The radius of a circle is 17.5 cm.
Question: The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44 cm in length. Solution: Given:Radius = 17.5 cmLength of the arc = 44 cmNow, Length of the arc $=\frac{2 \pi \mathrm{r} \theta}{360}$ $\Rightarrow 44=2 \times \frac{22}{7} \times 17.5 \times \frac{\theta}{360}$ $\Rightarrow \theta=\frac{44 \times 7 \times 360}{44 \times 17.5}$ $\Rightarrow \theta=144^{\circ}$ Also, Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $=\frac{22}{7} \times 17...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $x-\frac{(x-1)}{2}=1-\frac{(x-2)}{3}$ Solution: $x-\frac{(x-1)}{2}=1-\frac{(x-2)}{3}$ or $\frac{2 \mathrm{x}-\mathrm{x}+1}{2}=\frac{3-\mathrm{x}+2}{3}$ or $\frac{\mathrm{x}+1}{2}=\frac{5-\mathrm{x}}{3}$ or $3 \mathrm{x}+3=10-2 \mathrm{x}$ or $5 \mathrm{x}=10-3$ or $\mathrm{x}=\frac{7}{5}$ Check : L. H.S. $=\frac{7}{5}-\frac{\frac{7}{5}-1}{2}=\frac{7}{5}-\frac{1}{5}=\frac{6}{5}$ R.H.S. $=1-\frac{\frac{7}{5}-2}...
Read More →The radius of a circle is 17.5 cm.
Question: The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44 cm in length. Solution: Given:Radius = 17.5 cmLength of the arc = 44 cmNow, Length of the arc $=\frac{2 \pi \mathrm{r} \theta}{360}$ $\Rightarrow 44=2 \times \frac{22}{7} \times 17.5 \times \frac{\theta}{360}$ $\Rightarrow \theta=\frac{44 \times 7 \times 360}{44 \times 17.5}$ $\Rightarrow \theta=144^{\circ}$ Also, Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $=\frac{22}{7} \times 17...
Read More →The perimeter of a certain sector of a circle of radius 6.5 cm in 31 cm.
Question: The perimeter of a certain sector of a circle of radius 6.5 cm in 31 cm. Find the area of the sector. Solution: Given:Radius = 6.5 cm Let O be the centre of the circle with radius 6.5 cm and OACBObe its sector with perimeter 31 cm.Thus, we have:OA + OB + arc AB = 31 cm $\Rightarrow 6.5+6.5+\operatorname{arc} A B=31$ $\Rightarrow \operatorname{arc} A B=31-13$ $\Rightarrow \operatorname{arc} A B=18 \mathrm{~cm}$ Now, Area of the sector $\mathrm{OACBO}=\frac{1}{2} \times$ Radius $\times$ ...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{3 a-2}{3}+\frac{2 a+3}{2}=a+\frac{7}{6}$ Solution: $\frac{3 a-2}{3}+\frac{2 a+3}{2}=a+\frac{7}{6}$ or $\frac{6 \mathrm{a}-4+6 \mathrm{a}+9}{6}=\mathrm{a}+\frac{7}{6}$ or $12 \mathrm{a}+5=6 \mathrm{a}+7$ or $6 \mathrm{a}=7-5$ or $\mathrm{a}=\frac{2}{6}=\frac{1}{3}$ Check : L.H.S. $=\frac{3 \times \frac{1}{3}-2}{3}+\frac{2 \times \frac{1}{3}+3}{2}=\frac{-1}{3}+\frac{11}{6}=\frac{9}{6}=\frac{3}{2}$ R.H. S...
Read More →The diagonals of a parallelogram ABCD
Question: The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area. Solution: Given in a parallelogram $A B C D$, diagonals interect at $O$ and draw a line $P Q$, which intersects $A D$ and $B C$. To prove $P Q$ divides the parallelogram $A B C D$ into two parts of equal area. i.e. ar $(A B Q P)=\operatorname{ar}(C D P Q)$ Proof We know that, diagonals of a parall...
Read More →The area of the sector of a circle of radius 10.5 cm is 69.3 cm2.
