) = d/2 = 28/2= 14cm Height of the cone = ? Now, Volume of the cone $(v)=1 / 3 \pi r^{2} h=9856 \mathrm{~cm}^{3}$ $\Rightarrow 1 / 3 * 3.14 * 14^{2} * h=9856$ $\Rightarrow \mathrm{h}=\frac{9856 * 3}{...
Read More →(i) If tan A=
Question: (i) If $\tan A=\frac{5}{6}$ and $\tan B=\frac{1}{11}$, prove that $A+B=\frac{\pi}{4}$. (ii) If $\tan A=\frac{m}{m-1}$ and $\tan B=\frac{1}{2 m-1}$, then prove that $A-B=\frac{\pi}{4}$. Solution: (i) We have: $\tan A=\frac{5}{6}$ and $\tan B=\frac{1}{11}$ Therefore, $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ $\Rightarrow \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}$ $\Rightarrow \tan (A+B)=\frac{\frac{5}{6}+\frac{1}{11}}{1-\frac{5}{6} \times \frac{1}{11}}$ $\Rightarrow \tan ...
Read More →In a right angled triangle with sides a and b and hypotenuse c,
Question: In a right angled triangle with sidesaandband hypotenusec, the altitude drawn on the hypotenuse isx. Prove thatab=cx. Solution: Let $\triangle \mathrm{ABC}$ be a right angle triangle having sides $a$ and $b$; and hypotenuse $c . \mathrm{BD}$ is the altitude drawn on the hypotenuse AC. Since the altitude is perpendicular on the hypotenuse, both the triangles are similar $\frac{A B}{B D}=\frac{A C}{B C}$ $\frac{a}{x}=\frac{c}{b}$ $x c=a b$ Hence, $a b=c x$....
Read More →Show that
Question: $\frac{d y}{d x}+\sec x y=\tan x\left(0 \leq x\frac{\pi}{2}\right)$ Solution: The given differential equation is: $\frac{d y}{d x}+p y=Q$ (where $p=\sec x$ and $Q=\tan x$ ) Now, I.F $=e^{\int p d x}=e^{\int \sec x d x}=e^{\log (\sec x+\tan x)}=\sec x+\tan x$. The general solution of the given differential equation is given by the relation, $y($ I.F. $)=\int($ Q $\times$ I.F. $) d x+\mathrm{C}$ $\Rightarrow y(\sec x+\tan x)=\int \tan x(\sec x+\tan x) d x+\mathrm{C}$ $\Rightarrow y(\sec ...
Read More →In the given figure, XY || BC. Find the length of XY.
Question: In the given figure, XY || BC. Find the length of XY. Solution: It is given that $X Y \| B C$. $A X=1 \mathrm{~cm}, X B=3 \mathrm{~cm}$ and $B C=6 \mathrm{~cm}$ We have to find $X Y$. Since $\triangle A X Y \sim \triangle A B C$ $\Rightarrow \frac{X Y}{B C}=\frac{A X}{A B}(A B=A X+X B=4)$ So $\frac{X Y}{6 \mathrm{~cm}}=\frac{1 \mathrm{~cm}}{4 \mathrm{~cm}}$ $X Y=\frac{6 \mathrm{~cm}}{4 \mathrm{~cm}}$ $=1.5 \mathrm{~cm}$ Hence, $X Y=1.5 \mathrm{~cm}$...
Read More →Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.
Question: Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm. Solution: Radius of the base of the largest cone = 1/2 * edge of the cube = 1/2 *14 = 7 cm Height of the cone = Edge of the cube = 14 cm Therefore, volume of cone $(v)=1 / 3 \pi r^{2} h$ $=1 / 3 * 3.14 * 7^{2} * 14=718.66 \mathrm{~cm}^{3}$...
Read More →In the given figure, AB || QR. Find the length of PB.
Question: In the given figure, AB || QR. Find the length of PB. Solution: It is given that $A B \| Q R$ $A B=3 \mathrm{~cm}, Q R=9 \mathrm{~cm}$ and $P R=6 \mathrm{~cm}$ We have to find $P B$. Since $\triangle P R Q \sim \triangle P A B$ $\Rightarrow \frac{A B}{Q R}=\frac{P B}{P R}$ So $\frac{A B}{O R}=\frac{P B}{P R}$ $\frac{3 \mathrm{~cm}}{9 \mathrm{~cm}}=\frac{P B}{6 \mathrm{~cm}}$ $P B=2 \mathrm{~cm}$ Hence, $P B=2 \mathrm{~cm}$...
Read More →Prove that
Question: Prove that $\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}=-1$. Solution: $\mathrm{LHS}=\frac{\tan 69^{\circ}+\tan 66^{\circ}}{1-\tan 69^{\circ} \tan 66^{\circ}}$ $=\tan \left(69^{\circ}+66^{\circ}\right) \quad\left[\mathrm{U} \operatorname{sing}\right.$ the formula $\left.\frac{\tan A+\tan B}{1-\tan A \tan B}=\tan (A+B)\right]$ $=\tan 135^{\circ}$ $=\tan \left(180^{\circ}-45^{\circ}\right)$ $=-\tan 45^{\circ} \quad[\tan (180-A)=-\tan A]$ $=-1$ = RHS Hence pr...
Read More →Show that
Question: $\frac{d y}{d x}+\frac{y}{x}=x^{2}$ Solution: The given differential equation is: $\frac{d y}{d x}+p y=Q\left(\right.$ where $p=\frac{1}{x}$ and $\left.Q=x^{2}\right)$ Now, I.F $=e^{\int p d x}=e^{\int \frac{1}{x} d x}=e^{\log x}=x$ The solution of the given differential equation is given by the relation, $y(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} . \mathrm{F} .) d x+\mathrm{C}$ $\Rightarrow y(x)=\int\left(x^{2} \cdot x\right) d x+\mathrm{C}$ $\Rightarrow x y=\int ...
Read More →A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length,
Question: A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area. Solution: It is given that Radius of cone(r) = 6.3 cm Height of the cone (h) = 10 cm We know that Slant height $(\mathrm{l})=\mathrm{l}=\sqrt{\mathrm{r}^{2}+\mathrm{h}^{2}}$ $=1=\sqrt{6.3^{2}+10^{2}}=11.819 \mathrm{~cm}$ Therefore volume of cone $(v)=1 / 3 \pi r^{...
Read More →