If the circumference of a circle and the perimeter of a square are equal, then
Question: If the circumference of a circle and the perimeter of a square are equal, then(a) area of the circle = area of the square(b) (area of the circle) (area of the square)(c) (area of the circle) (area of the square)(d) none of these Solution: (b) Area of the circle Area of the squareLetrbe the radius of the circle.We know: Circumference of the circle $=2 \pi r$ Now,Letabe the side of the square.We know:Perimeter of the square = 4aNow, $2 \pi r=4 a$ $\Rightarrow r=\frac{4 a}{2 \pi}$ $\there...
Read More →Prove the following
Question: In figure, $\angle A C B=40^{\circ}$. Find $\angle O A B$. Solution: Given, $\angle A C B=40^{\circ}$ We know that, a segment subtends an angle to the circle is half the angle subtends to the centre. $\therefore \quad \angle A O B=2 \angle A C B$ $\Rightarrow \quad \angle A C B=\frac{\angle A O B}{2}$ $40^{\circ}=\frac{1}{2} \angle A O B$ $\Rightarrow \quad \angle A O B=80^{\circ}$ $\ldots$ (i) In $\triangle A O B$, $A O=B O$ [both are the radius of a circle] $\Rightarrow$ $\angle O B ...
Read More →If the circumference of a circle and the perimeter of a square are equal, then
Question: If the circumference of a circle and the perimeter of a square are equal, then(a) area of the circle = area of the square(b) (area of the circle) (area of the square)(c) (area of the circle) (area of the square)(d) none of these Solution: (b) Area of the circle Area of the squareLetrbe the radius of the circle.We know: Circumference of the circle $=2 \pi r$ Now,Letabe the side of the square.We know:Perimeter of the square = 4aNow, $2 \pi r=4 a$ $\Rightarrow r=\frac{4 a}{2 \pi}$ $\there...
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Question: If $A=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$, find the value of $\lambda$ so that $A^{2}=\lambda A-2 I$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{ll}3 -2\end{array}\right.$ $4-2]$ $\therefore \mathrm{A}^{2}=\left[\begin{array}{ll}1 -2\end{array}\right.$ $4-4]$ Given : $A^{2}=\lambda A-2 I$ ....(1) $\Rightarrow[1-2$ $4-4]=\lambda\left[\begin{array}{ll}3 -2\end{array}\right.$ $4-2]-2\left[\begin{array}{ll}1 0\end{array}\right.$ $\left.\begin{array}{ll}0 1\en...
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Question: If $A=\left[\begin{array}{ll}3 -2 \\ 4 -2\end{array}\right]$, find the value of $\lambda$ so that $A^{2}=\lambda A-2 I$. Hence, find $A^{-1}$. Solution: $A=\left[\begin{array}{ll}3 -2\end{array}\right.$ $4-2]$ $\therefore \mathrm{A}^{2}=\left[\begin{array}{ll}1 -2\end{array}\right.$ $4-4]$ Given : $A^{2}=\lambda A-2 I$ ....(1) $\Rightarrow[1-2$ $4-4]=\lambda\left[\begin{array}{ll}3 -2\end{array}\right.$ $4-2]-2\left[\begin{array}{ll}1 0\end{array}\right.$ $\left.\begin{array}{ll}0 1\en...
Read More →The distance between two stations is 340 km.
Question: The distance between two stations is 340 km. Two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/hr. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train. Solution: Let, the speed of the first train be $x \mathrm{~km} / \mathrm{h}$. Then, the speed of the other train will be $(\mathrm{x}+5) \mathrm{km} / \mathrm{h}$. 2 hours af...
Read More →In a rational number, twice the numerator is 2 more than the denominator.