Question: The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector. Solution: Given:Area of the sector = 63 cm2Radius = 10.5 cmNow, Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $\Rightarrow 69.3=\frac{22}{7} \times 10.5 \times 10.5 \times \frac{\theta}{360}$ $\Rightarrow \theta=\frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5}$ $\Rightarrow \theta=72^{\circ}$ $\therefore$ Central angle of the sector $=72^{\circ}$...
Read More →The area of the sector of a circle of radius 10.5 cm is 69.3 cm2.
Question: The area of the sector of a circle of radius 10.5 cm is 69.3 cm2. Find the central angle of the sector. Solution: Given:Area of the sector = 63 cm2Radius = 10.5 cmNow, Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $\Rightarrow 69.3=\frac{22}{7} \times 10.5 \times 10.5 \times \frac{\theta}{360}$ $\Rightarrow \theta=\frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5}$ $\Rightarrow \theta=72^{\circ}$ $\therefore$ Central angle of the sector $=72^{\circ}$...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{6 x+1}{2}+1=\frac{7 x-3}{3}$ Solution: $\frac{6 x+1}{2}+1=\frac{7 x-3}{3}$ or $\frac{6 \mathrm{x}+1+2}{2}=\frac{7 \mathrm{x}-3}{3}$ or $18 \mathrm{x}+9=14 \mathrm{x}-6$ or $18 \mathrm{x}-14 \mathrm{x}=-6-9$ or $4 \mathrm{x}=-15$ or $\mathrm{x}=\frac{-15}{4}$ Check : L. H.S. $=\frac{6 \times \frac{-15}{4}+1}{2}+1=\frac{-45+2+4}{4}=\frac{-39}{4}$ R. H.S. $=\frac{7 \times \frac{-15}{4}-3}{3}=\frac{-105-12...
Read More →A sector of 56°, cut out from a circle, contains 17.6 cm2.
Question: A sector of 56, cut out from a circle, contains 17.6 cm2. Find the radius of the circle. Solution: Area of the sector =17.6 cm2 Area of the sector $=\frac{\pi r^{2} \theta}{360}$ $\Rightarrow 17.6=\frac{22}{7} \times r^{2} \times \frac{56}{360}$ $\Rightarrow r^{2}=\frac{17.6 \times 7 \times 360}{22 \times 56}$ $\Rightarrow r^{2}=36$ $\Rightarrow r=6 \mathrm{~cm}$ Radius of the circle = 6 cm...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{7}{2} x-\frac{5}{2} x=\frac{20}{3} x+10$ Solution: $\frac{7}{2} x-\frac{5}{2} x=\frac{20}{3} x+10$ or $\frac{7 \mathrm{x}-5 \mathrm{x}}{2}=\frac{20 \mathrm{x}+30}{3}$ or $40 \mathrm{x}+60=6 \mathrm{x}$ or $34 \mathrm{x}=-60$ or $\mathrm{x}=-\frac{60}{34}=-\frac{30}{17}$ Check : L.H. S. $=\frac{7}{2} \times \frac{-30}{17}-\frac{5}{2} \times-\frac{30}{17}=-\frac{30}{17}$ R.H.S. $=\frac{20}{3} \times \fra...
Read More →The minute hand of a clock is 15 cm long.
Question: The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes. Solution: Angle inscribed by the minute hand in 60 minutes $=360^{\circ}$ Angle inscribed by the minute hand in 20 minutes $=\frac{360}{60} \times 20=120^{\circ}$ We have: $\theta=120^{\circ}$ and $r=15 \mathrm{~cm}$ $\therefore$ Required area swept by the minute hand in 20 minutes $=$ Area of the sector with $r=15 \mathrm{~cm}$ and $\theta=120^{\circ}$ $=\frac{\pi r^{2} \theta}{360}$ $=3.14 \times ...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{3}{4} x+4 x=\frac{7}{8}+6 x-6$ Solution: $\frac{3}{4} x+4 x=\frac{7}{8}+6 x-6$ or $\frac{3}{4} \mathrm{x}-2 \mathrm{x}=\frac{7}{8}-6$ or $\frac{3 x-8 x}{4}=\frac{7-48}{8}$ or $\frac{-5 x}{4}=\frac{-41}{8}$ or $-40 \mathrm{x}=-164$ or $\mathrm{x}=\frac{-164}{-40}=\frac{41}{10}$ Check : L.H.S. $=\frac{3}{4} \times \frac{41}{10}+4 \times \frac{41}{10}=\frac{123}{40}+\frac{164}{10}=\frac{123+656}{40}=\frac...
