ABCD is a quadrilateral in which AD = BC. If P, Q, R, S
Question: ABCD is a quadrilateral in which AD = BC. If P, Q, R, S be the mid-points of AB, AC, CD and BD respectively, show that PQRS is a rhombus. Solution: Given:ABCD is quadrilateral in which AD = BC and P, Q, R, S are the mid points of AB, AC, CD, BD respectively. To Prove:PQRS is a rhombus. Proof:In∆ABC, P and Q are the midpoints of the sides AB and AC respectively.By the Mid point theorem, we getPQ || BC and PQ =12BC ...(1)In∆ADC, Q and R are the midpoints of the sides AC and DC respective...
Read More →If tan x + tan
Question: If $\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(x+\frac{2 \pi}{3}\right)=3$, then prove that $\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x}=1$. Solution: Given: $\tan x+\tan \left(x+\frac{\pi}{3}\right)+\tan \left(\mathrm{x}+\frac{2 \pi}{3}\right)=3$ $\Rightarrow \tan x+\frac{\tan x+\tan \frac{\pi}{3}}{1-\tan x \tan \frac{\pi}{3}}+\frac{\tan x+\tan \frac{2 \pi}{3}}{1-\tan x \tan \frac{2 \pi}{3}}=3$ $\Rightarrow \tan x+\frac{\tan x+\sqrt{3}}{1-\sqrt{3} \tan x}+\frac{\tan x-\sqrt{...
Read More →In the given figure, we have AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 10 cm, BD = 4 cm and DE = y cm, Calculate the values of x and y.
Question: In the given figure, we have AB || CD || EF. If AB = 6 cm, CD =xcm, EF = 10 cm, BD = 4 cm and DE =ycm, Calculate the values ofxandy. Solution: It is given that $A B\|C D\| E F$. $A B=6 \mathrm{~cm}, C D=x \mathrm{~cm}$ and $E F=10 \mathrm{~cm}$ We have to calculate the values of $x$ and $y$. In $\triangle A D B$ and $\triangle D E F$, we have $\angle A D B=\angle E D F \quad$ (Vertically opposite angles) $\Rightarrow \angle A B D=\angle D E F \quad$ (Alternate interior angles) So $\tri...
Read More →If x lies in the first quadrant and cos x
Question: If $x$ lies in the first quadrant and $\cos x=\frac{8}{17}$, then prove that: $\cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{2 \pi}{3}-x\right)=\left(\frac{\sqrt{3}-1}{2}+\frac{1}{\sqrt{2}}\right) \frac{23}{17}$ Solution: Given : $0x\frac{\pi}{2}$ Now, $\sin x=\sqrt{1-\cos ^{2} x}=\sqrt{1-\frac{64}{289}}=\frac{15}{17}$ $\mathrm{LHS}=\cos \left(\frac{\pi}{6}+x\right)+\cos \left(\frac{\pi}{4}-x\right)+\cos \left(\frac{2 \pi}{3}-x\right)$ $=\cos (30+...
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Question: $y d x+\left(x-y^{2}\right) d y=0$ Solution: $y d x+\left(x-y^{2}\right) d y=0$ $\Rightarrow y d x=\left(y^{2}-x\right) d y$ $\Rightarrow \frac{d x}{d y}=\frac{y^{2}-x}{y}=y-\frac{x}{y}$ $\Rightarrow \frac{d x}{d y}+\frac{x}{y}=y$ This is a linear differential equation of the form: $\frac{d y}{d x}+p x=Q$ ( where $p=\frac{1}{y}$ and $Q=y$ ) The general solution of the given differential equation is given by the relation, $x(\mathrm{I} . \mathrm{F} .)=\int(\mathrm{Q} \times \mathrm{I} ....
Read More →A hemispherical bowl made of brass has inner diameter 10.5 cm.
Question: A hemispherical bowl made of brass has inner diameter $10.5 \mathrm{~cm}$. Find the cost of tin plating it on the inside at the rate of Rs.4 per $100 \mathrm{~cm}^{2}$. Solution: Inner diameter of hemispherical bowl = 10.5cm Radius = 10.5/2 = 5.25 cm Surface area of hemispherical bowl $=2 \pi r^{2}$ $=2 \times 3.14 \times(5.25)^{2}$ $=173.25 \mathrm{~cm}^{2}$ Cost of tin plating $100 \mathrm{~cm}^{2}$ area $=$ Rs. 4 Cost of tin plating $173.25 \mathrm{~cm}^{2}$ area $=$ Rs. $\frac{4 \t...
