The area of a circle is 49 π cm2. Its circumference is
Question: The area of a circle is 49 cm2. Its circumference is(a) 7 cm(b) 14 cm(c) 21 cm(d) 28 cm Solution: (b) 14 cmLet the radius bercm.We know: Area of a circle $=\pi r^{2}$ Thus, we have: $\pi r^{2}=49 \pi$ $\Rightarrow r^{2}=49$ $\Rightarrow r=\sqrt{49}$ $\Rightarrow r=7$ Now, Circumference of the circle $=2 \pi \mathrm{r}$ $=\left(2 \times \frac{22}{7} \times 7\right) \mathrm{cm}$ $=14 \pi \mathrm{cm}$...
Read More →In figure, if AOB is a diameter
Question: In figure, if AOB is a diameter and ADC = 120, then CAB = 30. Solution: TrueJoin CA and CB. Since, ADCB is a cyclic quadrilateral. ADC + CBA = 180. [sum of opposite angles of cyclic quadrilateral is 180] =CBA = 180 -120 = 60 [ ADC = 120] In ΔACB, CAB + CBA + ACB = 180 [by angle sum property of a triangle] CAB + 60+ 90= 180 [triangle formed from diameter to the circle is 90 i.e., ACB = 90) = CAB = 180 150 = 30....
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{2 x-(7-5 x)}{9 x-(3+4 x)}=\frac{7}{6}$ Solution: $\frac{2 x-(7-5 x)}{9 x-(3+4 x)}=\frac{7}{6}$ or $\frac{7 x-7}{5 x-3}=\frac{7}{6}$ or $42 \mathrm{x}-42=35 \mathrm{x}-21$ [After c ross multiplication] or $42 \mathrm{x}-35 \mathrm{x}=-21+42$ or $7 \mathrm{x}=21$ or $\mathrm{x}=\frac{21}{7}$ or $\mathrm{x}=3$ Thus, $\mathrm{x}=3$ is the solution of the given equation. Check: Substituting $x=3$ in the given equation, we get : L. ...
Read More →Solve this problem
Question: Given $A=\left[\begin{array}{cc}2 -3 \\ -4 7\end{array}\right]$, compute $A^{-1}$ and show that $2 A^{-1}=9 I-A$ Solution: We have, $A=\left[\begin{array}{cc}2 -3 \\ -4 7\end{array}\right]$ Now, $\operatorname{adj}(A)=\left[\begin{array}{ll}7 3 \\ 4 2\end{array}\right]$ and $|A|=2$ $\therefore A^{-1}=\frac{1}{2}\left[\begin{array}{ll}7 3 \\ 4 2\end{array}\right]$ Now, $2 A^{-1}=9 I-A$ LHS $=2 A^{-1}=\left[\begin{array}{ll}7 3 \\ 4 2\end{array}\right]$ $\mathrm{RHS}=9 I-A=9\left[\begin{...
Read More →The area of a circle is 38.5 cm2. The circumference of the circle is
Question: The area of a circle is 38.5 cm2. The circumference of the circle is(a) 6.2 cm(b) 12.1 cm(c) 11 cm(d) 22 cm Solution: (d) 22 cmLet the radius bercm.We know: Area of a circle $=\pi r^{2} \mathrm{~cm}^{2}$ Thus, we have: $\pi r^{2}=38.5$ $\Rightarrow \frac{22}{7} \times r^{2}=38.5$ $\Rightarrow r^{2}=\left(38.5 \times \frac{7}{22}\right)$ $\Rightarrow r^{2}=\left(\frac{385}{10} \times \frac{7}{22}\right)$ $\Rightarrow r^{2}=\frac{49}{4}$ $\Rightarrow r=\frac{7}{2}$ Now, Circumference of ...
Read More →If A, B, C and D are four points such that
Question: If A, B, C and D are four points such that BAC = 45 and BDC = 45, then A, B, C and D are concyclic. Solution: True Since, BAC = 45 and BDC = 45 As we know, angles in the same segment of a circle are equal. Hence, A, B, C and D are concyclic....
Read More →The area of a circle is 38.5 cm2. The circumference of the circle is
Question: The area of a circle is 38.5 cm2. The circumference of the circle is(a) 6.2 cm(b) 12.1 cm(c) 11 cm(d) 22 cm Solution: (d) 22 cmLet the radius bercm.We know: Area of a circle $=\pi r^{2} \mathrm{~cm}^{2}$ Thus, we have: $\pi r^{2}=38.5$ $\Rightarrow \frac{22}{7} \times r^{2}=38.5$ $\Rightarrow r^{2}=\left(38.5 \times \frac{7}{22}\right)$ $\Rightarrow r^{2}=\left(\frac{385}{10} \times \frac{7}{22}\right)$ $\Rightarrow r^{2}=\frac{49}{4}$ $\Rightarrow r=\frac{7}{2}$ Now, Circumference of ...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{x+2}{x+5}=\frac{x}{x+6}$ Solution: $\frac{x+2}{x+5}=\frac{x}{x+6}$ or $\mathrm{x}^{2}+2 \mathrm{x}+6 \mathrm{x}+12=\mathrm{x}^{2}+5 \mathrm{x}$ [After $c$ ross multiplication] or $\mathrm{x}^{2}-\mathrm{x}^{2}+8 \mathrm{x}-5 \mathrm{x}=-12$ or $3 \mathrm{x}=-12$ or $\mathrm{x}=\frac{-12}{3}$ or $\mathrm{x}=-4$ Thus, $x=-4$ is the solution of given equation. Check : Substituting $\mathrm{x}=-4$ in the given equation, we get : L...