Read More →A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length.
Question: A pendulum swings through an angle of 30 and describes an arc 8.8 cm in length. Find the length of the pendulum. Solution: Given:Length of the arc = 8.8 cmAnd, $\theta=30^{\circ}$ Now, Length of the $\operatorname{arc}=\frac{2 \pi \mathrm{r} \theta}{360}$ $\Rightarrow 8.8=\frac{2 \times \frac{22}{7} \times r \times 30}{360}$ $\Rightarrow r=\frac{8.8 \times 360 \times 7}{44 \times 30}$ $\Rightarrow r=16.8 \mathrm{~cm}$ Length of the pendulum = 16.8 cm...
Read More →A and B are square matrices of the same order 3,
Question: $A$ and $B$ are square matrices of the same order 3 , such that $A B=2 /$ and $|A|=2$, write the value of $|B|$. Solution: Given: $A$ and $B$ are square matrices of order 3 $|A|=2$ $A B=21$ Now, $A B=2 I \Rightarrow|A B|=|2 I \Rightarrow| A|| B|=| 2 I \mid$ $(\because|A B|=|A| B \mid$, if they are square matrices of same order $) \Rightarrow 2|B|=|2 I|$ $(\because|A|=2) \Rightarrow 2|B|=2^{3}|I|$ $(\because$ Order of $I$ is $3 \times 3) \Rightarrow 2|B|=8 \times 1$ Hence, the value of ...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{1}{2} x+7 x-6=7 x+\frac{1}{4}$ Solution: $\frac{1}{2} x+7 x-6=7 x+\frac{1}{4}$ or $\frac{1}{2} \mathrm{x}+7 \mathrm{x}-7 \mathrm{x}=\frac{1}{4}+6$ or $\frac{x}{2}=\frac{1+24}{4}$ or $\frac{\mathrm{x}}{2}=\frac{25}{4}$ or $\mathrm{x}=\frac{25}{2}$ Check : L.H.S. $=\frac{1}{2} \times \frac{25}{2}+7 \times \frac{25}{2}-6=\frac{351}{4}$ R.H.S. $=7 \times \frac{25}{2}+\frac{1}{4}=\frac{351}{4}$ $\therefore$...
Read More →A point E is taken on the side BC of a parallelogram ABCD.
Question: A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ΔADF) = ar (ΔBFC). Solution: Given $A B C D$ is a parallelogram and $E$ is a point on $B C$. AE and DC are produced to meet at F. $A B \| C D$ anti $B C \| A D$, ..(i) $\therefore$ $A B \| C D$ and $B C \| A D$ ...(i) To prove $\operatorname{ar}(\triangle A D F)=\operatorname{ar}(A B F C)$ Construction Join $A C$ and $D E$. Proof Since, $A C$ is a diagonal of parallelogram $A ...
Read More →The circumference of a circle is 8 cm.
Question: The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72. Solution: Let the radius of the circle ber.Now, Circumference $=8$ $\Rightarrow 2 \pi r=8$ $\Rightarrow r=\frac{14}{11} \mathrm{~cm}$ We have $r=\frac{14}{11} \mathrm{~cm}$ and $\theta=72^{\circ}$ Area of sector $=\frac{\theta}{360^{\circ}} \times \pi r^{2}=\frac{72^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(\frac{14}{11}\right)^{2}=1.02 \mathrm{~cm}^{2}$ Hence, the area of the sec...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $x-2 x+2-\frac{16}{3} x+5=3-\frac{7}{2} x$ Solution: $x-2 x+2-\frac{16}{3} x+5=3-\frac{7}{2} x$ or $\frac{3 x-6 x+6-16 x+15}{3}=\frac{6-7 x}{2}$ or $\frac{-19 x+21}{3}=\frac{6-7 x}{2}$ or $-38 \mathrm{x}+42=18-21 \mathrm{x}$ or $-21 \mathrm{x}+38 \mathrm{x}=42-18$ or $17 \mathrm{x}=24$ or $\mathrm{x}=\frac{24}{17}$ Check: L.H. S. $=\frac{24}{17}-2 \times \frac{24}{17}+7-\frac{16}{3} \times \frac{24}{17}=\frac...