Read More →If tan A + tan B = a and cot A + cot B = b,
Question: If $\tan A+\tan B=a$ and $\cot A+\cot B=b$, prove that $\cot (A+B) \frac{1}{a}-\frac{1}{b}$. Solution: Given: $\cot A+\cot B=b$ $\Rightarrow \frac{1}{\tan A}+\frac{1}{\tan B}=b$ $\Rightarrow \frac{\tan A+\tan B}{\tan A \tan B}=b$ NOW, $\mathrm{RHS}=\frac{1}{a}-\frac{1}{b}$ $=\frac{1}{\tan A+\tan B}-\frac{\tan A \tan B}{\tan A+\tan B}$ $=\frac{1-\tan A \tan B}{\tan A+\tan B}$ $=\cot (A+B)$ = RHS Hence proved....
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Question: $(x+y) \frac{d y}{d x}=1$ Solution: $(x+y) \frac{d y}{d x}=1$ $\Rightarrow \frac{d y}{d x}=\frac{1}{x+y}$ $\Rightarrow \frac{d x}{d y}=x+y$ $\Rightarrow \frac{d x}{d y}-x=y$ This is a linear differential equation of the form: $\frac{d y}{d x}+p x=Q$ (where $p=-1$ and $Q=y$ ) Now, I.F $=e^{\int p d y}=e^{\int-d y}=e^{-y}$. The general solution of the given differential equation is given by the relation, $x($ I.F. $)=\int($ Q $\times$ I.F. $) d y+\mathrm{C}$ $\Rightarrow x e^{-y}=\int\le...
Read More →ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q.
Question: ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. Solution: Given:ABCD is a parallelogram and APQ is a straight line meeting BC at P and DC produced at Q. To Prove: The rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC. We need to prove thatBPDQ=ABBC Proof:In∆ABP and∆QCP, we haveABP =QCP (Alternate angles as AB || DC)BP...
Read More →The surface area of a sphere is 5544 cm2,
Question: The surface area of a sphere is $5544 \mathrm{~cm}^{2}$ find its diameter. Solution: Surface area of a sphere is $5544 \mathrm{~cm}^{2}$ $4 \pi r^{2}=5544$ $4 \times 3.14 \times(r)^{2}=5544$ $r^{2}=\frac{5544 \times 7}{88}$ $\mathrm{r}=\sqrt{21 \mathrm{~cm} \times 21 \mathrm{~cm}}$ r = 21cm Diameter = 2(radius) = 2(21) = 42 cm...
Read More →If cos A + sin B = m and sin A + cos B = n,
Question: If cosA+ sinB=mand sinA+ cosB=n, prove that 2 sin (A+B) =m2+n2 2. Solution: $\mathrm{RHS}=m^{2}+n^{2}-2$ $=(\cos A+\sin B)^{2}+(\sin A+\cos B)^{2}-2$ $=\cos ^{2} A+\sin ^{2} B+2 \cos A \sin B+\sin ^{2} A+\cos ^{2} B+2 \sin A \cos B-2$ $=1+1+2 \cos A \sin B+2 \sin A \cos B-2$ $=2(\cos A \sin B+\sin A \cos B)$ $=2 \sin (A+B)$ = RHS Hence proved....
Read More →D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm,
Question: D and E are the points on the sides AB and AC respectively of a ∆ABC such that: AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm, Prove that BC = 5/2 DE. Solution: It is given that $A D=8 \mathrm{~cm}, D B=12 \mathrm{~cm}, A E=6 \mathrm{~cm}$ and $C E=9 \mathrm{~cm}$. We have to prove that $B C=\frac{5}{2} D E$ Since clearly $\frac{A D}{A B}=\frac{A E}{A C}=\frac{2}{5}$ Also, $\angle A$ is common in $\triangle A B C$ and $\triangle A D E$ So $\triangle A D E \sim \triangle A B C$ (SAS Si...