Read More →If A, B, C and D are four points such that
Question: If A, B, C and D are four points such that BAC = 30 and BDC = 60, then D is the centre of the circle through A, B and C. Solution: False Because, there can be many points D, such that BDC = 60 and each such point cannot be the centre of the circle through A, B and C....
Read More →Solve this
Question: Let $A=\left[\begin{array}{ll}3 2 \\ 7 5\end{array}\right]$ and $B=\left[\begin{array}{ll}6 7 \\ 8 9\end{array}\right] .$ Find $(A B)^{-1}$ Solution: Given : $A=\left[\begin{array}{ll}3 2 \\ 7 5\end{array}\right]$ $B=\left[\begin{array}{ll}6 7 \\ 8 9\end{array}\right]$ $A B=\left[\begin{array}{ll}34 39 \\ 82 94\end{array}\right]$ Now, $|A B|=-2$ Since, $|A B| \neq 0$ Hence, $A B$ is invertible. Let $C_{i j}$ be the cofactor of $a_{i n}$ in $A B=\left[a_{i j}\right]$ $C_{11}=94, C_{12}=...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6}$ Solution: $\frac{9 x-7}{3 x+5}=\frac{3 x-4}{x+6}$ or $9 \mathrm{x}^{2}-7 \mathrm{x}+54 \mathrm{x}-42=9 \mathrm{x}^{2}-12 \mathrm{x}+15 \mathrm{x}-20$ [After $c$ ross multiplication] or $9 x^{2}-9 x^{2}+47 x-3 x=-20+42$ or $44 \mathrm{x}=22$ or $\mathrm{x}=\frac{22}{44}$ or $\mathrm{x}=\frac{1}{2}$ Thus, $\mathrm{x}=\frac{1}{2}$ is the solution of the given equation. Check: Substituting $x=\frac...
Read More →ABCD is a cyclic quadrilateral such that
Question: ABCD is a cyclic quadrilateral such that A = 90, B = 70, C = 95 and D = 105. Solution: False In a cyclic quadrilateral, the sum of opposite angles is 180. Now, A + C = 90 + 95 = 185 180 and B+D = 70 + 105 = 175 180 Here, we see that, the sum of opposite angles is not equal to 180. So, it is not a cyclic quadrilateral....
Read More →Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°
Question: Find the area of the major segment APB of a circle of radius 35 cm and AOB = 90, as shown in the given figure. Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \times \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$ $=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$ $=962.5-612.5$ $=350 \...
Read More →Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°
Question: Find the area of the major segment APB of a circle of radius 35 cm and AOB = 90, as shown in the given figure. Solution: Area of minor segment = Area of sector AOBC Area of right triangle AOB $=\frac{90^{\circ}}{360^{\circ}} \times \pi(\mathrm{OA})^{2}-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}$ $=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$ $=\frac{1}{4} \times \frac{22}{7} \times(35)^{2}-\frac{1}{2} \times 35 \times 35$ $=962.5-612.5$ $=350 \...
Read More →If AOB is a diameter of a circle
Question: If AOB is a diameter of a circle and C is a point on the circle, then AC2+ BC2= AB2. Solution: True Since, any diameter of the circle subtends a right angle to any point on the circle. If AOB is a diameter of a circle and C is a point on the circle, then ΔACB is right angled at C. In right angled ΔACB, [use Pythagoras theorem] AC2+ BC2= AB2...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\left(\frac{x+1}{x-4}\right)^{2}=\frac{x+8}{x-2}$ Solution: $\left(\frac{x+1}{x-4}\right)^{2}=\frac{x+8}{x-2}$ or $\frac{\mathrm{x}^{2}+2 \mathrm{x}+1}{\mathrm{x}^{2}-8 \mathrm{x}+16}=\frac{\mathrm{x}+8}{\mathrm{x}-2} \quad\left[(a+b)^{2}=a^{2}+b^{2}+2 a b\right.$ and $\left.(a-b)^{2}=a^{2}+b^{2}-2 a b\right]$ or $\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}-2 \mathrm{x}^{2}-4 \mathrm{x}-2=\mathrm{x}^{3}-8 \mathrm{x}^{2}+16 \mathrm{x}...
