In which of the following tables x and y vary inversely:
Question: In which of the following tablesxandyvary inversely: (i) (ii) (iii) (iv) Solution: (i) Since $x$ and $y \operatorname{var} y$ inversely, we have : $y=\frac{k}{x}$ $\Rightarrow x y=k$ $\therefore$ The product of $x$ and $y$ is constant. In all cases, the product $x y$ is constant $($ i.e., 24). Thus, in this case, $x$ and $y$ var $y$ inversely. (ii) In all ca $s$ es, the product $x y$ is constant for any two pairs of values for $x$ and $y$. Here, $x y=100$ for all cases Thus, in this ca...
Read More →In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm.
Question: In the given figure,ABCDis a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If = 3.14, then the area of the shaded region is (a) 264 cm2(b) 266 cm2(c) 272 cm2(d) 254 cm2 Solution: All options are incorrect; the correct answer is 30.5 cm.JoinAC.Now,ACis the diameter of the circle. We have : $A C^{2}=A B^{2}+B C^{2} \quad[$ By Pythagoras' theorem $]$ $\Rightarrow A C^{2}=\left\{(8)^{2}+(6)^{2}\right\} \mathrm{cm}^{2}$ $\Rightarrow A C^{2}=(64+36) \mathrm{cm}^{2}$ $\...
Read More →In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm.
Question: In the given figure,ABCDis a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If = 3.14, then the area of the shaded region is (a) 264 cm2(b) 266 cm2(c) 272 cm2(d) 254 cm2 Solution: All options are incorrect; the correct answer is 30.5 cm.JoinAC.Now,ACis the diameter of the circle. We have : $A C^{2}=A B^{2}+B C^{2} \quad[$ By Pythagoras' theorem $]$ $\Rightarrow A C^{2}=\left\{(8)^{2}+(6)^{2}\right\} \mathrm{cm}^{2}$ $\Rightarrow A C^{2}=(64+36) \mathrm{cm}^{2}$ $\...
Read More →The area of a sector of a circle with radius r, making an angle of x° at the centre is
Question: The area of a sector of a circle with radiusr, making an angle ofxat the centre is (a) $\frac{x}{180} \times 2 \pi r$ (b) $\frac{x}{180} \times \pi r^{2}$ (c) $\frac{x}{360} \times 2 \pi r$ (d) $\frac{x}{360} \times \pi r^{2}$ Solution: (d) $\frac{x}{360} \times \pi r^{2}$ The area of a sector of a circle with radius $r$ making an angle of $x^{\circ}$ at the centre is $\frac{x}{360} \times \pi r^{2}$....
Read More →The diameter of a wheel is 84 cm.
Question: The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?(a) 200(b) 250(c) 300(d) 350 Solution: (c) 300Letdcm be the diameter of the wheel.We know: Circumference of the whee $=\pi \times d$ $=\left(\frac{22}{7} \times 84\right) \mathrm{cm}$ $=264 \mathrm{~cm}$ Now, Number of revolutions to cover $792 \mathrm{~m}=\left(\frac{792 \times 1000}{264}\right)$ $=300$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{2}{3}(x-5)-\frac{1}{4}(x-2)=\frac{9}{2}$ Solution: $\frac{2}{3}(\mathrm{x}-5)-\frac{1}{4}(\mathrm{x}-2)=\frac{9}{2}$ or $\frac{2 \mathrm{x}-10}{3}-\frac{\mathrm{x}-2}{4}=\frac{9}{2}$ or $\frac{8 \mathrm{x}-40-3 \mathrm{x}+6}{12}=\frac{9}{2}$ or $\frac{5 \mathrm{x}-34}{12}=\frac{9}{2}$ or $10 \mathrm{x}-68=108$ or $10 \mathrm{x}=108+68$ or $\mathrm{x}=\frac{176}{10}=\frac{88}{5}$ Verification : L. H. S. $=\frac{2...
Read More →In the given figure, a square OABC has been inscribed in the quadrant OPBQ.