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Question: $x \frac{d y}{d x}+y-x+x y \cot x=0(x \neq 0)$ Solution: $x \frac{d y}{d x}+y-x+x y \cot x=0$ $\Rightarrow x \frac{d y}{d x}+y(1+x \cot x)=x$ $\Rightarrow \frac{d y}{d x}+\left(\frac{1}{x}+\cot x\right) y=1$ This equation is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q\left(\right.$ where $p=\frac{1}{x}+\cot x$ and $\left.Q=1\right)$ Now, I.F $=e^{\int p d x}=e^{\int\left(\frac{1}{x}+\cot x\right) d x}=e^{\log x+\log (\sin x)}=e^{\log (x \sin x)}=x \sin x$ The gen...
Read More →If tan (A + B) = x and tan (A − B) = y,
Question: If tan (A+B) =xand tan (AB) =y, find the values of tan 2Aand tan 2B. Solution: $\tan (2 A)=\tan (A+A)$ $=\tan (A+B+A-B)$ $=\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)}$ $=\frac{x+y}{1-x y}$ $\tan 2 B=\tan (B+B)$ $=\tan (B+A+B-A)$ $=\frac{\tan (A+B)+\tan (B-A)}{1-\tan (A+B) \tan (B-A)}$ $=\frac{\tan (A+B)-\tan (A-B)}{1+\tan (A+B) \tan (A-B)} \quad[\tan (-\theta)=-\tan \theta]$ $=\frac{x-y}{1+x y}$...
Read More →Find the total surface area of a hemisphere and a solid hemisphere each of radius 10cm (π = 3.14)
Question: Find the total surface area of a hemisphere and a solid hemisphere each of radius 10cm ( = 3.14) Solution: The surface area of the hemisphere $=2 \pi r^{2}$ $=2 \times 3.14 \times(10)^{2}$ $=628 \mathrm{~cm}^{2}$ The surface area of solid hemisphere $=2 \pi r^{2}$ $=3 \times 3.14 \times(10)^{2}$ $=942 \mathrm{~cm}^{2}$...
Read More →In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm;
Question: In ∆ABC and ∆DEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL BC and DM EF, find AL : DM. Solution: It is given that $A B=5 \mathrm{~cm}, B C=4 \mathrm{~cm}, C A=4.2 \mathrm{~cm}, D E=10 \mathrm{~cm}, E F=8 \mathrm{~cm}$ and $F D=8.4 \mathrm{~cm}$. We have to find $A L: D M$ Since both triangle are similar So, $\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{1}{2}$ Here, we use the result that in similar triangle the ...
Read More →If tan A = x tan B,
Question: If $\tan A=x \tan B$, prove that $\frac{\sin (A-B)}{\sin (A+B)}=\frac{x-1}{x+1}$. Solution: LHS $=\frac{\sin (A-B)}{\sin (A+B)}$ $=\frac{\sin A \cos B-\cos A \sin B}{\sin A \cos B+\cos A \sin B}$ Dividing numerator and denominator by $\cos A \cos B$ : $\frac{\tan A-\tan B}{\tan A+\tan B}$ $=\frac{x \tan B-\tan B}{x \tan B+\tan B} \quad($ Since $\tan A=x \tan B)$ $=\frac{\tan B(x-1)}{\tan B(x+1)}$ $=\frac{x-1}{x+1}$ = RHS Hence proved....
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Question: $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$ Solution: $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x$ $\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$ This equation is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q\left(\right.$ where $p=\frac{2 x}{1+x^{2}}$ and $\left.Q=\frac{\cot x}{1+x^{2}}\right)$ Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$ The general solution of the g...
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Question: $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x(x \neq 0)$ Solution: $\left(1+x^{2}\right) d y+2 x y d x=\cot x d x$ $\Rightarrow \frac{d y}{d x}+\frac{2 x y}{1+x^{2}}=\frac{\cot x}{1+x^{2}}$ This equation is a linear differential equation of the form: $\frac{d y}{d x}+p y=Q\left(\right.$ where $p=\frac{2 x}{1+x^{2}}$ and $\left.Q=\frac{\cot x}{1+x^{2}}\right)$ Now, I.F $=e^{\int p d x}=e^{\int \frac{2 x}{1+x^{2}} d x}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$ The general solution of the g...