Read More →For the following pairs of matrices
Question: For the following pairs of matrices verity that $(A B)^{-1}=B^{-1} A^{-1}$ : (i) $A=\left[\begin{array}{ll}3 2 \\ 7 5\end{array}\right]$ and $B\left[\begin{array}{ll}4 6 \\ 3 2\end{array}\right]$ (ii) $A=\left[\begin{array}{ll}2 1 \\ 5 3\end{array}\right]$ and $B\left[\begin{array}{ll}4 5 \\ 3 4\end{array}\right]$ Solution: (i) We have, $A=\left[\begin{array}{ll}3 2 \\ 7 5\end{array}\right]$ and $B=\left[\begin{array}{ll}4 6 \\ 3 2\end{array}\right]$ $\therefore A B=\left[\begin{array}...
Read More →A circle of radius 3 cm can be
Question: A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm. Solution: True Suppose, we consider diameter of a circle is AB = 66m. AB Then, radius of a circle = AB/2 = 6/2 = 3 cm, which is true....
Read More →From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°,
Question: From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and BCD = 90, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet. Solution: Since, BFEC is a quarter of a circle.Hence, BC = EC = 3.5 cmNow, DC = DE + EC = 2 + 3.5 = 5.5 cmArea of shaded region = Area of the trapezium ABCD Area of the quadrant BFEC $=\frac{1}{2} \times(\mathrm{AB}+\mathrm{DC}) \times \mathrm{BC}-\frac{1}{4} \ti...
Read More →Through three collinear points a circle can be drawn.
Question: Through three collinear points a circle can be drawn. Solution: False Because, circle can pass through only two collinear points but not through three collinear points....
Read More →From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°,
Question: From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and BCD = 90, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet. Solution: Since, BFEC is a quarter of a circle.Hence, BC = EC = 3.5 cmNow, DC = DE + EC = 2 + 3.5 = 5.5 cmArea of shaded region = Area of the trapezium ABCD Area of the quadrant BFEC $=\frac{1}{2} \times(\mathrm{AB}+\mathrm{DC}) \times \mathrm{BC}-\frac{1}{4} \ti...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\left(\frac{x+1}{x+2}\right)^{2}=\frac{x+2}{x+4}$ Solution: $\left(\frac{x+1}{x+2}\right)^{2}=\frac{x+2}{x+4}$ or $\frac{x^{2}+2 x+1}{x^{2}+4 x+4}=\frac{x+2}{x+4}$ or $\mathrm{x}^{3}+2 \mathrm{x}^{2}+\mathrm{x}+4 \mathrm{x}^{2}+8 \mathrm{x}+4=\mathrm{x}^{3}+4 \mathrm{x}^{2}+4 \mathrm{x}+2 \mathrm{x}^{2}+8 \mathrm{x}+8[$ After c ross multiplication $]$ or $\mathrm{x}^{3}-\mathrm{x}^{3}+6 \mathrm{x}^{2}-6 \mathrm{x}^{2}+9 \mathrm{x}-1...
Read More →The congruent circles with centres O
Question: The congruent circles with centres O and O intersect at two points A and B. Then, AOB = AOB. Solution: True Join AB, OA and OB, OA and BO.In ΔAOB and ΔAOB, OA = AO [both circles have same radius] OB = BO[both circles have same radius] and AB= AB [common chord] ΔAOB = ΔAOB [by SSC congruence rule] = AOB = AOB [by CPCT]...
Read More →In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn.
Question: In the given figure,Ois the centre of the bigger circle, andACis its diameter. Another circle with AB as diameter is drawn. IfAC= 54 cm andBC= 10, find the area of the shaded region. Solution: We have:OA = OC = 27 cm $A B=A C-B C$ $=54-10$ $=44$ AB is the diameter of the smaller circle.Thus, we have: Radius of the smaller circle $=\frac{\mathrm{AB}}{2}=\frac{44}{2}=22 \mathrm{~cm}$ Area of the smaller circle $=\pi r^{2}$ $=\frac{22}{7} \times 22 \times 22$ $=1521.14 \mathrm{~cm}^{2}$ R...
Read More →Solve the following equation and verify your answer:
Question: Solve the following equation and verify your answer: $\frac{7 x-2}{5 x-1}=\frac{7 x+3}{5 x+4}$ Solution: $\frac{7 x-2}{5 x-1}=\frac{7 x+3}{5 x+4}$ or $35 \mathrm{x}^{2}+28 \mathrm{x}-10 \mathrm{x}-8=35 \mathrm{x}^{2}+15 \mathrm{x}-7 \mathrm{x}-3[$ After cross multiplication $]$ or $35 x^{2}-35 x^{2}+18 x-8 x=-3+8$ or $10 x=5$ or $x=\frac{5}{10}$ or $x=\frac{1}{2}$ Thus, $x=\frac{1}{2}$ is the solution of the given equation. Check: Substituting $x=\frac{1}{2}$ in the given equation, we ...
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