Question: In the given figure, a squareOABChas been inscribed in the quadrantOPBQ. IfOA= 20 cm, then the area of the shaded region is (a) 214 cm2(b) 228 cm2(c) 242 cm2(d) 248 cm2 Solution: (b) 228 cm2JoinOB.Now,OBis the radius of the circle. We have : $O B^{2}=O A^{2}+A B^{2} \quad$ [By Pythagoras' theorem] $\Rightarrow O B^{2}=\left\{(20)^{2}+(20)^{2}\right\} \mathrm{cm}^{2}$ $\Rightarrow O B^{2}=(400+400) \mathrm{cm}^{2}$ $\Rightarrow O B^{2}=800 \mathrm{~cm}^{2}$ $\Rightarrow O B=20 \sqrt{2} ...
Read More →In the given figure, a square OABC has been inscribed in the quadrant OPBQ.
Question: In the given figure, a squareOABChas been inscribed in the quadrantOPBQ. IfOA= 20 cm, then the area of the shaded region is (a) 214 cm2(b) 228 cm2(c) 242 cm2(d) 248 cm2 Solution: (b) 228 cm2JoinOB.Now,OBis the radius of the circle. We have : $O B^{2}=O A^{2}+A B^{2} \quad$ [By Pythagoras' theorem] $\Rightarrow O B^{2}=\left\{(20)^{2}+(20)^{2}\right\} \mathrm{cm}^{2}$ $\Rightarrow O B^{2}=(400+400) \mathrm{cm}^{2}$ $\Rightarrow O B^{2}=800 \mathrm{~cm}^{2}$ $\Rightarrow O B=20 \sqrt{2} ...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution:13(y 4) 3(y 9) 5(y+ 4) = 0 Solution: $13(y-4)-3(y-9)-5(y+4)=0$ or $13 y-52-3 y+27-5 y-20=0$ or $5 y=45$ or $y=\frac{45}{5}=9$ Verification : L.H.S. $=13(9-4)-3(9-9)-5(9+4)$ $=13 \times 5-3 \times 0-5 \times 13=0=$ R. H.S....
Read More →For the matrix
Question: For the matrix $A=\left[\begin{array}{ccc}1 1 1 \\ 1 2 -3 \\ 2 -1 3\end{array}\right]$. Show that $A^{-3}-6 A^{2}+5 A+11 I_{3}=O$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{lll}1 1 1\end{array}\right.$ $\begin{array}{lll}1 2 -3\end{array}$ $\left.\begin{array}{lll}2 -1 3\end{array}\right]$ $\Rightarrow|A|=\mid \begin{array}{lll}1 1 1\end{array}$ $1 \quad 2-3$ $2-1 \quad 3 \mid=(1 \times 3)-(1 \times 9)+(1 \times-5)=3-9-5=-11$ Since, $|A| \neq 0$ Hence, $A^{-1}$ exists. Now...
Read More →If two equal chords of a circle intersect,
Question: If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord. Solution: Given Consider $A B$ and $C D$ are two equal chords of a circle, which meet at point $E$. To prove $A E=C E$ and $B E=D E$ Construction Draw $O M \perp A B$ and $O N \perp C D$ and join $O E$ where $O$ is the centre of circle. Proof in $\triangle O M E$ and $\triangle O N E$, $O M=O N$ [equal chords are equidistant from the centre] $O E=O E \quad$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{2 x-1}{3}-\frac{6 x-2}{5}=\frac{1}{3}$ Solution: $\frac{2 \mathrm{x}-1}{3}-\frac{6 \mathrm{x}-2}{5}=\frac{1}{3}$ or $\frac{10 \mathrm{x}-5-18 \mathrm{x}+6}{15}=\frac{1}{3}$ or $\frac{-8 \mathrm{x}+1}{15}=\frac{1}{3}$ or $-24 \mathrm{x}+3=15$ or $24 \mathrm{x}=3-15$ or $\mathrm{x}=\frac{-12}{24}=\frac{-1}{2}$ Verification : L.H.S. $=\frac{2 \times \frac{-1}{2}-1}{3}-\frac{6 \times \frac{-1}{2}-2}{5}$ $=\frac{-2}{...