Read More →Find the surface area of a sphere of diameter:
Question: Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 cm Solution: (i) Given Diameter = 14 cm Radius = Diameter/2 = 14/2 = 7 cm Surface area $=4 \pi r^{2}$ $=4 \times 22 / 7 \times(7)^{2}$ $=616 \mathrm{~cm}^{2}$ (ii) Given Diameter = 21cm Radius = Diameter/2 = 21/2 = 10.5 cm Surface area $=4 \pi r^{2}$ $=4 \times 22 / 7 \times(10.5)^{2}$ $=1386 \mathrm{~cm}^{2}$ (iii) Given diameter = 3.5 cm Radius = Diameter/2 = 3.5/2 = 1.75 cm Surface area $=4 \pi r^{2}$ $=4 ...
Read More →The perimeters of two similar triangles are 25 cm and 15 cm respectively.
Question: The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of first triangle is 9 cm, what is the corresponding side of the other triangle? Solution: It is given that perimeter of two similar triangle are $25 \mathrm{~cm}$ and $15 \mathrm{~cm}$ and one side $9 \mathrm{~cm}$. We have to find the other side.Let the corresponding side of the other triangle bexcm. Since ratio of perimeter $=$ ratio of corresponding side 25cm15cm=9cmx $25 \mathrm{~cm} \times x=9 \...
Read More →Prove that sin
Question: Prove that sin2(n+ 1)A sin2nA= sin (2n+ 1)AsinA. Solution: LHS $=\sin ^{2}(n+1) A-\sin ^{2} n A$ $=\sin [(n+1) A+n A] \sin [(n+1) A-n A]$ $\left[\right.$ Using the formula $\sin ^{2} X-\sin ^{2} Y=\sin (X+Y) \sin (X-Y)$ and taking $X=(n+1) A$ and $\left.Y=n A\right]$ $=\sin [(n+1+n) A] \sin [(n+1-n) A]$ $=\sin (2 n+1) A \sin A$ = RHS Hence proved....
Read More →Prove that:
Question: Prove that: $\frac{\tan ^{2} 2 x-\tan ^{2} x}{1-\tan ^{2} 2 x \tan ^{2} x}=\tan 3 x \tan x$ Solution: $\mathrm{LHS}=\frac{\tan ^{2} 2 x-\tan ^{2} x}{1-\tan ^{2} 2 x \tan ^{2} x}$ $=\frac{(\tan 2 x+\tan x)(\tan 2 x-\tan x)}{1-\tan ^{2} 2 x \tan ^{2} x} \quad\left\{\right.$ Using $\left.A^{2}-B^{2}=(A+B)(A-B)\right\}$ $\tan 3 x=\tan (2 x+x)$ and $\tan x=\tan (2 x-x)$ $=\frac{\tan 3 x(1-\tan 2 x \tan x) \times \tan x(1+\tan 2 x \tan x)}{1-\tan ^{2} 2 x \tan ^{2} x}$ $[\because \tan 2 x+\t...
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Question: $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$ Solution: The given differential equation is: $x \log x \frac{d y}{d x}+y=\frac{2}{x} \log x$ $\Rightarrow \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x^{2}}$ This equation is the form of a linear differential equation as: $\frac{d y}{d x}+p y=Q$ ( where $p=\frac{1}{x \log x}$ and $Q=\frac{2}{x^{2}}$ ) Now, I.F $=e^{\int \mu d x}=e^{\int \frac{1}{x \log } d x}=e^{\log (\log x)}=\log x$. The general solution of the given differential equat...
Read More →Find the surface area of a sphere of radius:
Question: Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm Solution: (i) Given Radius = 10.5 cm Surface area $=4 \pi r^{2}$ $=4 \times 22 / 7 \times(10.5)^{2}$ $=1386 \mathrm{~cm}^{2}$ (ii) Given radius = 5.6 cm Surface area $=4 \pi r^{2}$ $=4 \times 22 / 7 \times(5.6)^{2}$ $=394.24 \mathrm{~cm}^{2}$ (iii) Given radius $=14 \mathrm{~cm}$ Surface area $=4 \pi r^{2}$ $=4 \times 22 / 7 \times(14)^{2}$ $=2464 \mathrm{~cm}^{2}$...
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