Question: In a rational number, twice the numerator is 2 more than the denominator. If 3 is added to each, the numerator and the denominator, the new fraction is 2/3. Find the original number. Solution: Let the denominator be $\mathrm{x}$. $\therefore$ The numerator $=\frac{x+2}{2}$ $\therefore$ The rational number $=\frac{\mathrm{x}+2}{2 \mathrm{x}}$ According to the question, $\frac{\frac{x+2}{2}+3}{x+3}=\frac{2}{3}$ or $\frac{x+2+6}{2(x+3)}=\frac{2}{3}$ or $\frac{x+8}{2 x+6}=\frac{2}{3}$ or $...
Read More →If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then
Question: If the sum of the circumferences of two circles with radii R1and R2is equal to the circumference of a circle of radius R, then (a) $R_{1}+R_{2}=R$ (b) $R_{1}+R_{2}R$ (c) $R_{1}+R_{2}R$ (d) none of these Solution: (a) $R_{1}+R_{2}=R$ Because the sum of the circumferences of two circles with radii $R_{1}$ and $R_{2}$ is equal to the circumference of a circle with radius $R$, we have: $2 \pi R_{1}+2 \pi R_{2}=2 \pi R$ $\Rightarrow 2 \pi\left(R_{1}+R_{2}\right)=2 \pi R$ $\Rightarrow R_{1}+...
Read More →If the sum of the circumferences of two circles with radii R1 and R2 is equal to the circumference of a circle of radius R, then
Question: If the sum of the circumferences of two circles with radii R1and R2is equal to the circumference of a circle of radius R, then (a) $R_{1}+R_{2}=R$ (b) $R_{1}+R_{2}R$ (c) $R_{1}+R_{2}R$ (d) none of these Solution: (a) $R_{1}+R_{2}=R$ Because the sum of the circumferences of two circles with radii $R_{1}$ and $R_{2}$ is equal to the circumference of a circle with radius $R$, we have: $2 \pi R_{1}+2 \pi R_{2}=2 \pi R$ $\Rightarrow 2 \pi\left(R_{1}+R_{2}\right)=2 \pi R$ $\Rightarrow R_{1}+...
Read More →In figure, ∠ADC = 130° and chord
Question: In figure, ADC = 130 and chord BC = chord BE. Find CBE. Solution: We have, ADC = 130 and chord BC chord BE. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral. Since, the sum of opposite angles of a cyclic quadrilateral ΔDCB is 180. ADC + OBC = 180 = 130+ OBC = 180 = OBC = 180 -130 = 50 In ΔBOC and ΔBOE, BC = BE [given equal chord] OC =OE [both are the radius of the circle] and OB = OB [common side] ΔBOC ΔBOE OBC = OBE = 50 [by CPCT] CBE = CBO + EBO = 50 + 50 = 1...
Read More →The numerator of a rational number is 3 less than the denominator.
Question: The numerator of a rational number is 3 less than the denominator. If the denominator is increased by 5 and the numerator by 2, we get the rational number 1/2. Find the rational number. Solution: Let, the denominator of the rational number be $\mathrm{x}$. $\therefore$ The numerator of the rational number will be $\mathrm{x}-3$. $\therefore$ The rational number $=\frac{\mathrm{x}-3}{\mathrm{x}}$ According to the question, $\frac{x-3+2}{x+5}=\frac{1}{2}$ or $\frac{\mathrm{x}-1}{\mathrm{...
Read More →If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
Question: If the sum of the areas of two circles with radii R1and R2is equal to the area of a circle of radius R, then (a) $R_{1}+R_{2}=R$ (b) $R_{1}+R_{2}R$ (c) $R_{1}^{2}+R_{2}^{2}R^{2}$ (d) $R_{1}^{2}+R_{2}^{2}=R^{2}$ Solution: (d) $R_{1}^{2}+R_{2}^{2}=R^{2}$ Because the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle with radius $R$, we have: $\pi R_{1}^{2}+\pi R_{2}^{2}=\pi R^{2}$ $\Rightarrow \pi\left(\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}\rig...