Read More →In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre.
Question: In a circle of radius $14 \mathrm{~cm}$, an arc subtends an angle of $120^{\circ}$ at the centre. If $\sqrt{3}=1.73$ then the area of the segment of the circle is (a) 120.56 cm2(b) 124.63 cm2(c) 118.24 cm2(d) 130.57 cm2 Solution: Radius of the circle,r= 14 cmDraw a perpendicular OD to chord AB. It will bisect AB.A = 180 (90 + 60) = 30 $\cos 30^{\circ}=\frac{\mathrm{AD}}{\mathrm{OA}}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{14}$ $\Rightarrow \mathrm{AD}=7 \sqrt{3}$ $\Rightarr...
Read More →In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre.
Question: In a circle of radius $14 \mathrm{~cm}$, an arc subtends an angle of $120^{\circ}$ at the centre. If $\sqrt{3}=1.73$ then the area of the segment of the circle is (a) 120.56 cm2(b) 124.63 cm2(c) 118.24 cm2(d) 130.57 cm2 Solution: Radius of the circle,r= 14 cmDraw a perpendicular OD to chord AB. It will bisect AB.A = 180 (90 + 60) = 30 $\cos 30^{\circ}=\frac{\mathrm{AD}}{\mathrm{OA}}$ $\Rightarrow \frac{\sqrt{3}}{2}=\frac{\mathrm{AD}}{14}$ $\Rightarrow \mathrm{AD}=7 \sqrt{3}$ $\Rightarr...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{7}{\mathrm{x}}+35=\frac{1}{10}$ Solution: $\frac{7}{\mathrm{x}}+35=\frac{1}{10}$ or $\frac{7}{\mathrm{x}}=\frac{1}{10}-35$ or $\frac{7}{\mathrm{x}}=\frac{1-350}{10}$ or $\frac{\mathrm{x}}{7}=\frac{10}{{ }_{-349}}$ or $\mathrm{x}=\frac{-10 \times 7}{349}=\frac{-70}{349}$ Verification : L. H. S. $=\frac{7}{\frac{-70}{349}}+35$ $=7 \times \frac{349}{-70}+35$ $=\frac{349}{-10}+35=\frac{1}{10}=$ R.H.S....
Read More →In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre.
Question: In a circle of radius 21 cm, an arc subtends an angle of 60 at the centre. The length of the arc is(a) 21 cm(b) 22 cm(c) 18.16 cm(d) 23.5 cm Solution: We have $r=21 \mathrm{~cm}$ and $\theta=60^{\circ}$ Length of $\operatorname{arc}=\frac{\theta}{360^{\circ}} \times 2 \pi r=\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21=22 \mathrm{~cm}$ Hence, the correct answer is option (b)...
Read More →A chord of a circle of radius 10 cm subtends a right angle at the centre.
Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, = 3.14) is(a) 32.5 cm2(b) 34.5 cm2(c) 28.5 cm2(d) 30.5 cm2 Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times 3.14 \times(10)^{2}-\frac{1}{2} \times 10 \times 10$ $=78.5-50$ $=28.5 \mathrm{~cm}^{2}$ Hence, the correct ans...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}$ Solution: $\frac{x}{2}-\frac{4}{5}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}$ or $\frac{x}{2}+\frac{x}{5}+\frac{3 x}{10}=\frac{1}{5}+\frac{4}{5}$ or $\frac{5 x+2 x+3 x}{10}=\frac{5}{5}$ or $\frac{10 x}{10}=1$ Verification : L. H.S. $=\frac{1}{2}-\frac{4}{5}+\frac{1}{5}+\frac{3}{10}$ $=\frac{5-8+2+3}{10}=\frac{1}{5}=$ R. H.S....
Read More →A chord of a circle of radius 10 cm subtends a right angle at the centre.