Read More →If A is a 3 × 3 invertible matrix,
Question: If $A$ is a $3 \times 3$ invertible matrix, then what will be the value of $k$ if $\operatorname{det}\left(A^{-1}\right)=(\operatorname{det} A)^{k}$. Solution: As we know that $A^{-1}=\frac{\operatorname{Adj} A}{|\mathrm{~A}|}$ $\therefore\left|A^{-1}\right|=\frac{|\operatorname{Adj} A|}{|A|}$ $=\frac{|A|^{2-1}}{|A|} \quad\left[\because\right.$ If $A$ is a non singular matrix of order $n$, then $\left.|\operatorname{adj}(A)|=|\mathrm{A}|^{n-1}\right]$ $=\frac{|A|^{2}}{|A|}=|A|$ As we a...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{7 y+2}{5}=\frac{6 y-5}{11}$ Solution: $\frac{7 y+2}{5}=\frac{6 y-5}{11}$ or $77 \mathrm{y}+22=30 \mathrm{y}-25$ or $77 \mathrm{y}-30 \mathrm{y}=-25-22$ or $47 \mathrm{y}=-47 or $\mathrm{y}=\frac{-47}{47}=-1$ Verification : L. H. S. $=\frac{-7+2}{5}=\frac{-5}{5}=-1$ R. H. S. $=\frac{-6-5}{11}=\frac{-11}{11}=-1$ $\therefore$ L.H.S. $=$ R.H.S. for $y=-1$...
Read More →A square is inscribed in a circle.
Question: A square is inscribed in a circle. Find the ratio of the areas of the circle and the square. Solution: Let the side of the square beaand radius of the circle berWe know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square. $\therefore \sqrt{2} a=2 r$ $\Rightarrow a=\sqrt{2} r$ Now, $\frac{\text { Area of circle }}{\text { Area of square }}=\frac{\pi r^{2}}{a^{2}}$ $=\frac{\pi r^{2}}{(\sqrt{2} r)^{2}}$ $=\frac{\pi r^{2}}{2 r^{2}...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{a-8}{3}=\frac{a-3}{2}$ Solution: $\frac{a-8}{3}=\frac{a-3}{2}$ or $2 a-16=3 a-9$ or $3 a-2 a=-16+9$ or $a=-7$ Verification : L. H. S. $=\frac{-7-8}{3}=\frac{-15}{3}=-5$ R. H.S. $=\frac{-7-3}{2}=\frac{-10}{2}=-5$ $\therefore$ L.H.S. $=$ R.H.S. for $a=-7$...
Read More →Solve each of the following equation and also check your result in each case:
Question: Solve each of the following equation and also check your result in each case: $\frac{2 x+5}{3}=3 x-10$ Solution: $\frac{2 x+5}{3}=3 x-10$ or $2 \mathrm{x}+5=9 \mathrm{x}-30$ or $9 \mathrm{x}-2 \mathrm{x}=5+30$ or $7 \mathrm{x}=35$ or $\mathrm{x}=\frac{35}{7}$ or $\mathrm{x}=5$ Verification : L. H. S. $=\frac{10+5}{3}=\frac{15}{3}=5$ R. H.S. $=15-10=5$ $\therefore$ L.H.S. $=$ R.H.S. for $x=5$....
Read More →Solve this
Question: If $\left|\begin{array}{ccc}x \sin \theta \cos \theta \\ -\sin \theta -x 1 \\ \cos \theta 1 x\end{array}\right|=8$, write the value of $x$. Solution: $\left|\begin{array}{ccc}x \sin \theta \cos \theta \\ -\sin \theta -x 1 \\ \cos \theta 1 x\end{array}\right|=8$ Expanding along $R_{1}$, we get $x\left(-x^{2}-1\right)-\sin \theta(-x \sin \theta-\cos \theta)+\cos \theta(-\sin \theta+x \cos \theta)=8$ $\Rightarrow-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x...
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