Read More →If the sum of the areas of two circles with radii R1 and R2 is equal to the area of a circle of radius R, then
Question: If the sum of the areas of two circles with radii R1and R2is equal to the area of a circle of radius R, then (a) $R_{1}+R_{2}=R$ (b) $R_{1}+R_{2}R$ (c) $R_{1}^{2}+R_{2}^{2}R^{2}$ (d) $R_{1}^{2}+R_{2}^{2}=R^{2}$ Solution: (d) $R_{1}^{2}+R_{2}^{2}=R^{2}$ Because the sum of the areas of two circles with radii $R_{1}$ and $R_{2}$ is equal to the area of a circle with radius $R$, we have: $\pi R_{1}^{2}+\pi R_{2}^{2}=\pi R^{2}$ $\Rightarrow \pi\left(\mathrm{R}_{1}^{2}+\mathrm{R}_{2}^{2}\rig...
Read More →A chord of a circle is equal to its radius.
Question: A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment. Solution: Given, AB is a chord of a circle, which is equal to the radius of the circle, i.e., AB = BO (i) Join OA, AC and BC. Since, OA = OB= Radius of circle OA = AS = BO Thus, $\triangle O A B$ is an equilateral triangle. $\Rightarrow \quad \angle A O B=60^{\circ} \quad$ [each angle of an equilateral triangle is $60^{\circ}$ ] By using the theorem, in a circle, the angle su...
Read More →If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is
Question: If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is(a) 4 : (b) : 4(c) : 7(d) 7 : Solution: Let the side of the square beaand the radius of the circle ber.Now, Perimeter of circle = Circumference of the circle $\Rightarrow 4 a=2 \pi r$ $\Rightarrow \frac{a}{r}=\frac{\pi}{2}$ Now, $\frac{\text { Area of square }}{\text { Area of circle }}=\frac{a^{2}}{\pi r^{2}}$ $=\frac{1}{\pi} \times\left(\frac{a}{r}\right)^{2}$ $=\frac{1}{\pi} \times...
Read More →Ravish has three boxes whose total weight is 60
Question: Ravish has three boxes whose total weight is $60 \frac{1}{2} \mathrm{~kg} .$ Box $B$ weighs $3 \frac{1}{2} \mathrm{~kg}$ more than box $A$ and box $C$ weighs $5 \frac{1}{3} \mathrm{~kg}$ more than box $B$. Find the weight of box $A$. Solution: Let the weight of box A be $\mathrm{x} \mathrm{kg}$. Therefore, the weights of box B and box C will be $\left(\mathrm{x}+3 \frac{1}{2}\right) \mathrm{kg}$ and $\left(\mathrm{x}+3 \frac{1}{2}+5 \frac{1}{3}\right) \mathrm{kg}$, respectively. Accord...
Read More →If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is
Question: If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is(a) 4 : (b) : 4(c) : 7(d) 7 : Solution: Let the side of the square beaand the radius of the circle ber.Now, Perimeter of circle = Circumference of the circle $\Rightarrow 4 a=2 \pi r$ $\Rightarrow \frac{a}{r}=\frac{\pi}{2}$ Now, $\frac{\text { Area of square }}{\text { Area of circle }}=\frac{a^{2}}{\pi r^{2}}$ $=\frac{1}{\pi} \times\left(\frac{a}{r}\right)^{2}$ $=\frac{1}{\pi} \times...
Read More →The circumcentre of the ΔABC is 0.
Question: The circumcentre of the ΔABC is 0. Prove that OBC + BAC = 90. Solution: Given A circle is circumscribed on a ΔABC having centre O. To prove $\quad \angle O B C+\angle B A C=90^{\circ}$ Construction Join $B O$ and $C O$. Proof Let $\angle O B C=\angle O C B=\theta$$\ldots(i)$ In $\triangle O B C, \quad \angle B O C+\angle O C B+\angle C B O=180^{\circ}$ [by angle sum property of a triangle is $180^{\circ}$ ] $\Rightarrow \quad \angle B O C+\theta+\theta=180^{\circ}$ $\Rightarrow \quad \...