Question: A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, = 3.14) is(a) 32.5 cm2(b) 34.5 cm2(c) 28.5 cm2(d) 30.5 cm2 Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times 3.14 \times(10)^{2}-\frac{1}{2} \times 10 \times 10$ $=78.5-50$ $=28.5 \mathrm{~cm}^{2}$ Hence, the correct ans...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution:(x+ 2)(x+ 3) + (x 3)(x 2) 2x(x+ 1) = 0 Solution: $(\mathrm{x}+2)(\mathrm{x}+3)+(\mathrm{x}-3)(\mathrm{x}-2)-2 \mathrm{x}(\mathrm{x}+1)=0$ or $\mathrm{x}^{2}+5 \mathrm{x}+6+\mathrm{x}^{2}-5 \mathrm{x}+6-2 \mathrm{x}^{2}-2 \mathrm{x}=0$ or $12-2 \mathrm{x}=0$ or $\mathrm{x}=\frac{12}{2}=6$ Verification : L. H. S. $=(6+2)(6+3)+(6-3)(6-2)-2 \times 6(6+1)$ $=72+12-84=0=\mathrm{R} .$ H. S....
Read More →The length of the minute hand of a clock is 21 cm.
Question: The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is [CBSE 2012](a) 231 cm2(b) 210 cm2(c) 126 cm2(d) 252 cm2 Solution: Angle subtends by the minute hand in 1 minute = 6∘ Angle subtends by the minute hand in 10 minutes = 60∘Now, Area of the sector $=\frac{\theta}{360^{\circ}} \pi r^{2}=\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}(21)^{2}=231 \mathrm{~cm}^{2}$ Hence, the correct answer is option (a)....
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{2 x}{3}-\frac{3 x}{8}=\frac{7}{12}$ Solution: $\frac{2 \mathrm{x}}{3}-\frac{3 \mathrm{x}}{8}=\frac{7}{12}$ or $\frac{16 \mathrm{x}-9 \mathrm{x}}{24}=\frac{7}{12}$ or $\frac{7 \mathrm{x}}{24}=\frac{7}{12}$ or $\mathrm{x}=\frac{7}{12} \times \frac{24}{7}=2$ Verification : L.H. S. $=\frac{4}{3}-\frac{6}{8}=\frac{32-18}{24}=\frac{7}{12}$ R.H.S. $=\frac{7}{12}$ $\therefore$ R.H.S. $=$ L.H. S. for $\mathrm{x}=2$...
Read More →Prove the following
Question: In figure, OAB = 30 and OCB = 57. Find BOC and AOC. Solution: Given, $\angle O A B=30^{\circ}$ and $\angle O C B=57^{\circ}$ In $\triangle A O B$. $A O=O B$ [both are the radius of a circle] $\Rightarrow \quad \angle O B A=\angle B A O=30^{\circ}$ [angles opposite to equal sides are equal] In $\triangle A O B$, $\Rightarrow$ $\angle A O B+\angle O B A+\angle B A O=180^{\circ}$ [by angle sum property of a triangle] $\therefore \quad \angle A O B+30^{\circ}+30^{\circ}=180^{\circ}$ $\ther...
Read More →The length of an arc of the sector of angle θ° of a circle with radius R is
Question: The length of an arc of the sector of angleof a circle with radius R is (a) $\frac{2 \pi R \theta}{180}$ (b) $\frac{2 \pi R \theta}{360}$ (c) $\frac{\pi R^{2} \theta}{180}$ (d) $\frac{\pi R^{2} \theta}{360}$ Solution: (b) $\frac{2 \pi R \theta}{360}$...
Read More →Solve each of the following equation and also verify your solution:
Question: Solve each of the following equation and also verify your solution: $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$ Solution: $\frac{x}{2}+\frac{x}{8}=\frac{1}{8}$ or $\frac{4 \mathrm{x}+\mathrm{x}}{8}=\frac{1}{8}$ or $\frac{5 x}{8}=\frac{1}{8}$ or $\mathrm{x}=\frac{1}{8} \times \frac{8}{5}=\frac{1}{5}$ Verification : L.H.S. $=\frac{1}{2} \times \frac{1}{5}+\frac{1}{8} \times \frac{1}{5}=\frac{1}{10}+\frac{1}{40}=\frac{5}{40}=\frac{1}{8}=$ R.H. S....
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