Read More →Solve this
Question: If $A=\left[\begin{array}{ll}4 3 \\ 2 5\end{array}\right]$, find $x$ and $y$ such that $A^{2}=x A+y I=O$. Hence, evaluate $A^{-1}$. Solution: $A=\left[\begin{array}{ll}4 3 \\ 2 5\end{array}\right]$ $A=\left[\begin{array}{ll}4 3 \\ 2 5\end{array}\right]$ $\therefore A^{2}=\left[\begin{array}{ll}22 27 \\ 18 31\end{array}\right]$ Now, $A^{2}-x A+y I=O$ $\Rightarrow\left[\begin{array}{ll}22 27 \\ 18 31\end{array}\right]-\left[\begin{array}{ll}4 x 3 x \\ 2 x 5 x\end{array}\right]+\left[\beg...
Read More →A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day,
Question: A labourer is engaged for 20 days on the condition that he will receive Rs 60 for each day, he works and he will be fined Rs 5 for each day, he is absent. If he receives Rs 745 in all, for how many days he remained absent? Solution: Let the number of days for which the labour er is absent be $\mathrm{x}$. Therefore, the number of days for which he is present will be $(20-x)$. $\therefore$ Earnings $=$ Rs. $60(20-\mathrm{x})$ Fine $=$ Rs. $5 \mathrm{x}$ According to the question, $60(20...
Read More →The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm.
Question: The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is(a) 25 cm(b) 31 cm(c) 50 cm(d) 62 cm Solution: (c) 50 cmLetrcm be the radius of the new circle.Now,Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cmThus, we have: $\pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}$ $\Rightarrow \pi r^{2}=\left[\pi \times(24)^{2}+\pi \times(7)^{2}\right] \mathrm{cm}^{2}$ $\Rightarrow \pi r^{2}...
Read More →The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm.
Question: The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is(a) 25 cm(b) 31 cm(c) 50 cm(d) 62 cm Solution: (c) 50 cmLetrcm be the radius of the new circle.Now,Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cmThus, we have: $\pi r^{2}=\pi r_{1}^{2}+\pi r_{2}^{2}$ $\Rightarrow \pi r^{2}=\left[\pi \times(24)^{2}+\pi \times(7)^{2}\right] \mathrm{cm}^{2}$ $\Rightarrow \pi r^{2}...
Read More →The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm.
Question: The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm. The radius of the new circle is(a) 16 cm(b) 28 cm(c) 42 cm(d) 56 cm Solution: (b) 28 cmLetrcm be the radius of the new circle.We know:Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cmThus, we have: $2 \pi r=2 \pi r_{1}+2 \pi r_{2}$ $\Rightarrow 2 \pi r=(2 \pi \times 18)+(2 \pi \times 10)...
Read More →If a pair of opposite sides of a cyclic quadrilateral are equal,
Question: If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal. Thinking Process Firstly, prove that ΔACD is congruent to ΔBDC by SAS and then further prove the required result Solution: Given Let ABCD be a cyclic quadrilateral and AD = BC. Join AC and BD. To prove AC = BD Proof In ΔAOD and ΔBOC, OAD = OBC and ODA = OCB [since, same segments subtends equal angle to the circle] AB = BC [given] ΔAOD = ΔBOC [by ASA congruence rule] Adding is ...
Read More →The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm.
Question: The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm. The radius of the new circle is(a) 16 cm(b) 28 cm(c) 42 cm(d) 56 cm Solution: (b) 28 cmLetrcm be the radius of the new circle.We know:Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cmThus, we have: $2 \pi r=2 \pi r_{1}+2 \pi r_{2}$ $\Rightarrow 2 \pi r=(2 \pi \times 18)+(2 \pi \times 10)...
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